Author Archives: Dr. Darrin

Gradient Boosting With Random Forest Classification in R

In this blog we have already discussed and what gradient boosting is. However, for a brief recap, gradient boosting improves model performance by first developing an initial model called the base learner using whatever algorithm of your choice (linear, tree, etc.).

What follows next is that gradient boosting looks at the error in the first model and develops a second model using what is called the loss function. The loss function calculates the difference between the current accuracy and the desired prediction whether it’s accuracy for classification or error in regression. This process is repeated with the creation of additional models until a certain level of accuracy or reduction in error is attained.

This post what provide an example of the use of gradient boosting in random forest classification. Specifically, we will try to predict a person’s labor participation based on several independent variables.

library(randomForest);library(gbm);library(caret);library(Ecdat)
data("Participation")
str(Participation)
## 'data.frame':    872 obs. of  7 variables:
##  $ lfp    : Factor w/ 2 levels "no","yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ lnnlinc: num  10.8 10.5 11 11.1 11.1 ...
##  $ age    : num  3 4.5 4.6 3.1 4.4 4.2 5.1 3.2 3.9 4.3 ...
##  $ educ   : num  8 8 9 11 12 12 8 8 12 11 ...
##  $ nyc    : num  1 0 0 2 0 0 0 0 0 0 ...
##  $ noc    : num  1 1 0 0 2 1 0 2 0 2 ...
##  $ foreign: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

Data Preparation

We need to transform the ‘age’ variable by multiplying by ten so that the ages are realistic. In addition, we need to convert “lnnlinc” from the log of salary to regular salary. Below is the code to transform these two variables.

Participation$age<-10*Participation$age #normal age
Participation$lnnlinc<-exp(Participation$lnnlinc) #actual income not log

We can now create our train and test datasets

set.seed(502)
ind=sample(2,nrow(Participation),replace=T,prob=c(.7,.3))
train<-Participation[ind==1,]
test<-Participation[ind==2,]

We now need to create our grid and control. The grid allows us to create several different models with various parameter settings. This is important in determining what is the most appropriate model which is always determined by comparing. We are using random forest so we need to set the number of trees we desire, the depth of the trees, the shrinkage which controls the influence of each tree, and the minimum number of observations in a node. The control will allow us to set the cross-validation. Below is the code for the creation of the grid and control.

grid<-expand.grid(.n.trees=seq(200,500,by=200),.interaction.depth=seq(1,3,by=2),.shrinkage=seq(.01,.09,by=.04),
                   .n.minobsinnode=seq(1,5,by=2)) #grid features
control<-trainControl(method="CV",number = 10) #control

Parameter Selection

Now we set our seed and run the gradient boosted model.

set.seed(123)
gbm.lfp.train<-train(lfp~.,data=train,method='gbm',trControl=control,tuneGrid=grid)
gbm.lfp.train
## Stochastic Gradient Boosting 
## 
## 636 samples
##   6 predictors
##   2 classes: 'no', 'yes' 
## 
## No pre-processing
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 573, 573, 571, 572, 573, 572, ... 
## Resampling results across tuning parameters:
## 
##   shrinkage  interaction.depth  n.minobsinnode  n.trees  Accuracy 
##   0.01       1                  1               200      0.6666026
##   0.01       1                  1               400      0.6823306
##   0.01       1                  3               200      0.6588637
##   0.01       1                  3               400      0.6854804
##   0.01       1                  5               200      0.6792769
##   0.01       1                  5               400      0.6823306
##   0.01       3                  1               200      0.6730044
##   0.01       3                  1               400      0.6572051
##   0.01       3                  3               200      0.6793273
##   0.01       3                  3               400      0.6697787
##   0.01       3                  5               200      0.6682914
##   0.01       3                  5               400      0.6650416
##   0.05       1                  1               200      0.6759558
##   0.05       1                  1               400      0.6508040
##   0.05       1                  3               200      0.6681426
##   0.05       1                  3               400      0.6602286
##   0.05       1                  5               200      0.6680441
##   0.05       1                  5               400      0.6570788
##   0.05       3                  1               200      0.6493662
##   0.05       3                  1               400      0.6603518
##   0.05       3                  3               200      0.6540545
##   0.05       3                  3               400      0.6366911
##   0.05       3                  5               200      0.6712428
##   0.05       3                  5               400      0.6445299
##   0.09       1                  1               200      0.6461405
##   0.09       1                  1               400      0.6634768
##   0.09       1                  3               200      0.6571036
##   0.09       1                  3               400      0.6320765
##   0.09       1                  5               200      0.6554922
##   0.09       1                  5               400      0.6540755
##   0.09       3                  1               200      0.6523920
##   0.09       3                  1               400      0.6430140
##   0.09       3                  3               200      0.6430666
##   0.09       3                  3               400      0.6447749
##   0.09       3                  5               200      0.6540522
##   0.09       3                  5               400      0.6524416
##   Kappa    
##   0.3210036
##   0.3611194
##   0.3032151
##   0.3667274
##   0.3472079
##   0.3603046
##   0.3414686
##   0.3104335
##   0.3542736
##   0.3355582
##   0.3314006
##   0.3258459
##   0.3473532
##   0.2961782
##   0.3310251
##   0.3158762
##   0.3308353
##   0.3080692
##   0.2940587
##   0.3170198
##   0.3044814
##   0.2692627
##   0.3378545
##   0.2844781
##   0.2859754
##   0.3214156
##   0.3079460
##   0.2585840
##   0.3062307
##   0.3044324
##   0.3003943
##   0.2805715
##   0.2827956
##   0.2861825
##   0.3024944
##   0.3002135
## 
## Accuracy was used to select the optimal model using  the largest value.
## The final values used for the model were n.trees = 400,
##  interaction.depth = 1, shrinkage = 0.01 and n.minobsinnode = 3.

Gradient boosting provides us with the recommended parameters for our training model as shown above as well as the accuracy and kappa of each model. We also need to recode the dependent variable as 0 and 1 for the ‘gbm’ function.

Model Training

train$lfp=ifelse(train$lfp=="no",0,1)
gbm.lfp<-gbm(lfp~., distribution = 'bernoulli',data=train,n.trees = 400,interaction.depth = 1,shrinkage=.01,n.minobsinnode = 3)

You can see a summary of the most important variables for prediction as well as a plot by using the “summary” function.

summary(gbm.lfp)

1.png

##             var   rel.inf
## lnnlinc lnnlinc 28.680447
## age         age 27.451474
## foreign foreign 23.307932
## nyc         nyc 18.375856
## educ       educ  2.184291
## noc         noc  0.000000

Salary (lnnlinc), age and foreigner status are the most important predictors followed by number of younger children (nyc) and lastest education. Number of older children (noc) has no effect. We can now test our model on the test set.

Model Testing

gbm.lfp.test<-predict(gbm.lfp,newdata = test,type = 'response', n.trees = 400)

Our test model returns a set of probabilities. We need to convert this to a simple yes or no and this is done in the code below.

gbm.class<-ifelse(gbm.lfp.test<0.5,'no','yes')

We can now look at a table to see how accurate our model is as well as calculate the accuracy.

table(gbm.class,test$lfp)
##          
## gbm.class no yes
##       no  91  39
##       yes 39  67
(accuracy<-(91+67)/(91+67+39+39))
## [1] 0.6694915

The model is not great. However, you now have an example of how to use gradient boosting to develop a random forest classification model

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Washback

Washback is the effect that testing has on teaching and learning. This term is commonly used in used in language assessment but it is not limited to only that field. One of the primary concerns of many teachers is developing that provide washback or that enhances students learning and understanding of ideas in a class.

This post will discuss three ways in which washback can be improved in a class. The three ways are…

  • Written feedback on exams
  • Go over the results as a class
  • Meetings with students on exam performance

Written Feedback

Exams or assignments that are highly subjective (ie essays) require written feedback in order to provide washback. This means specific, personalized feedback for each student. This is a daunting task for most teachers especially as classes get larger. However, if your goal is to improve washback providing written comments is one way to achieve this.

The letter grade or numerical score a student receives on a test does not provide insights into how the student can improve. The reasoning behind what is right or wrong can be provided in the written feedback.

Go Over Answers in Class

Perhaps the most common way to enhance feedback is to go over the test in class. This allows the students to learn what the correct answer is, as well as why one answer is the answer. In addition, students are given time to ask questions and clarification of the reasoning behind the teacher’s marking.

If there were common points of confusion, going over the answers in this way allows for the teacher to reteach the confusing concepts. In many ways, the test revealed what was unclear and now the teacher is able to provide support to achieve mastery.

One-on-One Meetings

For highly complex and extremely subjective forms of assessments (ie research paper) one-on-one meetings may be the most appropriate. This may require a more personal touch and a greater deal of time.

During the meeting, students can have their questions addressed and learn what they need to do in order to improve. This is a useful method for assignments that require several rounds of feedback in order to be completed.

Conclusion

Washback, if done properly, can help with motivation, autonomy, and self-confidence of students. What this means is that assessment should not only be used for grades but also to develop learning skills.

Understanding Testing

Testing is standard practice in most educational context. A teacher needs a way to determine what level of knowledge the students currently have or have gained through the learning experience. However, identifying what testing is and is not has not always been clear.

In this post, we will look at exactly what testing as. In general, testing is a way of measuring a person’s ability and or knowledge in a given are of study. Specifically, there are five key characteristics of a test, and they are…

  • Systematic
  • Quantifiable
  • Individualistic
  • Competence
  • Domain specific

Systematic

A test must be well organized and structured. For example, the multiple choice are in one section while the short answers are in a different section. If an essay is required there is a rubric for grading. Directions for all sections are in the test to explain the expectations to the students.

This is not as easy or as obvious as some may believe. Developing a test takes a great deal of planning for the actual creation of the test.

Quantifiable

Test are intended to measure something. A test can measure general knowledge such as proficiency test of English or a test can be specific such as a test that only looks at vocabulary memorization. Either way, it is important for both the student and teacher to know what is being measured.

Another obvious but sometimes mistake by test makers is the reporting of results. How many points each section and even each question is important for students to know when taking a test. This information is also critical for the person who is responsible for grading the tests.

Individualistic 

Test are primarily designed to assess a student’s individual knowledge/performance. This is a Western concept of the responsibility of a person to have an individual expertise in a field of knowledge.

There are examples of groups working together on tests. However, group work is normally left to projects and not formal modes of assessment such as testing.

Competence

As has already been alluded too, tests assess competence either through the knowledge a person has about a subject or their performance doing something. For example, a vocabulary test assesses knowledge of words while a speaking test would assess a person ability to use words or their performance.

Generally, a test is either knowledge or performance based.  it is possible to blend the two, however, mixing styles raises the complexity not only for the student but also for the person who s responsible for marking the results.

Domain Specific

A test needs to be focused on a specific area of knowledge. A language test is specific to language as an example. A teacher needs to know in what specific area they are trying to assess students knowledge/performance. This not always easy to define as not only are there domains but sub-domains and many other ways to divide up the information in a given course.

Therefore, a teacher needs to identify what students need to know as well as what they should know and assess this information when developing a test. This helps to focus the test on relevant content for the students.

Conclusion

There is art and science to testing. There is no simple solution to how to setup tests to help students. However, the five concepts here provides a framework that can help a teacher to get started in developing tests.

Gradient Boosting Of Regression Trees in R

Gradient boosting is a machine learning tool for “boosting” or improving model performance. How this works is that you first develop an initial model called the base learner using whatever algorithm of your choice (linear, tree, etc.).

Gradient boosting looks at the error and develops a second model using what is called da loss function. the loss function is the difference between the current accuracy and the desired prediction whether it’s accuracy for classification or error in regression. This process of making additional models based only on the misclassified ones continues until the level of accuracy is reached.

Gradient boosting is also stochastic. This means that it randomly draws from the sample as it iterates over the data. This helps to improve accuracy and or reduce error.

In this post, we will use gradient boosting for regression trees. In particular, we will use the “Sacramento” dataset from the “caret” package. Our goal is to predict a house’s price based on the available variables. Below is some initial code

library(caret);library(gbm);library(corrplot)
data("Sacramento")
str(Sacramento)
## 'data.frame':    932 obs. of  9 variables:
##  $ city     : Factor w/ 37 levels "ANTELOPE","AUBURN",..: 34 34 34 34 34 34 34 34 29 31 ...
##  $ zip      : Factor w/ 68 levels "z95603","z95608",..: 64 52 44 44 53 65 66 49 24 25 ...
##  $ beds     : int  2 3 2 2 2 3 3 3 2 3 ...
##  $ baths    : num  1 1 1 1 1 1 2 1 2 2 ...
##  $ sqft     : int  836 1167 796 852 797 1122 1104 1177 941 1146 ...
##  $ type     : Factor w/ 3 levels "Condo","Multi_Family",..: 3 3 3 3 3 1 3 3 1 3 ...
##  $ price    : int  59222 68212 68880 69307 81900 89921 90895 91002 94905 98937 ...
##  $ latitude : num  38.6 38.5 38.6 38.6 38.5 ...
##  $ longitude: num  -121 -121 -121 -121 -121 ...

Data Preparation

Already there are some actions that need to be made. We need to remove the variables “city” and “zip” because they both have a large number of factors. Next, we need to remove “latitude” and “longitude” because these values are hard to interpret in a housing price model. Let’s run the correlations before removing this information

corrplot(cor(Sacramento[,c(-1,-2,-6)]),method = 'number')

1

There also appears to be a high correlation between “sqft” and beds and bathrooms. As such, we will remove “sqft” from te model. Below is the code for the revised variables remaining for the model.

sacto.clean<-Sacramento
sacto.clean[,c(1,2,5)]<-NULL
sacto.clean[,c(5,6)]<-NULL
str(sacto.clean)
## 'data.frame':    932 obs. of  4 variables:
##  $ beds : int  2 3 2 2 2 3 3 3 2 3 ...
##  $ baths: num  1 1 1 1 1 1 2 1 2 2 ...
##  $ type : Factor w/ 3 levels "Condo","Multi_Family",..: 3 3 3 3 3 1 3 3 1 3 ...
##  $ price: int  59222 68212 68880 69307 81900 89921 90895 91002 94905 98937 ...

