Using the aggregate function in R.

Advertisements

Leave a reply

Using the aggregate function in R.

Advertisements

Create subgroups in R

Subsetting data in R

Getting data out of R

Importing data into R

for loops in R

Anybody who has ever had to do any writing for academic purposes or in industry has had to deal with APA formatting. The rules and expectations seem to be endless and always changing. If you are able to maneuver the endless list of rules you still have to determine what to report and how when writing an article.

There is a package in R that can at least take away the mystery of how to report ANOVA, correlation, and regression tables. This package is called “apaTables”. In this post, we will look at how to use this package for making tables that are formatted according to APA.

We are going to create examples of ANOVA, correlation, and regression tables using the ‘mtcars’ dataset. Below is the initial code that we need to begin.

```
library(apaTables)
data("mtcars")
```

**ANOVA**

We will begin with the results of ANOVA. In order for this to be successful, you have to use the “lm” function to create the model. If you are familiar with ANOVA and regression this should not be surprising as they both find the same answer using different approaches. After the “lm” function you must use the “filename” argument and give the output a name in quotations. This file will be saved in your R working directory. You can also provide other information such as the table number and confidence level if you desire.

There will be two outputs in our code. The output to the console is in R. A second output will be in a word doc. Below is the code.

`apa.aov.table(lm(mpg~cyl,mtcars),filename = "Example1.doc",table.number = 1)`

```
##
##
## Table 1
##
## ANOVA results using mpg as the dependent variable
##
##
## Predictor SS df MS F p partial_eta2
## (Intercept) 3429.84 1 3429.84 333.71 .000
## cyl 817.71 1 817.71 79.56 .000 .73
## Error 308.33 30 10.28
## CI_90_partial_eta2
##
## [.56, .80]
##
##
## Note: Values in square brackets indicate the bounds of the 90% confidence interval for partial eta-squared
```

Here is the word doc output

Perhaps you are beginning to see the beauty of using this package and its functions. The “apa.aov.table”” function provides a nice table that requires no formatting by the researcher.

You can even make a table of the means and standard deviations of ANOVA. This is similar to what you would get if you used the “aggregate” function. Below is the code.

`apa.1way.table(cyl, mpg,mtcars,filename = "Example2.doc",table.number = 2)`

```
##
##
## Table 2
##
## Descriptive statistics for mpg as a function of cyl.
##
## cyl M SD
## 4 26.66 4.51
## 6 19.74 1.45
## 8 15.10 2.56
##
## Note. M and SD represent mean and standard deviation, respectively.
##
```

Here is what it looks like in word

**Correlation **

We will now look at an example of a correlation table. The function for this is “apa.cor.table”. This function works best with only a few variables. Otherwise, the table becomes bigger than a single sheet of paper. In addition, you probably will want to suppress the confidence intervals to save space. There are other arguments that you can explore on your own. Below is the code

`apa.cor.table(mtcars,filename = "Example3.doc",table.number = 3,show.conf.interval = F)`

```
##
##
## Table 3
##
## Means, standard deviations, and correlations
##
##
## Variable M SD 1 2 3 4 5 6 7
## 1. mpg 20.09 6.03
##
## 2. cyl 6.19 1.79 -.85**
##
## 3. disp 230.72 123.94 -.85** .90**
##
## 4. hp 146.69 68.56 -.78** .83** .79**
##
## 5. drat 3.60 0.53 .68** -.70** -.71** -.45**
##
## 6. wt 3.22 0.98 -.87** .78** .89** .66** -.71**
##
## 7. qsec 17.85 1.79 .42* -.59** -.43* -.71** .09 -.17
##
## 8. vs 0.44 0.50 .66** -.81** -.71** -.72** .44* -.55** .74**
##
## 9. am 0.41 0.50 .60** -.52** -.59** -.24 .71** -.69** -.23
##
## 10. gear 3.69 0.74 .48** -.49** -.56** -.13 .70** -.58** -.21
##
## 11. carb 2.81 1.62 -.55** .53** .39* .75** -.09 .43* -.66**
##
## 8 9 10
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
## .17
##
## .21 .79**
##
## -.57** .06 .27
##
##
## Note. * indicates p < .05; ** indicates p < .01.
## M and SD are used to represent mean and standard deviation, respectively.
##
```

Here is the word doc results

If you run this code at home and open the word doc in Word you will not see variables 9 and 10 because the table is too big by itself for a single page. I hade to resize it manually. One way to get around this is to delate the M and SD column and place those as rows below the table.

**Regression**

Our final example will be a regression table. The code is as follows

`apa.reg.table(lm(mpg~disp,mtcars),filename = "Example4",table.number = 4)`

```
##
##
## Table 4
##
## Regression results using mpg as the criterion
##
##
## Predictor b b_95%_CI beta beta_95%_CI sr2 sr2_95%_CI
## (Intercept) 29.60** [27.09, 32.11]
## disp -0.04** [-0.05, -0.03] -0.85 [-1.05, -0.65] .72 [.51, .81]
##
##
##
## r Fit
##
## -.85**
## R2 = .718**
## 95% CI[.51,.81]
##
##
## Note. * indicates p < .05; ** indicates p < .01.
## A significant b-weight indicates the beta-weight and semi-partial correlation are also significant.
## b represents unstandardized regression weights; beta indicates the standardized regression weights;
## sr2 represents the semi-partial correlation squared; r represents the zero-order correlation.
## Square brackets are used to enclose the lower and upper limits of a confidence interval.
##
```

Here are the results in word

You can also make regression tables that have multiple blocks or models. Below is an example