We will now develop our training and testing sets

set.seed(502)
ind=sample(2,nrow(sacto.clean),replace=T,prob=c(.7,.3))
train<-sacto.clean[ind==1,]
test<-sacto.clean[ind==2,]

We need to create a grid in order to develop the many different potential models available. We have to tune three different parameters for gradient boosting, These three parameters are number of trees, interaction depth, and shrinkage. Number of trees is how many trees gradient boosting g will make, interaction depth is the number of splits, shrinkage controls the contribution of each tree and stump to the final model. We also have to determine the type of cross-validation using the “trainControl”” function. Below is the code for the grid.

grid<-expand.grid(.n.trees=seq(100,500,by=200),.interaction.depth=seq(1,4,by=1),.shrinkage=c(.001,.01,.1),
                  .n.minobsinnode=10)
control<-trainControl(method = "CV")

Model Training

We now can train our model

gbm.train<-train(price~.,data=train,method='gbm',trControl=control,tuneGrid=grid)
gbm.train

 

Stochastic Gradient Boosting 

685 samples
  4 predictors

No pre-processing
Resampling: Cross-Validated (25 fold) 
Summary of sample sizes: 659, 657, 658, 657, 657, 657, ... 
Resampling results across tuning parameters:

  shrinkage  interaction.depth  n.trees  RMSE       Rsquared 
  0.001      1                  100      128372.32  0.4850879
  0.001      1                  300      120272.16  0.4965552
  0.001      1                  500      113986.08  0.5064680
  0.001      2                  100      127197.20  0.5463527
  0.001      2                  300      117228.42  0.5524074
  0.001      2                  500      109634.39  0.5566431
  0.001      3                  100      126633.35  0.5646994
  0.001      3                  300      115873.67  0.5707619
  0.001      3                  500      107850.02  0.5732942
  0.001      4                  100      126361.05  0.5740655
  0.001      4                  300      115269.63  0.5767396
  0.001      4                  500      107109.99  0.5799836
  0.010      1                  100      103554.11  0.5286663
  0.010      1                  300       90114.05  0.5728993
  0.010      1                  500       88327.15  0.5838981
  0.010      2                  100       97876.10  0.5675862
  0.010      2                  300       88260.16  0.5864650
  0.010      2                  500       86773.49  0.6007150
  0.010      3                  100       96138.06  0.5778062
  0.010      3                  300       87213.34  0.5975438
  0.010      3                  500       86309.87  0.6072987
  0.010      4                  100       95260.93  0.5861798
  0.010      4                  300       86962.20  0.6011429
  0.010      4                  500       86380.39  0.6082593
  0.100      1                  100       86808.91  0.6022690
  0.100      1                  300       86081.65  0.6100963
  0.100      1                  500       86197.52  0.6081493
  0.100      2                  100       86810.97  0.6036919
  0.100      2                  300       87251.66  0.6042293
  0.100      2                  500       88396.21  0.5945206
  0.100      3                  100       86649.14  0.6088309
  0.100      3                  300       88565.35  0.5942948
  0.100      3                  500       89971.44  0.5849622
  0.100      4                  100       86922.22  0.6037571
  0.100      4                  300       88629.92  0.5894188
  0.100      4                  500       91008.39  0.5718534

Tuning parameter 'n.minobsinnode' was held constant at a value of 10
RMSE was used to select the optimal model using  the smallest value.
The final values used for the model were n.trees = 300, interaction.depth = 1, shrinkage = 0.1 and n.minobsinnode = 10.

The print out shows you the values for each potential model. At the bottom of the printout are the recommended parameters for our model. We take the values at the bottom to create our model for the test data.

gbm.price<-gbm(price~.,data=train,n.trees = 300,interaction.depth = 1,
              shrinkage = .1,distribution = 'gaussian')

Test Model

Now we use the test data, below we predict as well as calculate the error and make a plot.

gbm.test<-predict(gbm.price,newdata = test,n.trees = 300)
gbm.resid<-gbm.test-test$price
mean(gbm.resid^2)
## [1] 8721772767
plot(gbm.test,test$price)

1

The actual value for the mean squared error is relative and means nothing by its self. The plot, however, looks good and indicates that our model may be doing well. The mean squared error is only useful when comparing one model to another it does not mean much by its self.

English Language and the Church

The English language during the middle ages had a serious struggle with the church of its time. Church officials supported that the bible should only be published in Latin. This led to a large number of people having no idea what was happening during a worship service. Even though church attendance was mandatory.

One response to this problem was the development of “mystery plays.” These were theatrical performances based on the bible. The topics ranged from Genesis to Revelation and were performed in local languages. However, watching pseudo-movies and reading the text for yourself are widely different experiences.

This post will look at the role of several prominent people’s response to the suppression of English in religious text.

John Wycliffe

The lack of scripture in the English language led to John Wycliffe translating the Latin Vulgate into English. Naturally, this was illegal and Wycliffe faced significant trouble over doing this. Despite this, his translation was one of the first translations of the bible into what was called at the time a “vulgar” language.

Wycliffe’s translation was not from the original text but rather from the Latin. This means it was a translation of a translation which nearly destroys the comprehensibility of the text.

William Tyndale

William Tyndale attempted to deal with the challenges of the Wycliff translation by translating the bible from the original greek and Hebrew. Tyndale’s translation heavily influences the English language as he literally had to create words to capture the meaning of the text. Such phrases as “scapegoat”, “sea-shore”, and “my brother’s  keeper” were developed by Tyndale to communicate ideas within the bible.  For his work, Tyndale was put to death.  It took him about

Naturally, many were not happen with what Tyndale had accomplished. For his work, Tyndale was put to death.  It took him about four years to complete his work

King James Bible

However, the move away from Latin to English was made complete with the development of the 1611 King James bible. The KJV is named as King James the I of England who sponsored the translation of the bible for political reasons.  By the 17th century, there were so many versions of the bible that scholars wanted a definitive translation and King James I sponsored this.

Over fifty scholars worked on this translation for five years. Despite all this work, the 1611 KJV is 60-80% based on Tyndale’s work a century prior. This makes Tyndale’s work all the more amazing that he did the work of 50 scholars in the same amount of time. From this moment English became know as the language of the preacher

Conclusion

The role of English in religious matters today is due in part to the work of Wycliffe, Tyndale, and the scholars of the KJV. Their efforts led to supplanting Latin as the language of worship while also contributing many idioms to the English language

Random Forest Classification in R

This post will cover the use of random forest for classification. Random forest involves the use of many decision trees in the development of a classification or regression tree. The results of each individual tree is added together and the mean is used in the final classification of an example. The use of an ensemble helps in dealing with the bias-variance tradeoff.

In the example of random forest classification, we will use the “Participation” dataset from the “ecdat” package. We want to classify people by their labor participation based on the other variables available in the dataset. Below is some initial code

library(randomForest);library(Ecdat)
data("Participation")
str(Participation)
## 'data.frame':    872 obs. of  7 variables:
##  $ lfp    : Factor w/ 2 levels "no","yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ lnnlinc: num  10.8 10.5 11 11.1 11.1 ...
##  $ age    : num  3 4.5 4.6 3.1 4.4 4.2 5.1 3.2 3.9 4.3 ...
##  $ educ   : num  8 8 9 11 12 12 8 8 12 11 ...
##  $ nyc    : num  1 0 0 2 0 0 0 0 0 0 ...
##  $ noc    : num  1 1 0 0 2 1 0 2 0 2 ...
##  $ foreign: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

For the data preparation, we need to multiple age by ten as the current values imply small children. Furthermore, we need to change the “lnnlinc” variable from the log of salary to just the regular salary. After completing these two steps, we need to split our data into training and testing sets. Below is the code

Participation$age<-10*Participation$age #normal age
Participation$lnnlinc<-exp(Participation$lnnlinc) #actual income not log
#split data
set.seed(502)
ind=sample(2,nrow(Participation),replace=T,prob=c(.7,.3))
train<-Participation[ind==1,]
test<-Participation[ind==2,]

We will now create our classification model using random forest.

set.seed(123)
rf.lfp<-randomForest(lfp~.,data = train)
rf.lfp
## 
## Call:
##  randomForest(formula = lfp ~ ., data = train) 
##                Type of random forest: classification
##                      Number of trees: 500
## No. of variables tried at each split: 2
## 
##         OOB estimate of  error rate: 32.39%
## Confusion matrix:
##      no yes class.error
## no  248  93   0.2727273
## yes 113 182   0.3830508

The output is mostly self-explanatory. It includes the number of trees, number of variables at each split, error rate, and the confusion matrix. In general, are error rate is poor and we are having a hard time distinguishing between those who work and do not work based on the variables in the dataset. However, this is based on having all 500 trees in the analysis. Having this many trees is probably not necessary but we need to confirm this.

We can also plot the error by tree using the “plot” function as shown below.

plot(rf.lfp)

1.png

It looks as though error lowest with around 400 trees. We can confirm this using the “which.min” function and call information from “err.rate” in our model.

which.min(rf.lfp$err.rate[,1])
## [1] 242

We need 395 trees in order to reduce the error rate to its most optimal level. We will now create a new model that contains 395 trees in it.

rf.lfp2<-randomForest(lfp~.,data = train,ntree=395)
rf.lfp2
## 
## Call:
##  randomForest(formula = lfp ~ ., data = train, ntree = 395) 
##                Type of random forest: classification
##                      Number of trees: 395
## No. of variables tried at each split: 2
## 
##         OOB estimate of  error rate: 31.92%
## Confusion matrix:
##      no yes class.error
## no  252  89   0.2609971
## yes 114 181   0.3864407

The results are mostly the same. There is a small decline in error but not much to get excited about. We will now run our model on the test set.

rf.lfptest<-predict(rf.lfp2,newdata=test,type = 'response')
table(rf.lfptest,test$lfp)
##           
## rf.lfptest no yes
##        no  93  48
##        yes 37  58
(92+63)/(92+63+43+38) #calculate accuracy
## [1] 0.6567797

Still disappointing, there is one last chart we should examine and that is the importance of each variable plot. It shows which variables are most useful in the prediction process. Below is the code.

varImpPlot(rf.lfp2)

1.png

This plot clearly indicates that salary (“lnnlinc”), age, and education are the strongest features for classifying by labor activity. However, the overall model is probably not useful.

Conclusion

This post explained and demonstrated how to conduct a random forest analysis. This form of analysis is powerful in dealing with large datasets with nonlinear relationships among the variables.

Post Norman Conquest Decline of the French Language

After the Norman Conquest of England in 1066, French dominated England for three hundred years. The decline French can be traced to at least two main reasons, which are…

  • War/politics
  • Disease

This post will examine briefly the role of these two phenomena in shaping the decline of the French language in England as well as the reemergence of English.

War/Politics

The King of Normandy was also the King of England. In 1204, John, King of Normandy and England, lost his Norman territory to the King of France. This left a large number of Norman nobles living in England with any property back in France unless the swore allegiance to the King of France, Philipp II. The consequence of forced loyalty was the development of an English identity among the elite.

In 1295, Philip IV, King of France, threaten to invade England. Edward I, King of England, communicated with the people in English in order to unite the people. While speaking to the people in English, Edward I stated that Philip IV intended to destroy the English language. When the French invasion never came, Edward set aside his use of English

Disease

In the mid-1300’s, the Bubonic plague spread through England and wipe out 1/3 of the population. The plague was particular hard on the clergy killing almost 1/2 of them and removing the influence of Latin on English. The replacement clergy used English.

The loss of so many people allowed English-speaking peasants to take over empty homes and demand higher wages. The price of land fell as there was no one to work the fields nor was there as much demand for products with so many dead. The bonds of serfdom were severely broken.

When the nobility tried to push the peasants back onto the lands as serfs, it led several revolts. When communicating both the nobility and peasants used English. The nobility used English to make promises that were not kept and destroy resistance their rule.

Aftermath

Through war and disease, English rose to prominence again. By the 1400’s English was the language of education and official business. In 1399, Henry IV was sworn in as king with the use of the English language. After three centuries of oppression, the English language emerged as the language of the elite as well as the commoner again.

Norman Conquest and the English Language

The year 1066 is highly significant in the English language. This is the year that William, the Duke of Normandy, conquer most of what today is known as Great Britain. The effects of this upon the English language was significant.

Background

As a background, when the King of England, Edward the Confessor died, he named William, the duke of Normandy, as King of England. Edward was childless but his mother was from Normandy, which is located in France.  As such, the English court was already full of French speaking Normans as Edward’s supporters.

Naming a Norman to the throne of England did not sit well with one Edward’s biggest rivals, Earl Harold Godwineson. Harold quickly led a rebellion against Willam but was defeated and William of Normandy became known as William the Conqueror and was crowned King of England Christmas day of 1066

Aftermath

Over the next three centuries under French rule, the English language was invaded by as many as 10,000 French words. Such words as “city”, “bacon”, “biscuit”, and “felony” to name a few. The English court quickly became a French court.

The English court quickly became a French court. All positions of power were taken by Normans. This was not only because of conquest but also because most of the English nobility and leadership were killed in the Battle of Hastings.

The only way to get ahead in this context was to learn French and leave English in the home. In many ways, French became a high language and English was relegated to a low language almost as a diglossia situation. English was the language of the poor and French of the elite. Most documents during this time were produced in French and even written English was pushed aside.

The division by class has led some to allege that this kept English alive. This is to say that the rich and the poor had their own separate languages and both work to preserve their own manner of communication.

Conclusion

War is yet another factor to consider when looking at the development of a language. Even without intending to do so William the Conqueror made a major impact on the English language simply by sticking to his mother tongue of French when he took the English throne. To this day, loan words from French play a major role in communication in the English language.

Random Forest Regression Trees in R

Random forest involves the process of creating multiple decision trees and the combing of their results. How this is done is through r using 2/3 of the data set to develop decision tree. This is done dozens, hundreds, or more times. Every tree made is created with a slightly different sample. The results of all these trees are then averaged together. This process of sampling is called bootstrap aggregation or bagging for short.

While the random forest algorithm is developing different samples it also randomly selects which variables to be use din each tree that is developed. By randomizing the sample and the features used in the tree, random forest is able to reduce both bias and variance in a model. In addition, random forest is robust against outliers and collinearity. Lastly, keep in mind that random forest can be used for regression and classification trees

In our example, we will use the “Participation” dataset from the “Ecdat” package. We will create a random forest regression tree to predict income of people. Below is some initial code

library(randomForest);library(rpart);library(Ecdat)
data("Participation")
str(Participation)
## 'data.frame':    872 obs. of  7 variables:
##  $ lfp    : Factor w/ 2 levels "no","yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ lnnlinc: num  10.8 10.5 11 11.1 11.1 ...
##  $ age    : num  3 4.5 4.6 3.1 4.4 4.2 5.1 3.2 3.9 4.3 ...
##  $ educ   : num  8 8 9 11 12 12 8 8 12 11 ...
##  $ nyc    : num  1 0 0 2 0 0 0 0 0 0 ...
##  $ noc    : num  1 1 0 0 2 1 0 2 0 2 ...
##  $ foreign: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

We now need to prepare the data. We need to transform the lnnlinc from a log of salary to the actual salary. In addition, we need to multiply “age” by ten as 3.4 & 4.5 do not make any sense. Below is the code

Participation$age<-10*Participation$age #normal age
Participation$lnnlinc<-exp(Participation$lnnlinc) #actual income not log

Now we create our training and testing sets.

set.seed(123)
ind=sample(2,nrow(Participation),replace=T,prob=c(.7,.3))
train<-Participation[ind==1,]
test<-Participation[ind==2,]

We are now ready to create our model. Below is the code

set.seed(123)
rf.pros<-randomForest(lnnlinc~.,data = train)
rf.pros
## 
## Call:
##  randomForest(formula = lnnlinc ~ ., data = train) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 2
## 
##           Mean of squared residuals: 529284177
##                     % Var explained: 13.74

As you can see from calling “rf.pros” the variance explained is low at around 14%. The output also tells us how many trees were created. You have to be careful with making too many trees as this leads to overfitting. We can determine how many trees are optimal by looking at a plot and then using the “which.min”. Below is a plot of the number of trees by the mean squared error.

plot(rf.pros)

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As you can see, as there are more trees there us less error to a certain point. It looks as though about 50 trees is enough. To confirm this guess, we used the “which.min” function. Below is the code

which.min(rf.pros$mse)
## [1] 45

We need 45 trees to have the lowest error. We will now rerun the model and add an argument called “ntrees” to indicating the number of trees we want to generate.

set.seed(123)
rf.pros.45<-randomForest(lnnlinc~.,data = train,ntree=45)
rf.pros.45
## 
## Call:
##  randomForest(formula = lnnlinc ~ ., data = train, ntree = 45) 
##                Type of random forest: regression
##                      Number of trees: 45
## No. of variables tried at each split: 2
## 
##           Mean of squared residuals: 520705601
##                     % Var explained: 15.13

This model is still not great. We explain a little bit more of the variance and the error decreased slightly. We can now see which of the features in our model are the most useful by using the “varImpPlot” function. Below is the code.

varImpPlot(rf.pros.45)

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The higher the IncNodePurity the more important the variable. AS you can see, education is most important followed by age and then the number of older children. The raw scores for each variable can be examined using the “importance” function. Below is the code.

importance(rf.pros.45)
##         IncNodePurity
## lfp       16678498398
## age       66716765357
## educ      72007615063
## nyc        9337131671
## noc       31951386811
## foreign   10205305287

We are now ready to test our model with the test set. We will then calculate the residuals and the mean squared error

rf.pros.test<-predict(rf.pros.45,newdata = test)
rf.resid<-rf.pros.test-test$lnnlinc
mean(rf.resid^2)
## [1] 381850711

Remember that the mean squared error calculated here is only useful in comparison to other models. Random forest provides a way in which to remove the weaknesses of one decision tree by averaging the results of many. This form of ensemble learning is one of the more powerful algorithms in machine learning.

Understanding Classification Trees Using R

Classification trees are similar to regression trees except that the determinant of success is not the residual sum of squares but rather the error rate. The strange thing about classification trees is that you can you can continue to gain information in splitting the tree without necessarily improving the misclassification rate. This is done through calculating a measure of error called the Gini coefficient

Gini coefficient is calculated using the values of the accuracy and error in an equation. For example, if we have a model that is 80% accurate with a 20% error rate the Gini coefficient is calculated as follows for a single node

n0gini<- 1 - (((8/10)^2) -((2/10)^2)) 
n0gini
## [1] 0.4

Now if we split this into two nodes notice the change in the Gini coefficient

n1gini<-1-(((5/6)^2)-((1/7)^2))
n2gini<-1-(((3/4)^2))-((1/3)^2)
newgini<-(.8*n1gini) + (.2*n2gini)
newgini
## [1] 0.3260488

The lower the Gini coefficient the better as it measures purity. IN the example, there is no improvement in the accuracy yet there is an improvement in the Gini coefficient. Therefore, classification is about purity and not the residual sum of squares.

In this post, we will make a classification tree to predict if someone is participating in the labor market. We will do this using the “Participation” dataset from the “Ecdat” package. Below is some initial code to get started.

library(Ecdat);library(rpart);library(partykit)
data(Participation)
str(Participation)
## 'data.frame':    872 obs. of  7 variables:
##  $ lfp    : Factor w/ 2 levels "no","yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ lnnlinc: num  10.8 10.5 11 11.1 11.1 ...
##  $ age    : num  3 4.5 4.6 3.1 4.4 4.2 5.1 3.2 3.9 4.3 ...
##  $ educ   : num  8 8 9 11 12 12 8 8 12 11 ...
##  $ nyc    : num  1 0 0 2 0 0 0 0 0 0 ...
##  $ noc    : num  1 1 0 0 2 1 0 2 0 2 ...
##  $ foreign: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

The ‘age’ feature needs to be transformed. Since it is doubtful that the survey was conducted among 4 and 5-year-olds. We need to multiply this variable by ten. In addition, the “lnnlinc” feature is the log of income and we want the actual income so we will exponentiate this information. Below is the code for these two steps.

Participation$age<-10*Participation$age #normal age
Participation$lnnlinc<-exp(Participation$lnnlinc) #actual income not log

We will now create our training and testing datasets with the code below.

set.seed(502)
ind=sample(2,nrow(Participation),replace=T,prob=c(.7,.3))
train<-Participation[ind==1,]
test<-Participation[ind==2,]

We can now create our classification tree and take a look at the output

tree.pros<-rpart(lfp~.,data = train)
tree.pros
## n= 636 
## 
## node), split, n, loss, yval, (yprob)
##       * denotes terminal node
## 
##   1) root 636 295 no (0.5361635 0.4638365)  
##     2) foreign=no 471 182 no (0.6135881 0.3864119)  
##       4) nyc>=0.5 99  21 no (0.7878788 0.2121212) *
##       5) nyc< 0.5 372 161 no (0.5672043 0.4327957)  
##        10) age>=49.5 110  25 no (0.7727273 0.2272727) *
##        11) age< 49.5 262 126 yes (0.4809160 0.5190840)  
##          22) lnnlinc>=46230.43 131  50 no (0.6183206 0.3816794)  
##            44) noc>=0.5 102  34 no (0.6666667 0.3333333) *
##            45) noc< 0.5 29  13 yes (0.4482759 0.5517241)  
##              90) lnnlinc>=47910.86 22  10 no (0.5454545 0.4545455)  
##               180) lnnlinc< 65210.78 12   3 no (0.7500000 0.2500000) *
##               181) lnnlinc>=65210.78 10   3 yes (0.3000000 0.7000000) *
##              91) lnnlinc< 47910.86 7   1 yes (0.1428571 0.8571429) *
##          23) lnnlinc< 46230.43 131  45 yes (0.3435115 0.6564885) *
##     3) foreign=yes 165  52 yes (0.3151515 0.6848485)  
##       6) lnnlinc>=56365.39 16   5 no (0.6875000 0.3125000) *
##       7) lnnlinc< 56365.39 149  41 yes (0.2751678 0.7248322) *

In the text above, the first split is made on the feature “foreign” which is a yes or no possibility. 471 were not foreigners will 165 were foreigners. The accuracy here is not great at 61% for those not classified as foreigners and 31% for those classified as foreigners. For the 165 that are classified as foreigners, the next split is by their income, etc. This is hard to understand. Below is an actual diagram of the text above.

plot(as.party(tree.pros))

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We now need to determining if pruning the tree is beneficial. We do this by looking at the cost complexity. Below is the code.

tree.pros$cptable
##           CP nsplit rel error    xerror       xstd
## 1 0.20677966      0 1.0000000 1.0000000 0.04263219
## 2 0.04632768      1 0.7932203 0.7932203 0.04122592
## 3 0.02033898      4 0.6542373 0.6677966 0.03952891
## 4 0.01016949      5 0.6338983 0.6881356 0.03985120
## 5 0.01000000      8 0.6033898 0.6915254 0.03990308

The “rel error” indicates that our model is bad no matter how any splits. Even with 9 splits we have an error rate of 60%. Below is a plot of the table above

plotcp(tree.pros)

1.png

Based on the table, we will try to prune the tree to 5 splits. The plot above provides a visual as it has the lowest error. The table indicates that a tree of five splits (row number 4) has the lowest cross-validation error (xstd). Below is the code for pruning the tree followed by a plot of the modified tree.

cp<-min(tree.pros$cptable[4,])
pruned.tree.pros<-prune(tree.pros,cp=cp)
plot(as.party(pruned.tree.pros))

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IF you compare the two trees we have developed. One of the main differences is that the pruned.tree is missing the “noc” (number of older children) variable. There are also fewer splits on the income variable (lnnlinc). We can no use the pruned tree with the test data set.

party.pros.test<-predict(pruned.tree.pros,newdata=test,type="class")
table(party.pros.test,test$lfp)
##                
## party.pros.test no yes
##             no  90  41
##             yes 40  65

Now for the accuracy

(90+65) / (90+41+40+65)
## [1] 0.6567797

This is surprisingly high compared to the results for the training set but 65% is not great, However, this is fine for a demonstration.

Conclusion

Classification trees are one of many useful tools available for data analysis. When developing classification trees one of the key ideas to keep in mind is the aspect of prunning as this affects the complexity of the model.

Numeric Prediction with Support Vector Machines in R

In this post, we will look at support vector machines for numeric prediction. SVM is used for both classification and numeric prediction. The advantage of SVM for numeric prediction is that SVM will automatically create higher dimensions of the features and summarizes this in the output. In other words, unlike in regression where you have to decide for yourself how to modify your features, SVM does this automatically using different kernels.

Different kernels transform the features in different ways. And the cost function determines the penalty for an example being on the wrong side of the margin developed by the kernel. Remember that SVM draws lines and separators to divide the examples. Examples on the wrong side are penalized as determined by the researcher.

Just like with regression, generally, the model with the least amount of error may be the best model. As such, the purpose of this post is to use SVM to predict income in the “Mroz” dataset from the “Ecdat” package. We will use several different kernels that will transformation the features different ways and calculate the mean-squared error to determine the most appropriate model. Below is some initial code.

library(caret);library(e1071);library(corrplot);library(Ecdat)
data(Mroz)
str(Mroz)
## 'data.frame':    753 obs. of  18 variables:
##  $ work      : Factor w/ 2 levels "yes","no": 2 2 2 2 2 2 2 2 2 2 ...
##  $ hoursw    : int  1610 1656 1980 456 1568 2032 1440 1020 1458 1600 ...
##  $ child6    : int  1 0 1 0 1 0 0 0 0 0 ...
##  $ child618  : int  0 2 3 3 2 0 2 0 2 2 ...
##  $ agew      : int  32 30 35 34 31 54 37 54 48 39 ...
##  $ educw     : int  12 12 12 12 14 12 16 12 12 12 ...
##  $ hearnw    : num  3.35 1.39 4.55 1.1 4.59 ...
##  $ wagew     : num  2.65 2.65 4.04 3.25 3.6 4.7 5.95 9.98 0 4.15 ...
##  $ hoursh    : int  2708 2310 3072 1920 2000 1040 2670 4120 1995 2100 ...
##  $ ageh      : int  34 30 40 53 32 57 37 53 52 43 ...
##  $ educh     : int  12 9 12 10 12 11 12 8 4 12 ...
##  $ wageh     : num  4.03 8.44 3.58 3.54 10 ...
##  $ income    : int  16310 21800 21040 7300 27300 19495 21152 18900 20405 20425 ...
##  $ educwm    : int  12 7 12 7 12 14 14 3 7 7 ...
##  $ educwf    : int  7 7 7 7 14 7 7 3 7 7 ...
##  $ unemprate : num  5 11 5 5 9.5 7.5 5 5 3 5 ...
##  $ city      : Factor w/ 2 levels "no","yes": 1 2 1 1 2 2 1 1 1 1 ...
##  $ experience: int  14 5 15 6 7 33 11 35 24 21 ...

We need to place the factor variables next to each other as it helps in having to remove them when we need to scale the data. We must scale the data because SVM is based on distance when making calculations. If there are different scales the larger scale will have more influence on the results. Below is the code

mroz.scale<-Mroz[,c(17,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18)]
mroz.scale<-as.data.frame(scale(mroz.scale[,c(-1,-2)])) #remove factor variables for scaling
mroz.scale$city<-Mroz$city # add factor variable back into the dataset
mroz.scale$work<-Mroz$work # add factor variable back into the dataset
#mroz.cor<-cor(mroz.scale[,-17:-18])
#corrplot(mroz.cor,method='number', col='black')

Below is the code for creating the train and test datasets.

set.seed(502)
ind=sample(2,nrow(mroz.scale),replace=T,prob=c(.7,.3))
train<-mroz.scale[ind==1,]
test<-mroz.scale[ind==2,]

Linear Kernel

Our first kernel is the linear kernel. Below is the code. We use the “tune.svm” function from the “e1071” package. We set the kernel to “linear” and we pick our own values for the cost function. The numbers for the cost function can be whatever you want. Also, keep in mind that r will produce six different models because we have six different values in the “cost” argument.