`apa.reg.table(lm(mpg~disp,mtcars),lm(mpg~disp+hp,mtcars),filename = "Example5",table.number = 5)`

```
##
##
## Table 5
##
## Regression results using mpg as the criterion
##
##
## Predictor b b_95%_CI beta beta_95%_CI sr2 sr2_95%_CI
## (Intercept) 29.60** [27.09, 32.11]
## disp -0.04** [-0.05, -0.03] -0.85 [-1.05, -0.65] .72 [.51, .81]
##
##
##
## (Intercept) 30.74** [28.01, 33.46]
## disp -0.03** [-0.05, -0.02] -0.62 [-0.94, -0.31] .15 [.00, .29]
## hp -0.02 [-0.05, 0.00] -0.28 [-0.59, 0.03] .03 [-.03, .09]
##
##
##
## r Fit Difference
##
## -.85**
## R2 = .718**
## 95% CI[.51,.81]
##
##
## -.85**
## -.78**
## R2 = .748** Delta R2 = .03
## 95% CI[.54,.83] 95% CI[-.03, .09]
##
##
## Note. * indicates p < .05; ** indicates p < .01.
## A significant b-weight indicates the beta-weight and semi-partial correlation are also significant.
## b represents unstandardized regression weights; beta indicates the standardized regression weights;
## sr2 represents the semi-partial correlation squared; r represents the zero-order correlation.
## Square brackets are used to enclose the lower and upper limits of a confidence interval.
##
```

Here is the word doc version

**Conculsion **

This is a real time saver for those of us who need to write and share statistical information.

Vectorization of function in R

If else statements in functions in R

If Statements in R

Arguments and functions in R

R functions intro

Introduction to using List

Manipulating Dataframes in R

Intro to dataframes in R

Local regression uses something similar to nearest neighbor classification to generate a regression line. In local regression, nearby observations are used to fit the line rather than all observations. It is necessary to indicate the percentage of the observations you want R to use for fitting the local line. The name for this hyperparameter is the span. The higher the span the smoother the line becomes.

Local regression is great one there are only a handful of independent variables in the model. When the total number of variables becomes too numerous the model will struggle. As such, we will only fit a bivariate model. This will allow us to process the model and to visualize it.

In this post, we will use the “Clothing” dataset from the “Ecdat” package and we will examine innovation (inv2) relationship with total sales (tsales). Below is some initial code.

`library(Ecdat)`

```
data(Clothing)
str(Clothing)
```

```
## 'data.frame': 400 obs. of 13 variables:
## $ tsales : int 750000 1926395 1250000 694227 750000 400000 1300000 495340 1200000 495340 ...
## $ sales : num 4412 4281 4167 2670 15000 ...
## $ margin : num 41 39 40 40 44 41 39 28 41 37 ...
## $ nown : num 1 2 1 1 2 ...
## $ nfull : num 1 2 2 1 1.96 ...
## $ npart : num 1 3 2.22 1.28 1.28 ...
## $ naux : num 1.54 1.54 1.41 1.37 1.37 ...
## $ hoursw : int 76 192 114 100 104 72 161 80 158 87 ...
## $ hourspw: num 16.8 22.5 17.2 21.5 15.7 ...
## $ inv1 : num 17167 17167 292857 22207 22207 ...
## $ inv2 : num 27177 27177 71571 15000 10000 ...
## $ ssize : int 170 450 300 260 50 90 400 100 450 75 ...
## $ start : num 41 39 40 40 44 41 39 28 41 37 ...
```

There is no data preparation in this example. The first thing we will do is fit two different models that have different values for the span hyperparameter. “fit” will have a span of .41 which means it will use 41% of the nearest examples. “fit2” will use .82. Below is the code.

```
fit<-loess(tsales~inv2,span = .41,data = Clothing)
fit2<-loess(tsales~inv2,span = .82,data = Clothing)
```

In the code above, we used the “loess” function to fit the model. The “span” argument was set to .41 and .82.

We now need to prepare for the visualization. We begin by using the “range” function to find the distance from the lowest to the highest value. Then use the “seq” function to create a grid. Below is the code.

```
inv2lims<-range(Clothing$inv2)
inv2.grid<-seq(from=inv2lims[1],to=inv2lims[2])
```

The information in the code above is for setting our x-axis in the plot. We are now ready to fit our model. We will fit the models and draw each regression line.

```
plot(Clothing$inv2,Clothing$tsales,xlim=inv2lims)
lines(inv2.grid,predict(fit,data.frame(inv2=inv2.grid)),col='blue',lwd=3)
lines(inv2.grid,predict(fit2,data.frame(inv2=inv2.grid)),col='red',lwd=3)
```

Not much difference in the two models. For our final task, we will predict with our “fit” model using all possible values of “inv2” and also fit the confidence interval lines.

```
pred<-predict(fit,newdata=inv2.grid,se=T)
plot(Clothing$inv2,Clothing$tsales)
lines(inv2.grid,pred$fit,col='red',lwd=3)
lines(inv2.grid,pred$fit+2*pred$se.fit,lty="dashed",lwd=2,col='blue')
lines(inv2.grid,pred$fit-2*pred$se.fit,lty="dashed",lwd=2,col='blue')
```

**Conclusion**

Local regression provides another way to model complex non-linear relationships in low dimensions. The example here provides just the basics of how this is done is much more complicated than described here.

This post will provide information on smoothing splines. Smoothing splines are used in regression when we want to reduce the residual sum of squares by adding more flexibility to the regression line without allowing too much overfitting.

In order to do this, we must tune the parameter called the smoothing spline. The smoothing spline is essentially a natural cubic spline with a knot at every unique value of x in the model. Having this many knots can lead to severe overfitting. This is corrected for by controlling the degrees of freedom through the parameter called lambda. You can manually set this value or select it through cross-validation.

We will now look at an example of the use of smoothing splines with the “Clothing” dataset from the “Ecdat” package. We want to predict “tsales” based on the use of innovation in the stores. Below is some initial code.