The process we are using to develop the models is as follows

  1. Set the seed
  2. Develop the initial model by setting the formula, dataset, kernel, cost function, and other needed information.
  3. Select the best model for the test set
  4. Predict with the best model
  5. Plot the predicted and actual results
  6. Calculate the mean squared error

The first time we will go through this process step-by-step. However, all future models will just have the code followed by an interpretation.

linear.tune<-tune.svm(income~.,data=train,kernel="linear",cost = c(.001,.01,.1,1,5,10))
summary(linear.tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##    10
## 
## - best performance: 0.3492453 
## 
## - Detailed performance results:
##    cost     error dispersion
## 1 1e-03 0.6793025  0.2285748
## 2 1e-02 0.3769298  0.1800839
## 3 1e-01 0.3500734  0.1626964
## 4 1e+00 0.3494828  0.1618478
## 5 5e+00 0.3493379  0.1611353
## 6 1e+01 0.3492453  0.1609774

The best model had a cost = 10 with a performance of .35. We will select the best model and use this on our test data. Below is the code.

best.linear<-linear.tune$best.model
tune.test<-predict(best.linear,newdata=test)

Now we will create a plot so we can see how well our model predicts. In addition, we will calculate the mean squared error to have an actual number of our model’s performance

plot(tune.test,test$income)

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tune.test.resid<-tune.test-test$income
mean(tune.test.resid^2)
## [1] 0.215056

The model looks good in the plot. However, we cannot tell if the error number is decent until it is compared to other models

Polynomial Kernel

The next kernel we will use is the polynomial one. The kernel requires two parameters the degree of the polynomial (3,4,5, etc) as well as the kernel coefficient. Below is the code

set.seed(123)
poly.tune<-tune.svm(income~.,data = train,kernal="polynomial",degree = c(3,4,5),coef0 = c(.1,.5,1,2,3,4))
best.poly<-poly.tune$best.model
poly.test<-predict(best.poly,newdata=test)
plot(poly.test,test$income)

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poly.test.resid<-poly.test-test$income
mean(poly.test.resid^2)
## [1] 0.2453022

The polynomial has an insignificant additional amount of error.

Radial Kernel

Next, we will use the radial kernel. One thing that is new here is the need for a parameter in the code call gamma. Below is the code.

set.seed(123)
rbf.tune<-tune.svm(income~.,data=train,kernel="radial",gamma = c(.1,.5,1,2,3,4))
summary(rbf.tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  gamma
##    0.1
## 
## - best performance: 0.5225952 
## 
## - Detailed performance results:
##   gamma     error dispersion
## 1   0.1 0.5225952  0.4183170
## 2   0.5 0.9743062  0.5293211
## 3   1.0 1.0475714  0.5304482
## 4   2.0 1.0582550  0.5286129
## 5   3.0 1.0590367  0.5283465
## 6   4.0 1.0591208  0.5283059
best.rbf<-rbf.tune$best.model
rbf.test<-predict(best.rbf,newdata=test)
plot(rbf.test,test$income)

1.png

rbf.test.resid<-rbf.test-test$income
mean(rbf.test.resid^2)
## [1] 0.3138517

The radial kernel is worst than the linear and polynomial kernel. However, there is not much different in the performance of the models so far.

Sigmoid Kernel

Next, we will try the sigmoid kernel. Sigmoid kernel relies on a “gamma” parameter and a cost function. Below is the code

set.seed(123)
sigmoid.tune<-tune.svm(income~., data=train,kernel="sigmoid",gamma = c(.1,.5,1,2,3,4),coef0 = c(.1,.5,1,2,3,4))
summary(sigmoid.tune)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  gamma coef0
##    0.1     3
## 
## - best performance: 0.8759507 
## 
## - Detailed performance results:
##    gamma coef0        error  dispersion
## 1    0.1   0.1   27.0808221   6.2866615
## 2    0.5   0.1  746.9235624 129.0224096
## 3    1.0   0.1 1090.9660708 198.2993895
## 4    2.0   0.1 1317.4497885 214.7997608
## 5    3.0   0.1 1339.8455047 180.3195491
## 6    4.0   0.1 1299.7469190 201.6901577
## 7    0.1   0.5  151.6070833  38.8450961
## 8    0.5   0.5 1221.2396575 335.4320445
## 9    1.0   0.5 1225.7731007 190.7718103
## 10   2.0   0.5 1290.1784238 216.9249899
## 11   3.0   0.5 1338.1069460 223.3126800
## 12   4.0   0.5 1261.8861304 300.0001079
## 13   0.1   1.0  162.6041229  45.3216740
## 14   0.5   1.0 2276.4330973 330.1739559
## 15   1.0   1.0 2036.4791854 335.8051736
## 16   2.0   1.0 1626.4347749 290.6445164
## 17   3.0   1.0 1333.0626614 244.4424896
## 18   4.0   1.0 1343.7617925 194.2220729
## 19   0.1   2.0   19.2061993   9.6767496
## 20   0.5   2.0 2504.9271757 583.8943008
## 21   1.0   2.0 3296.8519140 542.7903751
## 22   2.0   2.0 2376.8169815 398.1458855
## 23   3.0   2.0 1949.9232179 319.6548059
## 24   4.0   2.0 1758.7879267 313.2581011
## 25   0.1   3.0    0.8759507   0.3812578
## 26   0.5   3.0 1405.9712578 389.0822797
## 27   1.0   3.0 3559.4804854 843.1905348
## 28   2.0   3.0 3159.9549029 492.6072149
## 29   3.0   3.0 2428.1144437 412.2854724
## 30   4.0   3.0 1997.4596435 372.1962595
## 31   0.1   4.0    0.9543167   0.5170661
## 32   0.5   4.0  746.4566494 201.4341061
## 33   1.0   4.0 3277.4331302 527.6037421
## 34   2.0   4.0 3643.6413379 604.2778089
## 35   3.0   4.0 2998.5102806 471.7848740
## 36   4.0   4.0 2459.7133632 439.3389369
best.sigmoid<-sigmoid.tune$best.model
sigmoid.test<-predict(best.sigmoid,newdata=test)
plot(sigmoid.test,test$income)

 

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sigmoid.test.resid<-sigmoid.test-test$income
mean(sigmoid.test.resid^2)
## [1] 0.8004045

The sigmoid performed much worst then the other models based on the metric of error. You can further see the problems with this model in the plot above.

Conclusion

The final results are as follows

  • Linear kernel .21
  • Polynomial kernel .24
  • Radial kernel .31
  • Sigmoid kernel .80

Which model to select depends on the goals of the study. However, it definitely looks as though you would be picking from among the first three models. The power of SVM is the ability to use different kernels to uncover different results without having to really modify the features yourself.

The Beginnings of English

What we now know as English today has a long and complex history. With any subject that is complex, it is necessary to pick a starting point and work from there. For this post, we will date the origins of English from the early 5th century.

Early History

English was not born in England.  Rather, it came to England through the invasion of Germanic warriors. These “barbarian” hoards push the indigenous Celts and Britons almost into the ocean.

However, it was not only war and conquest that brought English. The roots of English arrived also in the immigration of farmers. Either way, English slowly grew to be one of the prominent languages of England.

In the late sixth century, the Roman Catholic Church came to England. This left a mark on English in the various words taken from Latin and Greek. Such words as “angels”, “pope”, and “minister” all arrived through the Catholic Church.

Vikings and Alfred the Great

By the 8th and 9th century the Vikings were invading lands all over Europe. It was the Danes in particular that almost wiped out the inhabitants of England. However, thanks to the craftiness of Alfred the Great the Danes were defeated and their leader Guthrum was so shocked at Alfred’s comeback victory that he was baptized and became a Christian.

Alfred set to work using the English language to unite the people. He supported education in the English language and the use of language in general. Furthermore, to try and prevent future conflicts with the Danes, Alfred gave them some territory called “Dane Law” where they could live. Naturally, staying in the area meant that the Danish language had an effect on English as well.

Alfred also supported religion. Thanks to the Viking invasions, there was almost no priest left in the entire country. Alfred could barely find a priest who could read Latin. Without religious scholarship, there could be no passing on of religious teachings. This lead Alfred to encourage the translation books in other languages (like Latin) into English.

Conclusion

The story of English is not one continuous rise to prominence. There were several experiences of up and down as the language was in England. For example, there was a time when the French language almost overran the country. Yet this is a story for another day.

Regression Tree Development in R

In this post, we will take a look at regression trees. Regression trees use a concept called recursive partitioning. Recursive partitioning involves splitting features in a way that reduces the error the most.

The splitting is also greedy which means that the algorithm will partition the data at one point without considered how it will affect future partitions. Ignoring how a current split affects the future splits can lead to unnecessary branches with high variance and low bias.

One of the main strengths of regression trees is their ability ti deal with nonlinear relationships. However, predictive performance can be hurt when a particular example is assigned the mean of a node. This forced assignment is a loss of data such as turning continuous variables into categorical variables.

in this post, we will use the “participation” dataset from the “ecdat” package to predict income based on the other variables in the dataset. Below is some initial code.

library(rpart);library(partykit);library(caret);library(Ecdat)
data("Participation")
str(Participation)
## 'data.frame':    872 obs. of  7 variables:
##  $ lfp    : Factor w/ 2 levels "no","yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ lnnlinc: num  10.8 10.5 11 11.1 11.1 ...
##  $ age    : num  3 4.5 4.6 3.1 4.4 4.2 5.1 3.2 3.9 4.3 ...
##  $ educ   : num  8 8 9 11 12 12 8 8 12 11 ...
##  $ nyc    : num  1 0 0 2 0 0 0 0 0 0 ...
##  $ noc    : num  1 1 0 0 2 1 0 2 0 2 ...
##  $ foreign: Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...

There are several things we need to do to make the results easier to interpret. The “age” variable needs to be multiplied by ten as it currently shows such results as 4.5, 3, etc. Common sense indicates that a four-year-old and a three-year-old is not earning an income.

In addition, we need to convert or income variable (lnnlinc) from the log of income to regular income. This will also help to understand the results. Below is the code.

Participation$age<-10*Participation$age #normal age
Participation$lnnlinc<-exp(Participation$lnnlinc) #actual income not log

The next step is to create our training and testing data sets. Below is the code.

set.seed(502)
ind=sample(2,nrow(Participation),replace=T,prob=c(.7,.3))
train<-Participation[ind==1,]
test<-Participation[ind==2,]

We can now develop our model. We will also use the ‘print’ command

reg.tree<-rpart(lnnlinc~.,data = train)

Below is a printout of the current tree

reg.tree
## n= 636 
## 
## node), split, n, deviance, yval
##       * denotes terminal node
## 
##   1) root 636 390503700000  48405.08  
##     2) educ< 11.5 473 127460900000  43446.69  
##       4) educ< 9.5 335  70269440000  40758.25  
##         8) foreign=yes 129  10617380000  36016.12 *
##         9) foreign=no 206  54934520000  43727.84 *
##       5) educ>=9.5 138  48892370000  49972.98 *
##     3) educ>=11.5 163 217668400000  62793.52  
##       6) age< 34.5 79  34015680000  51323.86  
##        12) age< 25.5 12    984764800  34332.97 *
##        13) age>=25.5 67  28946170000  54367.01 *
##       7) age>=34.5 84 163486000000  73580.46  
##        14) lfp=yes 36  23888410000  58916.66 *
##        15) lfp=no 48 126050900000  84578.31  
##          30) educ< 12.5 29  86940400000  74425.51  
##            60) age< 38.5 8    763764600  57390.34 *
##            61) age>=38.5 21  82970650000  80915.10  
##             122) age>=44 14  34091840000  68474.57 *
##             123) age< 44 7  42378600000 105796.20 *
##          31) educ>=12.5 19  31558550000 100074.70 *

I will not interpret all of this but here is a brief description use numbers 2,4, and 8. If the person has less than 11.5 years of education (473 qualify) If the person has less than 9.5 years of education (335 of the 473 qualify) If the person is a foreigner (129 of the 335 qualify) then their average salary is 36,016.12 dollars.

Perhaps now you can see how some data is lost. The average salary for people in this node is 36,016.12 dollars but probably nobody earns exactly this amount

If what I said does not make sense. Here is an actual plot of the current regression tree.

plot(as.party(reg.tree))

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The little boxes at the bottom are boxplots of that node.

Tree modification

We now will make modifications to the tree. We will begin by examining the cptable. Below is the code

reg.tree$cptable
##           CP nsplit rel error    xerror      xstd
## 1 0.11619458      0 1.0000000 1.0026623 0.1666662
## 2 0.05164297      1 0.8838054 0.9139383 0.1434768
## 3 0.03469034      2 0.8321625 0.9403669 0.1443843
## 4 0.02125215      3 0.7974721 0.9387060 0.1433101
## 5 0.01933892      4 0.7762200 0.9260030 0.1442329
## 6 0.01242779      5 0.7568810 0.9097011 0.1434606
## 7 0.01208066      7 0.7320255 0.9166627 0.1433779
## 8 0.01046022      8 0.7199448 0.9100704 0.1432901
## 9 0.01000000      9 0.7094846 0.9107869 0.1427025

The cptable shares a lot of information. First, cp stands for cost complexity and this is the column furthest to the left. This number decreases as the tree becomes more complex. “nsplit” indicates the number of splits in the tree. “rel error” as another term for the residual sum of squares or RSS error. The ‘xerror’ and ‘xstd’ are the cross-validated average error and standard deviation of the error respectively.

One thing we can see from the cptable is that 9 splits has the lowest error but 2 splits has the lowest cross-validated error. Below we will look at a printout of the current table.

We will now make a plot of the complexity parameter to determine at what point to prune the tree. Pruning helps in removing unnecessary splits that do not improve the model much. Below is the code. The information in the plot is a visual of the “cptable”

plotcp(reg.tree)

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It appears that a tree of size 2 is the best but this is boring. The next lowest dip is a tree of size 8. Therefore, we will prune our tree to have a size of 8 or eight splits. First, we need to create an object that contains how many splits we want. Then we use the “prune” function to make the actually modified tree.

cp<-min(reg.tree$cptable[8,])
pruned.reg.tree<-prune(reg.tree,cp=cp)

We will now make are modified tree

plot(as.party(pruned.reg.tree))

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The only difference is the loss of the age nod for greater or less than 25.5.

Model Test

We can now test our tree to see how well it performs.

reg.tree.test<-predict(pruned.reg.tree,newdata=test)
reg.tree.resid<-reg.tree.test-test$lnnlinc
mean(reg.tree.resid^2)
## [1] 431928030

The number we calculated is the mean squared error. This number must be compared to models that are developed differently in order to assess the current model. By it’s self it means nothing.

Conclusion

This post exposed you to regression trees. This type of tree can be used to m ake numeric predictions in nonlinear data. However, with the classification comes a loss of data as the uniqueness of each example is lost when placed in a node.

Statistical Models

In research, the term ‘model’ is employed frequently. Normally, a model is some sort of a description or explanation of a real world phenomenon. In data science, we employ the use of statistical models. Statistical models used numbers to help us to understand something that happens in the real world.

A statistical model used numbers to help us to understand something that happens in the real world.

In the real world, quantitative research relies on numeric observations of some phenomenon, behavior, and or perception. For example, let’s say we have the quiz results of 20 students as show below.

32 60 95 15 43 22 45 14 48 98 79 97 49 63 50 11 26 52 39 97

This is great information but what if we want to go beyond how these students did and try to understand how students in the population would do on the quizzes. Doing this requires the development of a model.

A model is simply trying to describe how the data is generated in terms of whatever we are mesuring while allowing for randomness. It helps in summarizing a large collection of numbers while also providing structure to it.

One commonly used model is the normal model. This model is the famous bell-curve model that most of us are familiar with. To calculate this model we need to calculate the mean and standard deviation to get a plot similar to the one below

 1

 Now, this model is not completely perfect. For example, a student cannot normally get a score above 100 or below 0 on a quiz. Despite this weakness, normal distribution gives is an indication of what the population looks like.

With this, we can also calculate the probability of getting a specific score on the quiz. For example, if we want to calculate the probability that a student would get a score of 70  or higher we can do a simple test and find that it is about 26%.

Other Options

The normal model is not the only model. There are many different models to match different types of data. There are the gamma, student t, binomial, chi-square, etc. To determine which model to use requires examining the distribution of your data and match it to an appropriate model.

Another option is to transform the data. This is normally done to make data conform to a normal distribution. Which transformation to employ depends on how the data looks when it is plotted.

Conclusion

Modeling helps to bring order to data that has been collected for analysis. By using a model such as the normal distribution, you can begin to make inferences about what the population is like. This allows you to take a handful of data to better understand the world.

K Nearest Neighbor in R

K-nearest neighbor is one of many nonlinear algorithms that can be used in machine learning. By non-linear I mean that a linear combination of the features or variables is not needed in order to develop decision boundaries. This allows for the analysis of data that naturally does not meet the assumptions of linearity.

KNN is also known as a “lazy learner”. This means that there are known coefficients or parameter estimates. When doing regression we always had coefficient outputs regardless of the type of regression (ridge, lasso, elastic net, etc.). What KNN does instead is used K nearest neighbors to give a label to an unlabeled example. Our job when using KNN is to determine the number of K neighbors to use that is most accurate based on the different criteria for assessing the models.

In this post, we will develop a KNN model using the “Mroz” dataset from the “Ecdat” package. Our goal is to predict if someone lives in the city based on the other predictor variables. Below is some initial code.

library(class);library(kknn);library(caret);library(corrplot)
library(reshape2);library(ggplot2);library(pROC);library(Ecdat)
data(Mroz)
str(Mroz)
## 'data.frame':    753 obs. of  18 variables:
##  $ work      : Factor w/ 2 levels "yes","no": 2 2 2 2 2 2 2 2 2 2 ...
##  $ hoursw    : int  1610 1656 1980 456 1568 2032 1440 1020 1458 1600 ...
##  $ child6    : int  1 0 1 0 1 0 0 0 0 0 ...
##  $ child618  : int  0 2 3 3 2 0 2 0 2 2 ...
##  $ agew      : int  32 30 35 34 31 54 37 54 48 39 ...
##  $ educw     : int  12 12 12 12 14 12 16 12 12 12 ...
##  $ hearnw    : num  3.35 1.39 4.55 1.1 4.59 ...
##  $ wagew     : num  2.65 2.65 4.04 3.25 3.6 4.7 5.95 9.98 0 4.15 ...
##  $ hoursh    : int  2708 2310 3072 1920 2000 1040 2670 4120 1995 2100 ...
##  $ ageh      : int  34 30 40 53 32 57 37 53 52 43 ...
##  $ educh     : int  12 9 12 10 12 11 12 8 4 12 ...
##  $ wageh     : num  4.03 8.44 3.58 3.54 10 ...
##  $ income    : int  16310 21800 21040 7300 27300 19495 21152 18900 20405 20425 ...
##  $ educwm    : int  12 7 12 7 12 14 14 3 7 7 ...
##  $ educwf    : int  7 7 7 7 14 7 7 3 7 7 ...
##  $ unemprate : num  5 11 5 5 9.5 7.5 5 5 3 5 ...
##  $ city      : Factor w/ 2 levels "no","yes": 1 2 1 1 2 2 1 1 1 1 ...
##  $ experience: int  14 5 15 6 7 33 11 35 24 21 ...

We need to remove the factor variable “work” as KNN cannot use factor variables. After this, we will use the “melt” function from the “reshape2” package to look at the variables when divided by whether the example was from the city or not.

Mroz$work<-NULL
mroz.melt<-melt(Mroz,id.var='city')
Mroz_plots<-ggplot(mroz.melt,aes(x=city,y=value))+geom_boxplot()+facet_wrap(~variable, ncol = 4)
Mroz_plots

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From the plots, it appears there are no differences in how the variable act whether someone is from the city or not. This may be a flag that classification may not work.

We now need to scale our data otherwise the results will be inaccurate. Scaling might also help our box-plots because everything will be on the same scale rather than spread all over the place. To do this we will have to temporarily remove our outcome variable from the data set because it’s a factor and then reinsert it into the data set. Below is the code.

mroz.scale<-as.data.frame(scale(Mroz[,-16]))
mroz.scale$city<-Mroz$city

We will now look at our box-plots a second time but this time with scaled data.

mroz.scale.melt<-melt(mroz.scale,id.var="city")
mroz_plot2<-ggplot(mroz.scale.melt,aes(city,value))+geom_boxplot()+facet_wrap(~variable, ncol = 4)
mroz_plot2

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This second plot is easier to read but there is still little indication of difference.

We can now move to checking the correlations among the variables. Below is the code

mroz.cor<-cor(mroz.scale[,-17])
corrplot(mroz.cor,method = 'number')

1.png

There is a high correlation between husband’s age (ageh) and wife’s age (agew). Since this algorithm is non-linear this should not be a major problem.

We will now divide our dataset into the training and testing sets

set.seed(502)
ind=sample(2,nrow(mroz.scale),replace=T,prob=c(.7,.3))
train<-mroz.scale[ind==1,]
test<-mroz.scale[ind==2,]

Before creating a model we need to create a grid. We do not know the value of k yet so we have to run multiple models with different values of k in order to determine this for our model. As such we need to create a ‘grid’ using the ‘expand.grid’ function. We will also use cross-validation to get a better estimate of k as well using the “trainControl” function. The code is below.

grid<-expand.grid(.k=seq(2,20,by=1))
control<-trainControl(method="cv")

Now we make our model,

knn.train<-train(city~.,train,method="knn",trControl=control,tuneGrid=grid)
knn.train
## k-Nearest Neighbors 
## 
## 540 samples
##  16 predictors
##   2 classes: 'no', 'yes' 
## 
## No pre-processing
## Resampling: Cross-Validated (10 fold) 
## Summary of sample sizes: 487, 486, 486, 486, 486, 486, ... 
## Resampling results across tuning parameters:
## 
##   k   Accuracy   Kappa    
##    2  0.6000095  0.1213920
##    3  0.6368757  0.1542968
##    4  0.6424325  0.1546494
##    5  0.6386252  0.1275248
##    6  0.6329998  0.1164253
##    7  0.6589619  0.1616377
##    8  0.6663344  0.1774391
##    9  0.6663681  0.1733197
##   10  0.6609510  0.1566064
##   11  0.6664018  0.1575868
##   12  0.6682199  0.1669053
##   13  0.6572111  0.1397222
##   14  0.6719586  0.1694953
##   15  0.6571425  0.1263937
##   16  0.6664367  0.1551023
##   17  0.6719573  0.1588789
##   18  0.6608811  0.1260452
##   19  0.6590979  0.1165734
##   20  0.6609510  0.1219624
## 
## Accuracy was used to select the optimal model using  the largest value.
## The final value used for the model was k = 14.

R recommends that k = 16. This is based on a combination of accuracy and the kappa statistic. The kappa statistic is a measurement of the accuracy of a model while taking into account chance. We don’t have a model in the sense that we do not use the ~ sign like we do with regression. Instead, we have a train and a test set a factor variable and a number for k. This will make more sense when you see the code. Finally, we will use this information on our test dataset. We will then look at the table and the accuracy of the model.

knn.test<-knn(train[,-17],test[,-17],train[,17],k=16) #-17 removes the dependent variable 'city
table(knn.test,test$city)
##         
## knn.test  no yes
##      no   19   8
##      yes  61 125
prob.agree<-(15+129)/213
prob.agree
## [1] 0.6760563

Accuracy is 67% which is consistent with what we found when determining the k. We can also calculate the kappa. This done by calculating the probability and then do some subtraction and division. We already know the accuracy as we stored it in the variable “prob.agree” we now need the probability that this is by chance. Lastly, we calculate the kappa.

prob.chance<-((15+4)/213)*((15+65)/213)
kap<-(prob.agree-prob.chance)/(1-prob.chance)
kap
## [1] 0.664827

A kappa of .66 is actual good.

The example we just did was with unweighted k neighbors. There are times when weighted neighbors can improve accuracy. We will look at three different weighing methods. “Rectangular” is unweighted and is the one that we used. The other two are “triangular” and “epanechnikov”. How these calculate the weights is beyond the scope of this post. In the code below the argument “distance” can be set to 1 for euclidean and 2 for absolute distance.

kknn.train<-train.kknn(city~.,train,kmax = 25,distance = 2,kernel = c("rectangular","triangular",
                                                                      "epanechnikov"))
plot(kknn.train)

1.png

kknn.train
## 
## Call:
## train.kknn(formula = city ~ ., data = train, kmax = 25, distance = 2,     kernel = c("rectangular", "triangular", "epanechnikov"))
## 
## Type of response variable: nominal
## Minimal misclassification: 0.3277778
## Best kernel: rectangular
## Best k: 14

If you look at the plot you can see which value of k is the best by looking at the point that is the lowest on the graph which is right before 15. Looking at the legend it indicates that the point is the “rectangular” estimate which is the same as unweighted. This means that the best classification is unweighted with a k of 14. Although it recommends a different value for k our misclassification was about the same.

Conclusion

In this post, we explored both weighted and unweighted KNN. This algorithm allows you to deal with data that does not meet the assumptions of regression by ignoring the need for parameters. However, because there are no numbers really attached to the results beyond accuracy it can be difficult to explain what is happening in the model to people. As such, perhaps the biggest drawback is communicating results when using KNN.

Common Speech Functions

Functions of speech are different ways of communicating. The differences among the speech functions have to do with the intention of the communication. Different intention or goal leads to the use of a different function of speech. There are many different functions if speech but we will look at the six that are listed below.

  • Referential
  • Directive
  • Expressive
  • Phatic
  • Poetic
  • Metalinguistic

Referential

Referential speech provides information. For example, a person might share the time with someone (“It’s five o’clock” ). Referential speech can often provide information to a question (“what time is it?”).

Directive

Directives or commands that try to get someone to do something. Examples include “turn left” or “sit down”. The context of a directive is one in which something needs or should be done. As such, one person tries to make one or more other persons do something. Even children say directives towards their parents (“give me the ball”).

Expressive

Expressive speech shares a person’s feelings. An example would be “I feel happy today!”. Expressive communication can at times provide clear evidence of how someone is doing.

Phatic

Phatic speech is closely related to expressive speech. However, the main difference is that phatic speech is focused on the well-being of others while expressive speech focuses on the feelings of the person speaking.

An example of phatic speech is saying “how are you?”. This is clearly a question but it is focusing on how the person is doing. Another phrase might be “I hope you get well soon.” Again the focus on is on the welfare of someone else.

Poetic

Poetic speech is speech that is highly aesthetic. Songs and poetry are examples of language that is poetic in nature. An example would be the famous nursery rhyme “Roses are red, violets are blue…..). Poetic speech often has a powerful emotional effect as well.

Metalinguistic 

Metalinguistic speech is communication about language. For example, this entire blog post would be considered by many to be metalinguistic because I a talking about language and not really using language as described in the other functons of speech.

Exceptions

There are many more categories than the ones presented. In addition, the categories presented are not mutually exclusive. Many phrases can be correctly classified into many different categories. For example, if someone says “I love you” you could argue that it’s expressive, poetic, and or even phatic. What is missing is the context in which such a statement is made.

Conclusion

The ways in which we communicated have been briefly explained here. Understanding how people communicate will help others to better understand those around us and improve our style of communicating.

Elastic Net Regression in R

Elastic net is a combination of ridge and lasso regression. What is most unusual about elastic net is that it has two tuning parameters (alpha and lambda) while lasso and ridge regression only has 1.

In this post, we will go through an example of the use of elastic net using the “VietnamI” dataset from the “Ecdat” package. Our goal is to predict how many days a person is ill based on the other variables in the dataset. Below is some initial code for our analysis

library(Ecdat);library(corrplot);library(caret);library(glmnet)
data("VietNamI")
str(VietNamI)
## 'data.frame':    27765 obs. of  12 variables:
##  $ pharvis  : num  0 0 0 1 1 0 0 0 2 3 ...
##  $ lnhhexp  : num  2.73 2.74 2.27 2.39 3.11 ...
##  $ age      : num  3.76 2.94 2.56 3.64 3.3 ...
##  $ sex      : Factor w/ 2 levels "female","male": 2 1 2 1 2 2 1 2 1 2 ...
##  $ married  : num  1 0 0 1 1 1 1 0 1 1 ...
##  $ educ     : num  2 0 4 3 3 9 2 5 2 0 ...
##  $ illness  : num  1 1 0 1 1 0 0 0 2 1 ...
##  $ injury   : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ illdays  : num  7 4 0 3 10 0 0 0 4 7 ...
##  $ actdays  : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ insurance: num  0 0 1 1 0 1 1 1 0 0 ...
##  $ commune  : num  192 167 76 123 148 20 40 57 49 170 ...
##  - attr(*, "na.action")=Class 'omit'  Named int 27734
##   .. ..- attr(*, "names")= chr "27734"

We need to check the correlations among the variables. We need to exclude the “sex” variable as it is categorical. Code is below.

p.cor<-cor(VietNamI[,-4])
corrplot.mixed(p.cor)

1.png

No major problems with correlations. Next, we set up our training and testing datasets. We need to remove the variable “commune” because it adds no value to our results. In addition, to reduce the computational time we will only use the first 1000 rows from the data set.