`library(Ecdat)`

```
data(Clothing)
str(Clothing)
```

```
## 'data.frame': 400 obs. of 13 variables:
## $ tsales : int 750000 1926395 1250000 694227 750000 400000 1300000 495340 1200000 495340 ...
## $ sales : num 4412 4281 4167 2670 15000 ...
## $ margin : num 41 39 40 40 44 41 39 28 41 37 ...
## $ nown : num 1 2 1 1 2 ...
## $ nfull : num 1 2 2 1 1.96 ...
## $ npart : num 1 3 2.22 1.28 1.28 ...
## $ naux : num 1.54 1.54 1.41 1.37 1.37 ...
## $ hoursw : int 76 192 114 100 104 72 161 80 158 87 ...
## $ hourspw: num 16.8 22.5 17.2 21.5 15.7 ...
## $ inv1 : num 17167 17167 292857 22207 22207 ...
## $ inv2 : num 27177 27177 71571 15000 10000 ...
## $ ssize : int 170 450 300 260 50 90 400 100 450 75 ...
## $ start : num 41 39 40 40 44 41 39 28 41 37 ...
```

We are going to create three models. Model one will have 70 degrees of freedom, model two will have 7, and model three will have the number of degrees of freedom are determined through cross-validation. Below is the code.

```
fit1<-smooth.spline(Clothing$inv2,Clothing$tsales,df=57)
fit2<-smooth.spline(Clothing$inv2,Clothing$tsales,df=7)
fit3<-smooth.spline(Clothing$inv2,Clothing$tsales,cv=T)
```

```
## Warning in smooth.spline(Clothing$inv2, Clothing$tsales, cv = T): cross-
## validation with non-unique 'x' values seems doubtful
```

`(data.frame(fit1$df,fit2$df,fit3$df))`

```
## fit1.df fit2.df fit3.df
## 1 57 7.000957 2.791762
```

In the code above we used the “smooth.spline” function which comes with base r.Notice that we did not use the same coding syntax as the “lm” function calls for. The code above also indicates the degrees of freedom for each model. You can see that for “fit3” the cross-validation determine that 2.79 was the most appropriate degrees of freedom. In addition, if you type in the following code.

`sapply(data.frame(fit1$x,fit2$x,fit3$x),length)`

```
## fit1.x fit2.x fit3.x
## 73 73 73
```

You will see that there are only 73 data points in each model. The “Clothing” dataset has 400 examples in it. The reason for this reduction is that the “smooth.spline” function only takes unique values from the original dataset. As such, though there are 400 examples in the dataset only 73 of them are unique.

Next, we plot our data and add regression lines

```
plot(Clothing$inv2,Clothing$tsales)
lines(fit1,col='red',lwd=3)
lines(fit2,col='green',lwd=3)
lines(fit3,col='blue',lwd=3)
legend('topright',lty=1,col=c('red','green','blue'),c("df = 57",'df=7','df=CV 2.8'))
```

You can see that as the degrees of freedom increase so does the flexibility in the line. The advantage of smoothing splines is to have a more flexible way to assess the characteristics of a dataset.

Normally, when least squares regression is used, you fit one line to the model. However, sometimes you may want enough flexibility that you fit different lines over different regions of your independent variable. This process of fitting different lines over different regions of X is known as Regression Splines.

How this works is that there are different coefficient values based on the regions of X. As the researcher, you can set the cutoff points for each region. The cutoff point is called a “knot.” The more knots you use the more flexible the model becomes because there are fewer data points with each range allowing for more variability.

We will now go through an example of polynomial regression splines. Remeber that polynomial means that we will have a curved line as we are using higher order polynomials. Our goal will be to predict total sales based on the amount of innovation a store employs. We will use the “Ecdat” package and the “Clothing” dataset. In addition, we will need the “splines” package. The code is as follows.

`library(splines);library(Ecdat)`

`data(Clothing)`

We will now fit our model. We must indicate the number and placement of the knots. This is commonly down at the 25th 50th and 75th percentile. Below is the code

`fit<-lm(tsales~bs(inv2,knots = c(12000,60000,150000)),data = Clothing)`

In the code above we used the traditional “lm” function to set the model. However, we also used the “bs” function which allows us to create our spline regression model. The argument “knots” was set to have three different values. Lastly, the dataset was indicated.

Remember that the default spline model in R is a third-degree polynomial. This is because it is hard for the eye to detect the discontinuity at the knots.

We now need X values that we can use for prediction purposes. In the code below we first find the range of the “inv2” variable. We then create a grid that includes all the possible values of “inv2” in increments of 1. Lastly, we use the “predict” function to develop the prediction model. We set the “se” argument to true as we will need this information. The code is below.

```
inv2lims<-range(Clothing$inv2)
inv2.grid<-seq(from=inv2lims[1],to=inv2lims[2])
pred<-predict(fit,newdata=list(inv2=inv2.grid),se=T)
```

We are now ready to plot our model. The code below graphs the model and includes the regression line (red), confidence interval (green), as well as the location of each knot (blue)

```
plot(Clothing$inv2,Clothing$tsales,main="Regression Spline Plot")
lines(inv2.grid,pred$fit,col='red',lwd=3)
lines(inv2.grid,pred$fit+2*pred$se.fit,lty="dashed",lwd=2,col='green')
lines(inv2.grid,pred$fit-2*pred$se.fit,lty="dashed",lwd=2,col='green')
segments(12000,0,x1=12000,y1=5000000,col='blue' )
segments(60000,0,x1=60000,y1=5000000,col='blue' )
segments(150000,0,x1=150000,y1=5000000,col='blue' )
```

When this model was created it was essentially three models connected. Model on goes from the first blue line to the second. Model 2 goes form the second blue line to the third and model three was from the third blue line until the end. This kind of flexibility is valuable in understanding nonlinear relationship

Factors in R

Polynomial regression is used when you want to develop a regression model that is not linear. It is common to use this method when performing traditional least squares regression. However, it is also possible to use polynomial regression when the dependent variable is categorical. As such, in this post, we will go through an example of logistic polynomial regression.

Specifically, we will use the “Clothing” dataset from the “Ecdat” package. We will divide the “tsales” dependent variable into two categories to run the analysis. Below is the code to get started.

`library(Ecdat)`

`data(Clothing)`

There is little preparation for this example. Below is the code for the model

`fitglm<-glm(I(tsales>900000)~poly(inv2,4),data=Clothing,family = binomial)`

Here is what we did

1. We created an object called “fitglm” to save our results

2. We used the “glm” function to process the model

3. We used the “I” function. This told R to process the information inside the parentheses as is. As such, we did not have to make a new variable in which we split the “tsales” variable. Simply, if sales were greater than 900000 it was code 1 and 0 if less than this amount.

4. Next, we set the information for the independent variable. We used the “poly” function. Inside this function, we placed the “inv2” variable and the highest order polynomial we want to explore.

5. We set the data to “Clothing”

6. Lastly, we set the “family” argument to “binomial” which is needed for logistic regression

Below is the results

`summary(fitglm)`

```
##
## Call:
## glm(formula = I(tsales > 9e+05) ~ poly(inv2, 4), family = binomial,
## data = Clothing)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.5025 -0.8778 -0.8458 1.4534 1.5681
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 3.074 2.685 1.145 0.2523
## poly(inv2, 4)1 641.710 459.327 1.397 0.1624
## poly(inv2, 4)2 585.975 421.723 1.389 0.1647
## poly(inv2, 4)3 259.700 178.081 1.458 0.1448
## poly(inv2, 4)4 73.425 44.206 1.661 0.0967 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 521.57 on 399 degrees of freedom
## Residual deviance: 493.51 on 395 degrees of freedom
## AIC: 503.51
##
## Number of Fisher Scoring iterations: 13
```

It appears that only the 4th-degree polynomial is significant and barely at that. We will now find the range of our independent variable “inv2” and make a grid from this information. Doing this will allow us to run our model using the full range of possible values for our independent variable.

```
inv2lims<-range(Clothing$inv2)
inv2.grid<-seq(from=inv2lims[1],to=inv2lims[2])
```

The “inv2lims” object has two values. The lowest value in “inv2” and the highest value. These values serve as the highest and lowest values in our “inv2.grid” object. This means that we have values started at 350 and going to 400000 by 1 in a grid to be used as values for “inv2” in our prediction model. Below is our prediction model.

`predsglm<-predict(fitglm,newdata=list(inv2=inv2.grid),se=T,type="response")`

Next, we need to calculate the probabilities that a given value of “inv2” predicts a store has “tsales” greater than 900000. The equation is as follows.

`pfit<-exp(predsglm$fit)/(1+exp(predsglm$fit))`

Graphing this leads to interesting insights. Below is the code

`plot(pfit)`

You can see the curves in the line from the polynomial expression. As it appears. As inv2 increase the probability increase until the values fall between 125000 and 200000. This is interesting, to say the least.

We now need to plot the actual model. First, we need to calculate the confidence intervals. This is done with the code below.

```
se.bandsglm.logit<-cbind(predsglm$fit+2*predsglm$se.fit,predsglm$fit-2*predsglm$se.fit)
se.bandsglm<-exp(se.bandsglm.logit)/(1+exp(se.bandsglm.logit))
```

The ’se.bandsglm” object contains the log odds of each example and the “se.bandsglm” has the probabilities. Now we plot the results

```
plot(Clothing$inv2,I(Clothing$tsales>900000),xlim=inv2lims,type='n')
points(jitter(Clothing$inv2),I((Clothing$tsales>900000)),cex=2,pch='|',col='darkgrey')
lines(inv2.grid,pfit,lwd=4)
matlines(inv2.grid,se.bandsglm,col="green",lty=6,lwd=6)
```

In the code above we did the following.

1. We plotted our dependent and independent variables. However, we set the argument “type” to n which means nothing. This was done so we can add the information step-by-step.

2. We added the points. This was done using the “points” function. The “jitter” function just helps to spread the information out. The other arguments (cex, pch, col) our for aesthetics and our optional.

3. We add our logistic polynomial line based on our independent variable grid and the “pfit” object which has all of the predicted probabilities.

4. Last, we add the confidence intervals using the “matlines” function. Which includes the grid object as well as the “se.bandsglm” information.

You can see that these results are similar to when we only graphed the “pfit” information. However, we also add the confidence intervals. You can see the same dip around 125000-200000 were there is also a larger confidence interval. if you look at the plot you can see that there are fewer data points in this range which may be what is making the intervals wider.

**Conclusion**

Logistic polynomial regression allows the regression line to have more curves to it if it is necessary. This is useful for fitting data that is non-linear in nature.

Polynomial regression is one of the easiest ways to fit a non-linear line to a data set. This is done through the use of higher order polynomials such as cubic, quadratic, etc to one or more predictor variables in a model.

Generally, polynomial regression is used for one predictor and one outcome variable. When there are several predictor variables it is more common to use generalized additive modeling/ In this post, we will use the “Clothing” dataset from the “Ecdat” package to predict total sales with the use of polynomial regression. Below is some initial code.

`library(Ecdat)`

`data(Clothing) str(Clothing)`

```
## 'data.frame': 400 obs. of 13 variables:
## $ tsales : int 750000 1926395 1250000 694227 750000 400000 1300000 495340 1200000 495340 ...
## $ sales : num 4412 4281 4167 2670 15000 ...
## $ margin : num 41 39 40 40 44 41 39 28 41 37 ...
## $ nown : num 1 2 1 1 2 ...
## $ nfull : num 1 2 2 1 1.96 ...
## $ npart : num 1 3 2.22 1.28 1.28 ...
## $ naux : num 1.54 1.54 1.41 1.37 1.37 ...
## $ hoursw : int 76 192 114 100 104 72 161 80 158 87 ...
## $ hourspw: num 16.8 22.5 17.2 21.5 15.7 ...
## $ inv1 : num 17167 17167 292857 22207 22207 ...
## $ inv2 : num 27177 27177 71571 15000 10000 ...
## $ ssize : int 170 450 300 260 50 90 400 100 450 75 ...
## $ start : num 41 39 40 40 44 41 39 28 41 37 ...
```

We are going to use the “inv2” variable as our predictor. This variable measures the investment in automation by a particular store. We will now run our polynomial regression model.

```
fit<-lm(tsales~poly(inv2,5),data = Clothing)
summary(fit)
```

```
##
## Call:
## lm(formula = tsales ~ poly(inv2, 5), data = Clothing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -946668 -336447 -96763 184927 3599267
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 833584 28489 29.259 < 2e-16 ***
## poly(inv2, 5)1 2391309 569789 4.197 3.35e-05 ***
## poly(inv2, 5)2 -665063 569789 -1.167 0.2438
## poly(inv2, 5)3 49793 569789 0.087 0.9304
## poly(inv2, 5)4 1279190 569789 2.245 0.0253 *
## poly(inv2, 5)5 -341189 569789 -0.599 0.5497
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 569800 on 394 degrees of freedom
## Multiple R-squared: 0.05828, Adjusted R-squared: 0.04633
## F-statistic: 4.876 on 5 and 394 DF, p-value: 0.0002428
```

The code above should be mostly familiar. We use the “lm” function as normal for regression. However, we then used the “poly” function on the “inv2” variable. What this does is runs our model 5 times (5 is the number next to “inv2”). Each time a different polynomial is used from 1 (no polynomial) to 5 (5th order polynomial). The results indicate that the 4th-degree polynomial is significant.