VietNamI$commune<-NULL
VietNamI_reduced<-VietNamI[1:1000,]
ind<-sample(2,nrow(VietNamI_reduced),replace=T,prob = c(0.7,0.3))
train<-VietNamI_reduced[ind==1,]
test<-VietNamI_reduced[ind==2,]

We need to create a grid that will allow us to investigate different models with different combinations of alpha ana lambda. This is done using the “expand.grid” function. In combination with the “seq” function below is the code

grid<-expand.grid(.alpha=seq(0,1,by=.5),.lambda=seq(0,0.2,by=.1))

We also need to set the resampling method, which allows us to assess the validity of our model. This is done using the “trainControl” function” from the “caret” package. In the code below “LOOCV” stands for “leave one out cross-validation”.

control<-trainControl(method = "LOOCV")

We are no ready to develop our model. The code is mostly self-explanatory. This initial model will help us to determine the appropriate values for the alpha and lambda parameters

enet.train<-train(illdays~.,train,method="glmnet",trControl=control,tuneGrid=grid)
enet.train
## glmnet 
## 
## 694 samples
##  10 predictors
## 
## No pre-processing
## Resampling: Leave-One-Out Cross-Validation 
## Summary of sample sizes: 693, 693, 693, 693, 693, 693, ... 
## Resampling results across tuning parameters:
## 
##   alpha  lambda  RMSE      Rsquared 
##   0.0    0.0     5.229759  0.2968354
##   0.0    0.1     5.229759  0.2968354
##   0.0    0.2     5.229759  0.2968354
##   0.5    0.0     5.243919  0.2954226
##   0.5    0.1     5.225067  0.2985989
##   0.5    0.2     5.200415  0.3038821
##   1.0    0.0     5.244020  0.2954519
##   1.0    0.1     5.203973  0.3033173
##   1.0    0.2     5.182120  0.3083819
## 
## RMSE was used to select the optimal model using  the smallest value.
## The final values used for the model were alpha = 1 and lambda = 0.2.

The output list all the possible alpha and lambda values that we set in the “grid” variable. It even tells us which combination was the best. For our purposes, the alpha will be .5 and the lambda .2. The r-square is also included.

We will set our model and run it on the test set. We have to convert the “sex” variable to a dummy variable for the “glmnet” function. We next have to make matrices for the predictor variables and a for our outcome variable “illdays”

train$sex<-model.matrix( ~ sex - 1, data=train ) #convert to dummy variable 
test$sex<-model.matrix( ~ sex - 1, data=test )
predictor_variables<-as.matrix(train[,-9])
days_ill<-as.matrix(train$illdays)
enet<-glmnet(predictor_variables,days_ill,family = "gaussian",alpha = 0.5,lambda = .2)

We can now look at specific coefficient by using the “coef” function.

enet.coef<-coef(enet,lambda=.2,alpha=.5,exact=T)
enet.coef
## 12 x 1 sparse Matrix of class "dgCMatrix"
##                         s0
## (Intercept)   -1.304263895
## pharvis        0.532353361
## lnhhexp       -0.064754000
## age            0.760864404
## sex.sexfemale  0.029612290
## sex.sexmale   -0.002617404
## married        0.318639271
## educ           .          
## illness        3.103047473
## injury         .          
## actdays        0.314851347
## insurance      .

You can see for yourself that several variables were removed from the model. Medical expenses (lnhhexp), sex, education, injury, and insurance do not play a role in the number of days ill for an individual in Vietnam.

With our model developed. We now can test it using the predict function. However, we first need to convert our test dataframe into a matrix and remove the outcome variable from it

test.matrix<-as.matrix(test[,-9])
enet.y<-predict(enet, newx = test.matrix, type = "response", lambda=.2,alpha=.5)

Let’s plot our results

plot(enet.y)

1.png

This does not look good. Let’s check the mean squared error

enet.resid<-enet.y-test$illdays
mean(enet.resid^2)
## [1] 20.18134

We will now do a cross-validation of our model. We need to set the seed and then use the “cv.glmnet” to develop the cross-validated model. We can see the model by plotting it.

set.seed(317)
enet.cv<-cv.glmnet(predictor_variables,days_ill,alpha=.5)
plot(enet.cv)

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You can see that as the number of features are reduce (see the numbers on the top of the plot) the MSE increases (y-axis). In addition, as the lambda increases, there is also an increase in the error but only when the number of variables is reduced as well.

The dotted vertical lines in the plot represent the minimum MSE for a set lambda (on the left) and the one standard error from the minimum (on the right). You can extract these two lambda values using the code below.

enet.cv$lambda.min
## [1] 0.3082347
enet.cv$lambda.1se
## [1] 2.874607

We can see the coefficients for a lambda that is one standard error away by using the code below. This will give us an alternative idea for what to set the model parameters to when we want to predict.

coef(enet.cv,s="lambda.1se")
## 12 x 1 sparse Matrix of class "dgCMatrix"
##                      1
## (Intercept)   2.34116947
## pharvis       0.003710399       
## lnhhexp       .       
## age           .       
## sex.sexfemale .       
## sex.sexmale   .       
## married       .       
## educ          .       
## illness       1.817479480
## injury        .       
## actdays       .       
## insurance     .

Using the one standard error lambda we lose most of our features. We can now see if the model improves by rerunning it with this information.

enet.y.cv<-predict(enet.cv,newx = test.matrix,type='response',lambda="lambda.1se", alpha = .5)
enet.cv.resid<-enet.y.cv-test$illdays
mean(enet.cv.resid^2)
## [1] 25.47966

A small improvement.  Our model is a mess but this post served as an example of how to conduct an analysis using elastic net regression.

Exploratory Data Analyst

In data science, exploratory data analyst serves the purpose of assessing whether the data set that you have is suitable for answering the research questions of the project. As such, there are several steps that can be taken to make this process more efficient.

Therefore, the purpose of this post is to explain one process that can be used for exploratory data analyst. The steps include ethe following.

  • Consult your questions
  • Check the structure of the dataset
  • Use visuals

Consult Your Questions

Research questions give a project a sense of direction. They help you to know what you want to know. In addition, research questions help you to determine what type of analyst to conduct as well.

During the data exploration stage, the purpose of a research question is not for analyst but rather to determine if your data can actually provide answers to the questions. For example, if you want to know what the average height of men in America are and your data tells you the salary of office workers there is a problem,. Your question (average height) cannot be answered with the current data that you have (office workers salaries).

As such, the research questions need to be answerable and specific before moving forward. By answerable, we mean that the data can provide the solution. By specific, we mean a question moves away from generalities and deals with clearly define phenomenon. For example, “what is the average height of males age 20-30 in the United states?” This question clearly identifies the what we want to know (average height) and among who (20-30, male Americans).

Not can you confirm if your questions are answerable you can also decide if you need to be more or less specific with your questions. Returning to our average height question. We may find that we can be more specific and check average height by state if we want. Or, we might learn that we can only determine the average height for a region. All this depends on the type of data we have.

Check the Structure

Checking the structure involves determining how many rows and columns in the dataset, the sample size, as well as looking for missing data and erroneous data. Data sets in data science almost always need some sort of cleaning or data wrangling before analyst and checking the structure helps to determine what needs to be done.

You should have a priori expectations for the structure of the data set. If the stakeholders tell you that there should be several million rows in the data set and you check and there are only several thousand you know there is a problem. This concept also applies to the number of features you expect as well.

Make Visuals

Visuals, which can be plots or tables, help you further develop your expectations as well as to look for deviations or outliers. Tables are an excellent source for summarizing data. Plots, on the other hand, allow you to see deviations from your expectations in the data.What kind of tables and plots to make depends heavily on

What kind of tables and plots to make depends heavily on the type of data as well as the type of questions that you have. For example, for descriptive questions tables of summary statistics with bar plots might be sufficient. For comparison questions, summary stats and boxplots may be enough. For relationship question, summary stat tables with a scatterplot may be enough. Please keep in mind that it is much more complicated than this.

Conclusion

Before questions can be answered the data needs to be explored. This will help to make sure that the potential answers that are developed are appropriate.

Accommodation Theory in Language Communication

Often when people communicate, they will make a subconscious or even a conscious decision to adjust their speech so that it is more alike or less alike. This is known as accommodation.

In this post, we will look at the following concepts related to accommodation

  • Speech convergence
  • Speech divergence

Speech Convergence

Speech convergence is when people speech starts to sound similar to each other. Often, this is a sign that the speakers are being polite to each other, like each other, and or when one speaker has the interest to please another.

Speech convergence is not only for social reasons. Another reason that a person will modify their speech is for the sake of removing technical jargon when dealing with people who are not familiar with it. For example, when a mechanic speaks to a doctor about what is wrong with their car or when a medical doctor speaks to a patient about the patient’s health. The modification happens so that the other person can understand.

Speech convergence can be overdone in terms of the perceptions of the hearers. For example, if a foreigner sounds too much like a native it can raise suspicion. Furthermore, over convergence can be perceived as insulting and or making fun of others.  As such, some difference is probably wise.

Speech Divergence

Speech divergence happens when people deliberately choose not to mirror each other speaking styles. The message that is sent when doing this is that the people communicating do not want to accommodate, seem polite, or perhaps that they do not like the people they are communicating with.

Examples of this often involve minority groups who desire to maintain their own cultural identity. Such a group will use their language judiciously, especially around the local dominant culture, as a sign of independence.

Accent divergence is also possible. For example, two people from the same country but different socioeconomic standings may deliberately choose to maintain their specific style of communication to indicate the differences between them.

Conclusion

Convergence and divergence in communication can send many different messages to people. It is difficult to determine how people will respond to how a people convergence or divergences from their speaking style. However, the main motivations for accommodation appear to be how such behavior benefits the communicator.

Data Science Research Questions

Developing research questions is an absolute necessity in completing any research project. The questions you ask help to shape the type of analysis that you need to conduct.

The type of questions you ask in the context of analytics and data science are similar to those found in traditional quantitative research. Yet data science, like any other field, has its own distinct traits.

In this post, we will look at six different types of questions that are used frequently in the context of the field of data science. The six questions are…

  1. Descriptive
  2. Exploratory/Inferential
  3. Predictive
  4. Causal
  5. Mechanistic

Understanding the types of question that can be asked will help anyone involved in data science to determine what exactly it is that they want to know.

Descriptive

A descriptive question seeks to describe a characteristic of the dataset. For example, if I collect the GPA of 100 university student I may want to what the average GPA of the students is. Seeking the average is one example of a descriptive question.

With descriptive questions, there is no need for a hypothesis as you are not trying to infer, establish a relationship, or generalize to a broader context. You simply want to know a trait of the dataset.

Exploratory/Inferential

Exploratory questions seek to identify things that may be “interesting” in the dataset. Examples of things that may be interesting include trends, patterns, and or relationships among variables.

Exploratory questions generate hypotheses. This means that they lead to something that may be more formal questioned and tested. For example, if you have GPA and hours of sleep for university students. You may explore the potential that there is a relationship between these two variables.

 

Inferential questions are an extension of exploratory questions. What this means is that the exploratory question is formally tested by developing an inferential question. Often, the difference between an exploratory and inferential question is the following

  1. Exploratory questions are usually developed first
  2. Exploratory questions generate inferential questions
  3. Inferential questions are tested often on a different dataset from exploratory questions

In our example, if we find a relationship between GPA and sleep in our dataset. We may test this relationship in a different, perhaps larger dataset. If the relationship holds we can then generalize this to the population of the study.

Causal

Causal questions address if a change in one variable directly affects another. In analytics, A/B testing is one form of data collection that can be used to develop causal questions. For example, we may develop two version of a website and see which one generates more sales.

In this example, the type of website is the independent variable and sales is the dependent variable. By controlling the type of website people see we can see if this affects sales.

Mechanistic 

Mechanistic questions deal with how one variable affects another. This is different from causal questions that focus on if one variable affects another. Continuing with the website example, we may take a closer look at the two different websites and see what it was about them that made one more succesful in generating sales. It may be that one had more banners than another or fewer pictures. Perhaps there were different products offered on the home page.

All of these different features, of course, require data that helps to explain what is happening. This leads to an important point that the questions that can be asked are limited by the available data. You can’t answer a question that does not contain data that may answer it.

Conclusion

Answering questions is essential what research is about. In order to do this, you have to know what your questions are. This information will help you to decide on the analysis you wish to conduct. Familiarity with the types of research questions that are common in data science can help you to approach and complete analysis much faster than when this is unclear

Lasso Regression in R

In this post, we will conduct an analysis using the lasso regression. Remember lasso regression will actually eliminate variables by reducing them to zero through how the shrinkage penalty can be applied.

We will use the dataset “nlschools” from the “MASS” packages to conduct our analysis. We want to see if we can predict language test scores “lang” with the other available variables. Below is some initial code to begin the analysis

library(MASS);library(corrplot);library(glmnet)
data("nlschools")
str(nlschools)
## 'data.frame':    2287 obs. of  6 variables:
##  $ lang : int  46 45 33 46 20 30 30 57 36 36 ...
##  $ IQ   : num  15 14.5 9.5 11 8 9.5 9.5 13 9.5 11 ...
##  $ class: Factor w/ 133 levels "180","280","1082",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ GS   : int  29 29 29 29 29 29 29 29 29 29 ...
##  $ SES  : int  23 10 15 23 10 10 23 10 13 15 ...
##  $ COMB : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...

We need to remove the “class” variable as it is used as an identifier and provides no useful data. After this, we can check the correlations among the variables. Below is the code for this.

nlschools$class<-NULL
p.cor<-cor(nlschools[,-5])
corrplot.mixed(p.cor)

1.png

No problems with collinearity. We will now setup are training and testing sets.

ind<-sample(2,nrow(nlschools),replace=T,prob = c(0.7,0.3))
train<-nlschools[ind==1,]
test<-nlschools[ind==2,]

Remember that the ‘glmnet’ function does not like factor variables. So we need to convert our “COMB” variable to a dummy variable. In addition, “glmnet” function does not like data frames so we need to make two data frames. The first will include all the predictor variables and the second we include only the outcome variable. Below is the code

train$COMB<-model.matrix( ~ COMB - 1, data=train ) #convert to dummy variable 
test$COMB<-model.matrix( ~ COMB - 1, data=test )
predictor_variables<-as.matrix(train[,2:4])
language_score<-as.matrix(train$lang)

We can now run our model. We place both matrices inside the “glmnet” function. The family is set to “gaussian” because our outcome variable is continuous. The “alpha” is set to 1 as this indicates that we are using lasso regression.

lasso<-glmnet(predictor_variables,language_score,family="gaussian",alpha=1)

Now we need to look at the results using the “print” function. This function prints a lot of information as explained below.