We now will prepare a visual of the results but first, there are several things we need to prepare. First, we want to find what the range of our predictor variable “inv2” is and we will save this information in a grade. The code is below.

`inv2lims<-range(Clothing$inv2)`

Second, we need to create a grid that has all the possible values of “inv2” from the lowest to the highest broken up in intervals of one. Below is the code.

`inv2.grid<-seq(from=inv2lims[1],to=inv2lims[2])`

We now have a dataset with almost 400000 data points in the “inv2.grid” object through this approach. We will now use these values to predict “tsales.” We also want the standard errors so we se “se” to TRUE

`preds<-predict(fit,newdata=list(inv2=inv2.grid),se=TRUE)`

We now need to find the confidence interval for our regression line. This is done by making a dataframe that takes the predicted fit adds or subtracts 2 and multiples this number by the standard error as shown below.

`se.bands<-cbind(preds$fit+2*preds$se.fit,preds$fit-2*preds$se.fit)`

With these steps completed, we are ready to create our civilization.

To make our visual, we use the “plot” function on the predictor and outcome. Doing this gives us a plot without a regression line. We then use the “lines” function to add the polynomial regression line, however, this line is based on the “inv2.grid” object (40,000 observations) and our predictions. Lastly, we use the “matlines” function to add the confidence intervals we found and stored in the “se.bands” object.

```
plot(Clothing$inv2,Clothing$tsales)
lines(inv2.grid,preds$fit,lwd=4,col='blue')
matlines(inv2.grid,se.bands,lwd = 4,col = "yellow",lty=4)
```

**Conclusion**

You can clearly see the curvature of the line. Which helped to improve model fit. Now any of you can tell that we are fitting this line to mostly outliers. This is one reason we the standard error gets wider and wider it is because there are fewer and fewer observations on which to base it. However, for demonstration purposes, this is a clear example of the power of polynomial regression.

Searching text within an object in R

Partial least squares regression is a form of regression that involves the development of components of the original variables in a supervised way. What this means is that the dependent variable is used to help create the new components form the original variables. This means that when pls is used the linear combination of the new features helps to explain both the independent and dependent variables in the model.

In this post, we will use predict “income” in the “Mroz” dataset using pls. Below is some initial code.

`library(pls);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 ...
```

First, we must prepare our data by dividing it into a training and test set. We will do this by doing a 50/50 split of the data.

```
set.seed(777)
train<-sample(c(T,F),nrow(Mroz),rep=T) #50/50 train/test split
test<-(!train)
```

In the code above we set the “set.seed function in order to assure reduplication. Then we created the “train” object and used the “sample” function to make a vector with ‘T’ and ‘F’ based on the number of rows in “Mroz”. Lastly, we created the “test”” object base don everything that is not in the “train” object as that is what the exclamation point is for.

Now we create our model using the “plsr” function from the “pls” package and we will examine the results using the “summary” function. We will also scale the data since this the scale affects the development of the components and use cross-validation. Below is the code.

```
set.seed(777)
pls.fit<-plsr(income~.,data=Mroz,subset=train,scale=T,validation="CV")
summary(pls.fit)
```

```
## Data: X dimension: 392 17
## Y dimension: 392 1
## Fit method: kernelpls
## Number of components considered: 17
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 11218 8121 6701 6127 5952 5886 5857
## adjCV 11218 8114 6683 6108 5941 5872 5842
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 5853 5849 5854 5853 5853 5852 5852
## adjCV 5837 5833 5837 5836 5836 5835 5835
## 14 comps 15 comps 16 comps 17 comps
## CV 5852 5852 5852 5852
## adjCV 5835 5835 5835 5835
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps
## X 17.04 26.64 37.18 49.16 59.63 64.63 69.13
## income 49.26 66.63 72.75 74.16 74.87 75.25 75.44
## 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps
## X 72.82 76.06 78.59 81.79 85.52 89.55 92.14
## income 75.49 75.51 75.51 75.52 75.52 75.52 75.52
## 15 comps 16 comps 17 comps
## X 94.88 97.62 100.00
## income 75.52 75.52 75.52
```

The printout includes the root mean squared error for each of the components in the VALIDATION section as well as the variance explained in the TRAINING section. There are 17 components because there are 17 independent variables. You can see that after component 3 or 4 there is little improvement in the variance explained in the dependent variable. Below is the code for the plot of these results. It requires the use of the “validationplot” function with the “val.type” argument set to “MSEP” Below is the code

`validationplot(pls.fit,val.type = "MSEP")`

We will do the predictions with our model. We use the “predict” function, use our “Mroz” dataset but only those index in the “test” vector and set the components to three based on our previous plot. Below is the code.

```
set.seed(777)
pls.pred<-predict(pls.fit,Mroz[test,],ncomp=3)
```

After this, we will calculate the mean squared error. This is done by subtracting the results of our predicted model from the dependent variable of the test set. We then square this information and calculate the mean. Below is the code

`mean((pls.pred-Mroz$income[test])^2)`

`## [1] 63386682`

As you know, this information is only useful when compared to something else. Therefore, we will run the data with a tradition least squares regression model and compare the results.

```
set.seed(777)
lm.fit<-lm(income~.,data=Mroz,subset=train)
lm.pred<-predict(lm.fit,Mroz[test,])
mean((lm.pred-Mroz$income[test])^2)
```

`## [1] 59432814`

The least squares model is slightly better then our partial least squares model but if we look at the model we see several variables that are not significant. We will remove these see what the results are

`summary(lm.fit)`

```
##
## Call:
## lm(formula = income ~ ., data = Mroz, subset = train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -20131 -2923 -1065 1670 36246
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.946e+04 3.224e+03 -6.036 3.81e-09 ***
## workno -4.823e+03 1.037e+03 -4.651 4.59e-06 ***
## hoursw 4.255e+00 5.517e-01 7.712 1.14e-13 ***
## child6 -6.313e+02 6.694e+02 -0.943 0.346258
## child618 4.847e+02 2.362e+02 2.052 0.040841 *
## agew 2.782e+02 8.124e+01 3.424 0.000686 ***
## educw 1.268e+02 1.889e+02 0.671 0.502513
## hearnw 6.401e+02 1.420e+02 4.507 8.79e-06 ***
## wagew 1.945e+02 1.818e+02 1.070 0.285187
## hoursh 6.030e+00 5.342e-01 11.288 < 2e-16 ***
## ageh -9.433e+01 7.720e+01 -1.222 0.222488
## educh 1.784e+02 1.369e+02 1.303 0.193437
## wageh 2.202e+03 8.714e+01 25.264 < 2e-16 ***
## educwm -4.394e+01 1.128e+02 -0.390 0.697024
## educwf 1.392e+02 1.053e+02 1.322 0.186873
## unemprate -1.657e+02 9.780e+01 -1.694 0.091055 .
## cityyes -3.475e+02 6.686e+02 -0.520 0.603496
## experience -1.229e+02 4.490e+01 -2.737 0.006488 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5668 on 374 degrees of freedom
## Multiple R-squared: 0.7552, Adjusted R-squared: 0.744
## F-statistic: 67.85 on 17 and 374 DF, p-value: < 2.2e-16
```

```
set.seed(777)
lm.fit<-lm(income~work+hoursw+child618+agew+hearnw+hoursh+wageh+experience,data=Mroz,subset=train)
lm.pred<-predict(lm.fit,Mroz[test,])
mean((lm.pred-Mroz$income[test])^2)
```

`## [1] 57839715`

As you can see the error decreased even more which indicates that the least squares regression model is superior to the partial least squares model. In addition, the partial least squares model is much more difficult to explain because of the use of components. As such, the least squares model is the favored one.

This post will explain and provide an example of principal component regression (PCR). Principal component regression involves having the model construct components from the independent variables that are a linear combination of the independent variables. This is similar to principal component analysis but the components are designed in a way to best explain the dependent variable. Doing this often allows you to use fewer variables in your model and usually improves the fit of your model as well.

Since PCR is based on principal component analysis it is an unsupervised method, which means the dependent variable has no influence on the development of the components. As such, there are times when the components that are developed may not be beneficial for explaining the dependent variable.

Our example will use the “Mroz” dataset from the “Ecdat” package. Our goal will be to predict “income” based on the variables in the dataset. Below is the initial code

`library(pls);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 ...
```

Our first step is to divide our dataset into a train and test set. We will do a simple 50/50 split for this demonstration.

```
train<-sample(c(T,F),nrow(Mroz),rep=T) #50/50 train/test split
test<-(!train)
```

In the code above we use the “sample” function to create a “train” index based on the number of rows in the “Mroz” dataset. Basically, R is making a vector that randomly assigns different rows in the “Mroz” dataset to be marked as True or False. Next, we use the “train” vector and we assign everything or every number that is not in the “train” vector to the test vector by using the exclamation mark.

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

```
set.seed(777)
pcr.fit<-pcr(income~.,data=Mroz,subset=train,scale=T,validation="CV")
```

To make our model we use the “pcr” function from the “pls” package. The “subset” argument tells r to use the “train” vector to select examples from the “Mroz” dataset. The “scale” argument makes sure everything is measured the same way. This is important when using a component analysis tool as variables with different scale have a different influence on the components. Lastly, the “validation” argument enables cross-validation. This will help us to determine the number of components to use for prediction. Below is the results of the model using the “summary” function.

`summary(pcr.fit)`

```
## Data: X dimension: 381 17
## Y dimension: 381 1
## Fit method: svdpc
## Number of components considered: 17
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 12102 11533 11017 9863 9884 9524 9563
## adjCV 12102 11534 11011 9855 9878 9502 9596
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 9149 9133 8811 8527 7265 7234 7120
## adjCV 9126 9123 8798 8877 7199 7172 7100
## 14 comps 15 comps 16 comps 17 comps
## CV 7118 7141 6972 6992
## adjCV 7100 7123 6951 6969
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps
## X 21.359 38.71 51.99 59.67 65.66 71.20 76.28
## income 9.927 19.50 35.41 35.63 41.28 41.28 46.75
## 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps
## X 80.70 84.39 87.32 90.15 92.65 95.02 96.95
## income 47.08 50.98 51.73 68.17 68.29 68.31 68.34
## 15 comps 16 comps 17 comps
## X 98.47 99.38 100.00
## income 68.48 70.29 70.39
```

There is a lot of information here.The VALIDATION: RMSEP section gives you the root mean squared error of the model broken down by component. The TRAINING section is similar the printout of any PCA but it shows the amount of cumulative variance of the components, as well as the variance, explained for the dependent variable “income.” In this model, we are able to explain up to 70% of the variance if we use all 17 components.

We can graph the MSE using the “validationplot” function with the argument “val.type” set to “MSEP”. The code is below.

`validationplot(pcr.fit,val.type = "MSEP")`

How many components to pick is subjective, however, there is almost no improvement beyond 13 so we will use 13 components in our prediction model and we will calculate the means squared error.

```
set.seed(777)
pcr.pred<-predict(pcr.fit,Mroz[test,],ncomp=13)
mean((pcr.pred-Mroz$income[test])^2)
```

`## [1] 48958982`

MSE is what you would use to compare this model to other models that you developed. Below is the performance of a least squares regression model

```
set.seed(777)
lm.fit<-lm(income~.,data=Mroz,subset=train)
lm.pred<-predict(lm.fit,Mroz[test,])
mean((lm.pred-Mroz$income[test])^2)
```

`## [1] 47794472`

If you compare the MSE the least squares model performs slightly better than the PCR one. However, there are a lot of non-significant features in the model as shown below.