  • Df = number of variables including in the model (this is always the same number in a ridge model)
  • %Dev = Percent of deviance explained. The higher the better
  • Lambda = The lambda used to obtain the %Dev

When you use the “print” function for a lasso model it will print up to 100 different models. Fewer models are possible if the percent of deviance stops improving. 100 is the default stopping point. In the code below we will use the “print” function but, I only printed the first 5 and last 5 models in order to reduce the size of the printout. Fortunately, it only took 60 models to converge.

print(lasso)
## 
## Call:  glmnet(x = predictor_variables, y = language_score, family = "gaussian",      alpha = 1) 
## 
##       Df    %Dev  Lambda
##  [1,]  0 0.00000 5.47100
##  [2,]  1 0.06194 4.98500
##  [3,]  1 0.11340 4.54200
##  [4,]  1 0.15610 4.13900
##  [5,]  1 0.19150 3.77100
............................
## [55,]  3 0.39890 0.03599
## [56,]  3 0.39900 0.03280
## [57,]  3 0.39900 0.02988
## [58,]  3 0.39900 0.02723
## [59,]  3 0.39900 0.02481
## [60,]  3 0.39900 0.02261

The results from the “print” function will allow us to set the lambda for the “test” dataset. Based on the results we can set the lambda at 0.02 because this explains the highest amount of deviance at .39.

The plot below shows us lambda on the x-axis and the coefficients of the predictor variables on the y-axis. The numbers next to the coefficient lines refers to the actual coefficient of a particular variable as it changes from using different lambda values. Each number corresponds to a variable going from left to right in a dataframe/matrix using the “View” function. For example, 1 in the plot refers to “IQ” 2 refers to “GS” etc.

plot(lasso,xvar="lambda",label=T)

1.png

As you can see, as lambda increase the coefficient decrease in value. This is how regularized regression works. However, unlike ridge regression which never reduces a coefficient to zero, lasso regression does reduce a coefficient to zero. For example, coefficient 3 (SES variable) and coefficient 2 (GS variable) are reduced to zero when lambda is near 1.

You can also look at the coefficient values at a specific lambda values. The values are unstandardized and are used to determine the final model selection. In the code below the lambda is set to .02 and we use the “coef” function to do see the results

lasso.coef<-coef(lasso,s=.02,exact = T)
lasso.coef
## 4 x 1 sparse Matrix of class "dgCMatrix"
##                       1
## (Intercept)  9.35736325
## IQ           2.34973922
## GS          -0.02766978
## SES          0.16150542

Results indicate that for a 1 unit increase in IQ there is a 2.41 point increase in language score. When GS (class size) goes up 1 unit there is a .03 point decrease in language score. Finally, when SES (socioeconomic status) increase  1 unit language score improves .13 point.

The second plot shows us the deviance explained on the x-axis. On the y-axis is the coefficients of the predictor variables. Below is the code

plot(lasso,xvar='dev',label=T)

1.png

If you look carefully, you can see that the two plots are completely opposite to each other. increasing lambda cause a decrease in the coefficients. Furthermore, increasing the fraction of deviance explained leads to an increase in the coefficient. You may remember seeing this when we used the “print”” function. As lambda became smaller there was an increase in the deviance explained.

Now, we will assess our model using the test data. We need to convert the test dataset to a matrix. Then we will use the “predict”” function while setting our lambda to .02. Lastly, we will plot the results. Below is the code.

test.matrix<-as.matrix(test[,2:4])
lasso.y<-predict(lasso,newx = test.matrix,type = 'response',s=.02)
plot(lasso.y,test$lang)

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The visual looks promising. The last thing we need to do is calculated the mean squared error. By its self this number does not mean much. However, it provides a benchmark for comparing our current model with any other models that we may develop. Below is the code

lasso.resid<-lasso.y-test$lang
mean(lasso.resid^2)
## [1] 46.74314

Knowing this number, we can, if we wanted, develop other models using other methods of analysis to try to reduce it. Generally, the lower the error the better while keeping in mind the complexity of the model.

Theories on Language Change in Groups

As people interact with each other, it naturally leads to changes in how communication  takes place. Fortunately, there are several views that attempt to explain in a systematic way how language changes. In general, there are at least 3 viewpoints on how language changes. These viewpoints are

  • Group to group
  • Style to style
  • Word to word

In this post, we will look at each of these viewpoints on language change.

Group to Group

The group to group hypothesis sees language change like a wave in a lake. The changes originates from one or more groups and slowly spreads to other groups.  This happens because different groups interact with each other. Furthermore, many people are members of more than one group and bring the language they use in one group to another.

Style to Style

The style to style hypothesis suggest that language changes as there are shifts between language styles. For example, from a formal way of speaking to a colloquial way of speaking and vice versa.

A change in the language  that is seen as prestigious is usually from a higher more affluent section of society. Of course, the opposite is also true and un-prestigious language change comes from the least fortunate.

The style of a speaker also changes over time. The younger the person is the more they use vernacular and slang in general.

Word to Word 

There are times in which individual words will change within a language and this change will spread to other languages. This is known as lexical diffusion.

Such a change can take decades and even century to take place. It is also common when two languages interact through mutually changing each other pronunciation. Such as the role of French in England for several centuries.

Conclusion

It is not so much that any of the examples discussed here are exclusively responsible for change. Rather, all of these examples play varying roles in influencing changes in a language.

Ridge Regression in R

In this post, we will conduct an analysis using ridge regression. Ridge regression is a type of regularized regression. By applying a shrinkage penalty, we are able to reduce the coefficients of many variables almost to zero while still retaining them in the model. This allows us to develop models that have many more variables in them compared to models using best subset or stepwise regression.

In the example used in this post, we will use the “SAheart” dataset from the “ElemStatLearn” package. We want to predict systolic blood pressure (sbp) using all of the other variables available as predictors. Below is some initial code that we need to begin.

library(ElemStatLearn);library(car);library(corrplot)
library(leaps);library(glmnet);library(caret)
data(SAheart)
str(SAheart)
## 'data.frame':    462 obs. of  10 variables:
##  $ sbp      : int  160 144 118 170 134 132 142 114 114 132 ...
##  $ tobacco  : num  12 0.01 0.08 7.5 13.6 6.2 4.05 4.08 0 0 ...
##  $ ldl      : num  5.73 4.41 3.48 6.41 3.5 6.47 3.38 4.59 3.83 5.8 ...
##  $ adiposity: num  23.1 28.6 32.3 38 27.8 ...
##  $ famhist  : Factor w/ 2 levels "Absent","Present": 2 1 2 2 2 2 1 2 2 2 ...
##  $ typea    : int  49 55 52 51 60 62 59 62 49 69 ...
##  $ obesity  : num  25.3 28.9 29.1 32 26 ...
##  $ alcohol  : num  97.2 2.06 3.81 24.26 57.34 ...
##  $ age      : int  52 63 46 58 49 45 38 58 29 53 ...
##  $ chd      : int  1 1 0 1 1 0 0 1 0 1 ...

A look at the object using the “str” function indicates that one variable “famhist” is a factor variable. The “glmnet” function that does the ridge regression analysis cannot handle factors so we need to converts this to a dummy variable. However, there are two things we need to do before this. First, we need to check the correlations to make sure there are no major issues with multi-collinearity Second, we need to create our training and testing data sets. Below is the code for the correlation plot.

p.cor<-cor(SAheart[,-5])
corrplot.mixed(p.cor)

1

First we created a variable called “p.cor” the -5 in brackets means we removed the 5th column from the “SAheart” data set which is the factor variable “Famhist”. The correlation plot indicates that there is one strong relationship between adiposity and obesity. However, one common cut-off for collinearity is 0.8 and this value is 0.72 which is not a problem.

We will now create are training and testing sets and convert “famhist” to a dummy variable.

ind<-sample(2,nrow(SAheart),replace=T,prob = c(0.7,0.3))
train<-SAheart[ind==1,]
test<-SAheart[ind==2,]
train$famhist<-model.matrix( ~ famhist - 1, data=train ) #convert to dummy variable 
test$famhist<-model.matrix( ~ famhist - 1, data=test )

We are still not done preparing our data yet. “glmnet” cannot use data frames, instead, it can only use matrices. Therefore, we now need to convert our data frames to matrices. We have to create two matrices, one with all of the predictor variables and a second with the outcome variable of blood pressure. Below is the code

predictor_variables<-as.matrix(train[,2:10])
blood_pressure<-as.matrix(train$sbp)

We are now ready to create our model. We use the “glmnet” function and insert our two matrices. The family is set to Gaussian because “blood pressure” is a continuous variable. Alpha is set to 0 as this indicates ridge regression. Below is the code

ridge<-glmnet(predictor_variables,blood_pressure,family = 'gaussian',alpha = 0)

Now we need to look at the results using the “print” function. This function prints a lot of information as explained below.

  •  Df = number of variables including in the model (this is always the same number in a ridge model)
  •  %Dev = Percent of deviance explained. The higher the better
  • Lambda = The lambda used to attain the %Dev

When you use the “print” function for a ridge model it will print up to 100 different models. Fewer models are possible if the percent of deviance stops improving. 100 is the default stopping point. In the code below we have the “print” function. However, I have only printed the first 5 and last 5 models in order to save space.

print(ridge)
## 
## Call:  glmnet(x = predictor_variables, y = blood_pressure, family = "gaussian",      alpha = 0) 
## 
##        Df      %Dev    Lambda
##   [1,] 10 7.622e-37 7716.0000
##   [2,] 10 2.135e-03 7030.0000
##   [3,] 10 2.341e-03 6406.0000
##   [4,] 10 2.566e-03 5837.0000
##   [5,] 10 2.812e-03 5318.0000
................................
##  [95,] 10 1.690e-01    1.2290
##  [96,] 10 1.691e-01    1.1190
##  [97,] 10 1.692e-01    1.0200
##  [98,] 10 1.693e-01    0.9293
##  [99,] 10 1.693e-01    0.8468
## [100,] 10 1.694e-01    0.7716

The results from the “print” function are useful in setting the lambda for the “test” dataset. Based on the results we can set the lambda at 0.83 because this explains the highest amount of deviance at .20.

The plot below shows us lambda on the x-axis and the coefficients of the predictor variables on the y-axis. The numbers refer to the actual coefficient of a particular variable. Inside the plot, each number corresponds to a variable going from left to right in a data-frame/matrix using the “View” function. For example, 1 in the plot refers to “tobacco” 2 refers to “ldl” etc. Across the top of the plot is the number of variables used in the model. Remember this number never changes when doing ridge regression.

plot(ridge,xvar="lambda",label=T)

1.png

As you can see, as lambda increase the coefficient decrease in value. This is how ridge regression works yet no coefficient ever goes to absolute 0.

You can also look at the coefficient values at a specific lambda value. The values are unstandardized but they provide a useful insight when determining final model selection. In the code below the lambda is set to .83 and we use the “coef” function to do this

ridge.coef<-coef(ridge,s=.83,exact = T)
ridge.coef
## 11 x 1 sparse Matrix of class "dgCMatrix"
##                                   1
## (Intercept)            105.69379942
## tobacco                 -0.25990747
## ldl                     -0.13075557
## adiposity                0.29515034
## famhist.famhistAbsent    0.42532887
## famhist.famhistPresent  -0.40000846
## typea                   -0.01799031
## obesity                  0.29899976
## alcohol                  0.03648850
## age                      0.43555450
## chd                     -0.26539180

The second plot shows us the deviance explained on the x-axis and the coefficients of the predictor variables on the y-axis. Below is the code

plot(ridge,xvar='dev',label=T)

1.png

The two plots are completely opposite to each other. Increasing lambda cause a decrease in the coefficients while increasing the fraction of deviance explained leads to an increase in the coefficient. You can also see this when we used the “print” function. As lambda became smaller there was an increase in the deviance explained.

We now can begin testing our model on the test data set. We need to convert the test dataset to a matrix and then we will use the predict function while setting our lambda to .83 (remember a lambda of .83 explained the most of the deviance). Lastly, we will plot the results. Below is the code.

test.matrix<-as.matrix(test[,2:10])
ridge.y<-predict(ridge,newx = test.matrix,type = 'response',s=.83)
plot(ridge.y,test$sbp)

1

The last thing we need to do is calculated the mean squared error. By it’s self this number is useless. However, it provides a benchmark for comparing the current model with any other models you may develop. Below is the code

ridge.resid<-ridge.y-test$sbp
mean(ridge.resid^2)
## [1] 372.4431

Knowing this number, we can develop other models using other methods of analysis to try to reduce it as much as possible.

Regularized Linear Regression

Traditional linear regression has been a tried and true  model for making predictions for decades. However, with the growth of Big Data and datasets with 100’s of variables  problems have begun to arise. For example, using stepwise or best subset method with regression could take hours if not days to converge in even some of the best computers.

To deal with this problem, regularized regression has been developed to help to determine which features or variables to keep when developing models from large datasets with a huge number of variables. In this post, we will look at the following concepts

  • Definition of regularized regression
  • Ridge regression
  • Lasso regression
  • Elastic net regression

Regularization

Regularization involves the use of a shrinkage penalty in order to reduce the residual sum of squares (RSS). This is done through selecting a value for a tuning parameter called “lambda”. Tuning parameters are used in machine learning algorithms to control the behavior of the models that are developed.

The lambda is multiplied by the normalized coefficients of the model and added to the RSS. Below is an equation of what was just said

RSS + λ(normalized coefficients)

The benefits of regularization are at least three-fold. First, regularization is highly computationally efficient. Instead of fitting k-1 models when k is the number of variables available (for example, 50 variables would lead 49 models!), with regularization only one model is developed for each value of lambda you specify.

Second, regularization helps to deal with the bias-variance headache of model development. When small changes are made to data, such as switching from the training to testing data, there can be wild changes in the estimates. Regularization can often smooth this problem out substantially.