`summary(lm.fit)`

```
##
## Call:
## lm(formula = income ~ ., data = Mroz, subset = train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -27646 -3337 -1387 1860 48371
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.215e+04 3.987e+03 -5.556 5.35e-08 ***
## workno -3.828e+03 1.316e+03 -2.909 0.00385 **
## hoursw 3.955e+00 7.085e-01 5.582 4.65e-08 ***
## child6 5.370e+02 8.241e+02 0.652 0.51512
## child618 4.250e+02 2.850e+02 1.491 0.13673
## agew 1.962e+02 9.849e+01 1.992 0.04709 *
## educw 1.097e+02 2.276e+02 0.482 0.63013
## hearnw 9.835e+02 2.303e+02 4.270 2.50e-05 ***
## wagew 2.292e+02 2.423e+02 0.946 0.34484
## hoursh 6.386e+00 6.144e-01 10.394 < 2e-16 ***
## ageh -1.284e+01 9.762e+01 -0.132 0.89542
## educh 1.460e+02 1.592e+02 0.917 0.35982
## wageh 2.083e+03 9.930e+01 20.978 < 2e-16 ***
## educwm 1.354e+02 1.335e+02 1.014 0.31115
## educwf 1.653e+02 1.257e+02 1.315 0.18920
## unemprate -1.213e+02 1.148e+02 -1.057 0.29140
## cityyes -2.064e+02 7.905e+02 -0.261 0.79421
## experience -1.165e+02 5.393e+01 -2.159 0.03147 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6729 on 363 degrees of freedom
## Multiple R-squared: 0.7039, Adjusted R-squared: 0.69
## F-statistic: 50.76 on 17 and 363 DF, p-value: < 2.2e-16
```

Removing these and the MSE is almost the same for the PCR and least square models

```
set.seed(777)
lm.fit2<-lm(income~work+hoursw+hearnw+hoursh+wageh,data=Mroz,subset=train)
lm.pred2<-predict(lm.fit2,Mroz[test,])
mean((lm.pred2-Mroz$income[test])^2)
```

`## [1] 47968996`

**Conclusion**

Since the least squares model is simpler it is probably the superior model. PCR is strongest when there are a lot of variables involve and if there are issues with multicollinearity.

Combining and Splitting Strings in R

This post will provide an example of best subset regression. This is a topic that has been covered before in this blog. However, in the current post, we will approach this using a slightly different coding and a different dataset. We will be using the “HI” dataset from the “Ecdat” package. Our goal will be to predict the number of hours a women works based on the other variables in the dataset. Below is some initial code.

`library(leaps);library(Ecdat)`

```
data(HI)
str(HI)
```

```
## 'data.frame': 22272 obs. of 13 variables:
## $ whrswk : int 0 50 40 40 0 40 40 25 45 30 ...
## $ hhi : Factor w/ 2 levels "no","yes": 1 1 2 1 2 2 2 1 1 1 ...
## $ whi : Factor w/ 2 levels "no","yes": 1 2 1 2 1 2 1 1 2 1 ...
## $ hhi2 : Factor w/ 2 levels "no","yes": 1 1 2 2 2 2 2 1 1 2 ...
## $ education : Ord.factor w/ 6 levels "<9years"<"9-11years"<..: 4 4 3 4 2 3 5 3 5 4 ...
## $ race : Factor w/ 3 levels "white","black",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ hispanic : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
## $ experience: num 13 24 43 17 44.5 32 14 1 4 7 ...
## $ kidslt6 : int 2 0 0 0 0 0 0 1 0 1 ...
## $ kids618 : int 1 1 0 1 0 0 0 0 0 0 ...
## $ husby : num 12 1.2 31.3 9 0 ...
## $ region : Factor w/ 4 levels "other","northcentral",..: 2 2 2 2 2 2 2 2 2 2 ...
## $ wght : int 214986 210119 219955 210317 219955 208148 213615 181960 214874 214874 ...
```

To develop a model we use the “regsubset” function from the “leap” package. Most of the coding is the same as linear regression. The only difference is the “nvmax” argument which is set to 13. The default setting for “nvmax” is 8. This is good if you only have 8 variables. However, the results from the “str” function indicate that we have 13 functions. Therefore, we need to set the “nvmax” argument to 13 instead of the default value of 8 in order to be sure to include all variables. Below is the code

`regfit.full<-regsubsets(whrswk~.,HI, nvmax = 13)`

We can look at the results with the “summary” function. For space reasons, the code is shown but the results will not be shown here.

`summary(regfit.full)`

If you run the code above in your computer you will 13 columns that are named after the variables created. A star in a column means that that variable is included in the model. To the left is the numbers 1-13 which. One means one variable in the model two means two variables in the model etc.

Our next step is to determine which of these models is the best. First, we need to decide what our criteria for inclusion will be. Below is a list of available fit indices.

`names(summary(regfit.full))`

`## [1] "which" "rsq" "rss" "adjr2" "cp" "bic" "outmat" "obj"`

For our purposes, we will use “rsq” (r-square) and “bic” “Bayesian Information Criteria.” In the code below we are going to save the values for these two fit indices in their own objects.

```
rsq<-summary(regfit.full)$rsq
bic<-summary(regfit.full)$bic
```

Now let’s plot them

`plot(rsq,type='l',main="R-Square",xlab="Number of Variables")`

`plot(bic,type='l',main="BIC",xlab="Number of Variables")`

You can see that for r-square the values increase and for BIC the values decrease. We will now make both of these plots again but we will have r tell the optimal number of variables when considering each model index. For we use the “which” function to determine the max r-square and the minimum BIC

`which.max(rsq)`

`## [1] 13`

`which.min(bic)`

`## [1] 12`

The model with the best r-square is the one with 13 variables. This makes sense as r-square always improves as you add variables. Since this is a demonstration we will not correct for this. For BIC the lowest values was for 12 variables. We will now plot this information and highlight the best model in the plot using the “points” function, which allows you to emphasis one point in a graph

```
plot(rsq,type='l',main="R-Square with Best Model Highlighted",xlab="Number of Variables")
points(13,(rsq[13]),col="blue",cex=7,pch=20)
```

```
plot(bic,type='l',main="BIC with Best Model Highlighted",xlab="Number of Variables")
points(12,(bic[12]),col="blue",cex=7,pch=20)
```

Since BIC calls for only 12 variables it is simpler than the r-square recommendation of 13. Therefore, we will fit our final model using the BIC recommendation of 12. Below is the code.