Finally, regularization can help to reduce or eliminate any multicollenarity in a model. As such, the benefits of using regularization make it clear that this should be considering when working with larger data sets.

Ridge Regression

Ridge regression involves the normalization of the squared weights or as shown in the equation below

RSS + λ(normalized coefficients^2)

This is also refered to as the L2-norm. As lambda increase in value the coefficients in the model are shrunk towards 0 but never reach 0. This is how the error is shrunk. The higher the lambda the lower the value of the coefficients as they are reduce more and more thus reducing the RSS.

The benefit is that predictive accuracy is often increased. However, interpreting and communicating your results can become difficult because no variables are removed from the model. Instead the variables are reduced near to zero. This can be especially tough if you have dozens of variables remaining in your model to try to explain.

Lasso

Lasso is short for “Least Absolute Shrinkage and Selection Operator”. This approach uses the L1-norm which is the sum of the absolute value of the coefficients or as shown in the equation below

RSS + λ(Σ|normalized coefficients|)

This shrinkage penalty will reduce a coefficient to 0 which is another way of saying that variables will be removed from the model. One problem is that highly correlated variables that need to be  in your model my be removed when Lasso shrinks coefficients. This is one reason why ridge regression is still used.

Elastic Net

Elastic net is the best of ridge and Lasso without the weaknesses of either. It combines extracts variables like Lasso and Ridge does not while also group variables like Ridge does but Lasso does not.

This is done by including a second tuning parameter called “alpha”. If alpha is set to 0 it is the same as ridge regression and if alpha is set to 1 it is the same as lasso regression. If you are able to appreciate it below is the formula used for elastic net regression

(RSS + l[(1 – alpha)(S|normalized coefficients|2)/2 + alpha(S|normalized coefficients|)])/N)

As such when working with elastic net you have to set two different tuning parameters (alpha and lambda) in order to develop a model.

Conclusion

Regularized regressio was developed as an answer to the growth in the size and number of variables in a data set today. Ridge, lasso an elastic net all provide solutions to converging over large datasets and selecting features.

Social Networks and Language Habits

In this post, we will look at how relationships that people have can play a role in how they communicate with those around them. Understanding this can help people to comprehend differences in communication style.

In sociolinguistics, social networks  can refer to the pattern of informal relationships that people have and experience on a consistent basis. There are two dimensions that can be used to describe a persons social network. These two terms are density and plexity.

Density

The density of a social network refers to how well people in your network know each other. In other words, density is ow well your friends know each other. We all have friends, we have friends who know each other, and we have friends who do not know each other.

If many of your friends know each other then the density is high. If your friends do not know each other the density is low. An example of a high density network would be the typical family. Everybody knows each other. An example of a low density network would be employees at a large company. In such a situation it would not be hard to find a friend of a friend that you do not know.

Plexity

Plexity is a  measure of the various types of interactions that you are involved in with other people. Plexity can be uniplex, which involves one type of interaction with a person or multiplex, which involves many types of interactions with a person.

An example of a uniplex interaction may be a worker with their boss. They only interact at work. A multiplex interaction would again be with members of one’s family. When dealing with family interactions could include school, work, recreation, shopping, etc. In all these examples it is the same people interacting in a multitude of settings.

Language Use in Social Networks

A person’s speech almost always reflects the network that they belong too. If the group is homogeneous we will almost always speak the way everyone else does assuming we want to be a part of the group. For example, a group of local construction workers will more than likely use similar language patterns due to the homogeneous nature of the group while a group of ESL bankers would not as they come from many different countries.

When a person belongs to more than one social network they will almost always unconsciously change the way they communicate based on the context. For example, anybody who has moved away from home communicates differently where they live then when they communicate with family and friends back home. This is true even when moving from one place to another in the same province or state in your country.

Conclusion

The language that people employ is affected by the dynamics of the social network. We naturally will adjust our communication to accommodate who we are talking too.

Primary Tasks in Data Analysis

Performing a data analysis in the realm of data science is a difficult task due to the huge number of decisions that need to be made. For some people,  plotting the course to conduct an analysis is easy. However, for most of us, beginning a project leads to a sense of paralysis as we struggle to determine what to do.

In light of this challenge, there are at least 5 core task that you need to consider when preparing to analyze data. These five task are

  1. Developing  your question(s)
  2. Data exploration
  3. Developing a statistical model
  4. Interpreting the results
  5. Sharing the results

Developing Your Question(s)

You really cannot analyze data until you first determine what it is you want to know. It is tempting to just jump in and start looking for interesting stuff but you will not know if something you find is interesting unless it helps to answer your question(s).

There are several types of research questions. The point is you need to ask them in order to answer them.

Data Exploration

Data exploration allows you to determine if you can answer your questions with the data you have. In data science, the data is normally already collected by the time you are called upon to analyze it. As such, what you want to find may not be possible.

In addition, exploration of the data allows you to determine if there are any problems with the data set such as missing data, strange variables, and if necessary to develop a data dictionary so you know the characteristics of the variables.

Data exploration allows you to determine what kind of data wrangling needs to be done. This involves the preparation of the data for a more formal analysis when you develop your statistical models. This process takes up the majority of a data scientist time and is not easy at all.  Mastery of this in many ways means being a master of data science

Develop a Statistical Model

Your research questions  and the data exploration  process helps you to determine what kind of model to develop. The factors that can affect this is whether your data is supervised or unsupervised and whether you want to classify or predict numerical values.

This is probably the funniest part of data analysis and is much easier then having to wrangle with the data. Your goal is to determine if the model helps to answer your question(s)

Interpreting the Results

Once a model is developed it is time to explain what it means. Sometimes you can make a really cool model that nobody (including yourself) can explain. This is especially true of “black box” methods such as support vector machines and artificial neural networks. Models need to normally be explainable to non-technical stakeholders.

With interpretation you are trying to determine “what does this answer mean to the stakeholders?”  For example, if you find that people who smoke are 5 times more likely to die before the age of 50 what are the implications of this? How can the stakeholders use this information to achieve their own goals? In other words, why should they care about what you found out?

Communication of Results

Now  is the time to actually share the answer(s) to your question(s). How this is done varies but it can be written, verbal or both. Whatever the mode of communication it is necessary to consider the following

  • The audience or stakeholders
  • The actual answers to the questions
  • The benefits of knowing this

You must remember the stakeholders because this affects how you communicate. How you speak to business professionals would be  different from academics. Next, you must share the answers to the questions. This can be done with charts, figures, illustrations etc. Data visualization is an expertise of its own. Lastly, you explain how this information is useful in a practical way.

Conclusion

The process shared here is one way to approach the analysis of data. Think of this as a framework from which to develop your own method of analysis.

Linear VS Quadratic Discriminant Analysis in R

In this post we will look at linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA). Discriminant analysis is used when the dependent variable is categorical. Another commonly used option is logistic regression but there are differences between logistic regression and discriminant analysis. Both LDA and QDA are used in situations in which there is a clear separation between the classes you want to predict. If the categories are fuzzier logistic regression is often the better choice.

For our example, we will use the “Mathlevel” dataset found in the “Ecdat” package. Our goal will be to predict the sex of a respondent based on SAT math score, major, foreign language proficiency, as well as the number of math, physic, and chemistry classes a respondent took. Below is some initial code to start our analysis.

library(MASS);library(Ecdat)
data("Mathlevel")

The first thing we need to do is clean up the data set. We have to remove any missing data in order to run our model. We will create a dataset called “math” that has the “Mathlevel” dataset but with the “NA”s removed use the “na.omit” function. After this, we need to set our seed for the purpose of reproducibility using the “set.seed” function. Lastly, we will split the data using the “sample” function using a 70/30 split. The training dataset will be called “math.train” and the testing dataset will be called “math.test”. Below is the code

math<-na.omit(Mathlevel)
set.seed(123)
math.ind<-sample(2,nrow(math),replace=T,prob = c(0.7,0.3))
math.train<-math[math.ind==1,]
math.test<-math[math.ind==2,]

Now we will make our model and it is called “lda.math” and it will include all available variables in the “math.train” dataset. Next we will check the results by calling the modle. Finally, we will examine the plot to see how our model is doing. Below is the code.

lda.math<-lda(sex~.,math.train)
lda.math
## Call:
## lda(sex ~ ., data = math.train)
## 
## Prior probabilities of groups:
##      male    female 
## 0.5986079 0.4013921 
## 
## Group means:
##        mathlevel.L mathlevel.Q mathlevel.C mathlevel^4 mathlevel^5
## male   -0.10767593  0.01141838 -0.05854724   0.2070778  0.05032544
## female -0.05571153  0.05360844 -0.08967303   0.2030860 -0.01072169
##        mathlevel^6      sat languageyes  majoreco  majoross   majorns
## male    -0.2214849 632.9457  0.07751938 0.3914729 0.1472868 0.1782946
## female  -0.2226767 613.6416  0.19653179 0.2601156 0.1907514 0.2485549
##          majorhum mathcourse physiccourse chemistcourse
## male   0.05426357   1.441860    0.7441860      1.046512
## female 0.07514451   1.421965    0.6531792      1.040462
## 
## Coefficients of linear discriminants:
##                       LD1
## mathlevel.L    1.38456344
## mathlevel.Q    0.24285832
## mathlevel.C   -0.53326543
## mathlevel^4    0.11292817
## mathlevel^5   -1.24162715
## mathlevel^6   -0.06374548
## sat           -0.01043648
## languageyes    1.50558721
## majoreco      -0.54528930
## majoross       0.61129797
## majorns        0.41574298
## majorhum       0.33469586
## mathcourse    -0.07973960
## physiccourse  -0.53174168
## chemistcourse  0.16124610
plot(lda.math,type='both')

1.png

Calling “lda.math” gives us the details of our model. It starts be indicating the prior probabilities of someone being male or female. Next is the means for each variable by sex. The last part is the coefficients of the linear discriminants. Each of these values is used to determine the probability that a particular example is male or female. This is similar to a regression equation.

The plot provides us with densities of the discriminant scores for males and then for females. The output indicates a problem. There is a great deal of overlap between male and females in the model. What this indicates is that there is a lot of misclassification going on as the two groups are not clearly separated. Furthermore, this means that logistic regression is probably a better choice for distinguishing between male and females. However, since this is for demonstrating purposes we will not worry about this.

We will now use the “predict” function on the training set data to see how well our model classifies the respondents by gender. We will then compare the prediction of the model with thee actual classification. Below is the code.

math.lda.predict<-predict(lda.math)
math.train$lda<-math.lda.predict$class
table(math.train$lda,math.train$sex)
##         
##          male female
##   male    219    100
##   female   39     73
mean(math.train$lda==math.train$sex)
## [1] 0.6774942

As you can see, we have a lot of misclassification happening. A large amount of false negatives which is a lot of males being classified as female. The overall accuracy us only 59% which is not much better than chance.

We will now conduct the same analysis on the test data set. Below is the code.

lda.math.test<-predict(lda.math,math.test)
math.test$lda<-lda.math.test$class
table(math.test$lda,math.test$sex)
##         
##          male female
##   male     92     43
##   female   23     20
mean(math.test$lda==math.test$sex)
## [1] 0.6292135

As you can see the results are similar. To put it simply, our model is terrible. The main reason is that there is little distinction between males and females as shown in the plot. However, we can see if perhaps a quadratic discriminant analysis will do better

QDA allows for each class in the dependent variable to have it’s own covariance rather than a shared covariance as in LDA. This allows for quadratic terms in the development of the model. To complete a QDA we need to use the “qda” function from the “MASS” package. Below is the code for the training data set.

math.qda.fit<-qda(sex~.,math.train)
math.qda.predict<-predict(math.qda.fit)
math.train$qda<-math.qda.predict$class
table(math.train$qda,math.train$sex)
##         
##          male female
##   male    215     84
##   female   43     89
mean(math.train$qda==math.train$sex)
## [1] 0.7053364

You can see there is almost no difference. Below is the code for the test data.

math.qda.test<-predict(math.qda.fit,math.test)
math.test$qda<-math.qda.test$class
table(math.test$qda,math.test$sex)
##         
##          male female
##   male     91     43
##   female   24     20
mean(math.test$qda==math.test$sex)
## [1] 0.6235955

Still disappointing. However, in this post we reviewed linear discriminant analysis as well as learned about the use of quadratic linear discriminant analysis. Both of these statistical tools are used for predicting categorical dependent variables. LDA assumes shared covariance in the dependent variable categories will QDA allows for each category in the dependent variable to have it’s own variance.

Sociolinguistic Insights into Female Communication

In general, women tend to prefer to use the most standard or prestige form of a language regardless of cultural background or geography. Linguist have proposed several potential reasons for this. This post will share some of the most common ideas on why women often used the standard form of their language.

Social Status

There is a belief among many linguist that women use the most prestigious forms of their language because they are more status-conscious than men. By using the standard version of their language a women is able to claim a higher status.

The implication of this is that women have a lower status in society and try to elevate themselves through their use of language. However, this conclusion has been refuted as women who work outside the home use more of the standard form of their language then women who work in their home.

If the social status hypothesis was correct women who work at home, and thus have the lowest status, should use more of the standard form then women who work. Currently, this is not the case.

Women as Protector of Society’s Values

The women as protector of values view see social pressure as a constraint on how women communicate. Simply, women use more standard forms of their language then men because women are expected to behave better. It is thrust upon women to serve as an example for their community and especially for their children.

This answer is considered correct but depends highly on context. For example, this idea falls a part most frequently when women communicate with their children. The informal and intimate setting often leads to most women using the vernacular aspects of their language.

Women as Subordinate Group

A third suggestion is that women, who are often a subordinate group, use the more standard version of their language to show deference to those over them. In other words, women use the most polite forms of their language to avoid offending men.

However, this suggestion also fails because it equates politeness with the standard form of a language. People can be polite using vernacular and they can be rude using the most prestigious form of their language possible.

Vernacular as Masculine

A final common hypothesis on women’s use of standard forms is the perception that the use of the vernacular is masculine and tough. Women choose the standard form as a way of demonstrating behaviors traditionally associated with gender in their culture. Men on the other hand, use vernacular forms to show traits that are traditionally associated with male behaviors.

The problem with this belief is the informal settings. As mentioned previously, women and men use more vernacular forms of their language in informal settings. As such, it seems that context is one of the strongest factors in how language is used and not necessarily gender.