`coef(regfit.full,12)`

```
## (Intercept) hhiyes whiyes
## 30.31321796 1.16940604 18.25380263
## education.L education^4 education^5
## 6.63847641 1.54324869 -0.77783663
## raceblack hispanicyes experience
## 3.06580207 -1.33731802 -0.41883100
## kidslt6 kids618 husby
## -6.02251640 -0.82955827 -0.02129349
## regionnorthcentral
## 0.94042820
```

So here is our final model. This is what we would use for our test set.

**Conclusion**

Best subset regression provides the researcher with insights into every possible model as well as clues as to which model is at least statistically superior. This knowledge can be used for developing models for data science applications.

Understanding character vectors in R

There are times when least squares regression is not able to provide accurate predictions or explanation in an object. One example in which least scares regression struggles with a small sample size. By small, we mean when the total number of variables is greater than the sample size. Another term for this is high dimensions which means more variables than examples in the dataset

This post will explain the consequences of what happens when high dimensions is a problem and also how to address the problem.

**Inaccurate measurements**

One problem with high dimensions in regression is that the results for the various metrics are overfitted to the data. Below is an example using the “attitude” dataset. There are 2 variables and 3 examples for developing a model. This is not strictly high dimensions but it is an example of a small sample size.

```
data("attitude")
reg1 <- lm(complaints[1:3]~rating[1:3],data=attitude[1:3])
summary(reg1)
```

```
##
## Call:
## lm(formula = complaints[1:3] ~ rating[1:3], data = attitude[1:3])
##
## Residuals:
## 1 2 3
## 0.1026 -0.3590 0.2564
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 21.95513 1.33598 16.43 0.0387 *
## rating[1:3] 0.67308 0.02221 30.31 0.0210 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4529 on 1 degrees of freedom
## Multiple R-squared: 0.9989, Adjusted R-squared: 0.9978
## F-statistic: 918.7 on 1 and 1 DF, p-value: 0.021
```

With only 3 data points the fit is perfect. You can also examine the mean squared error of the model. Below is a function for this followed by the results

```
mse <- function(sm){
mean(sm$residuals^2)}
mse(reg1)
```

`## [1] 0.06837607`

Almost no error. Lastly, let’s look at a visual of the model

```
with(attitude[1:3],plot(complaints[1:3]~ rating[1:3]))
title(main = "Sample Size 3")
abline(lm(complaints[1:3]~rating[1:3],data = attitude))
```

You can see that the regression line goes almost perfectly through each data point. If we tried to use this model on the test set in a real data science problem there would be a huge amount of bias. Now we will rerun the analysis this time with the full sample.

```
reg2<- lm(complaints~rating,data=attitude)
summary(reg2)
```

```
##
## Call:
## lm(formula = complaints ~ rating, data = attitude)
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.3880 -6.4553 -0.2997 6.1462 13.3603
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.2445 7.6706 1.075 0.292
## rating 0.9029 0.1167 7.737 1.99e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.65 on 28 degrees of freedom
## Multiple R-squared: 0.6813, Adjusted R-squared: 0.6699
## F-statistic: 59.86 on 1 and 28 DF, p-value: 1.988e-08
```

You can clearly see a huge reduction in the r-square from .99 to .68. Next, is the mean-square error

`mse(reg2)`

`## [1] 54.61425`

The error has increased a great deal. Lastly, we fit the regression line

```
with(attitude,plot(complaints~ rating))
title(main = "Full Sample Size")
abline(lm(complaints~rating,data = attitude))
```

Naturally, the second model is more likely to perform better with a test set. The problem is that least squares regression is too flexible when the number of features is greater than or equal to the number of examples in a dataset.

**What to Do?**

If least squares regression must be used. One solution to overcoming high dimensionality is to use some form of regularization regression such as ridge, lasso, or elastic net. Any of these regularization approaches will help to reduce the number of variables or dimensions in the final model through the use of shrinkage.

However, keep in mind that no matter what you do as the number of dimensions increases so does the r-square even if the variable is useless. This is known as the curse of dimensionality. Again, regularization can help with this.

Remember that with a large number of dimensions there are normally several equally acceptable models. To determine which is most useful depends on understanding the problem and context of the study.

**Conclusion**

With the ability to collect huge amounts of data has led to the growing problem of high dimensionality. One there are more features than examples it can lead to statistical errors. However, regularization is one tool for dealing with this problem.

Logical vectors in r studio

Leave one out cross validation. (LOOCV) is a variation of the validation approach in that instead of splitting the dataset in half, LOOCV uses one example as the validation set and all the rest as the training set. This helps to reduce bias and randomness in the results but unfortunately, can increase variance. Remember that the goal is always to reduce the error rate which is often calculated as the mean-squared error.

In this post, we will use the “Hedonic” dataset from the “Ecdat” package to assess several different models that predict the taxes of homes In order to do this, we will also need to use the “boot” package. Below is the code.

`library(Ecdat);library(boot)`

`data(Hedonic) str(Hedonic)`

```
## 'data.frame': 506 obs. of 15 variables:
## $ mv : num 10.09 9.98 10.45 10.42 10.5 ...
## $ crim : num 0.00632 0.02731 0.0273 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 ...
## $ chas : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
## $ nox : num 28.9 22 22 21 21 ...
## $ rm : num 43.2 41.2 51.6 49 51.1 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 ...
## $ dis : num 1.41 1.6 1.6 1.8 1.8 ...
## $ rad : num 0 0.693 0.693 1.099 1.099 ...
## $ tax : int 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 ...
## $ blacks : num 0.397 0.397 0.393 0.395 0.397 ...
## $ lstat : num -3 -2.39 -3.21 -3.53 -2.93 ...
## $ townid : int 1 2 2 3 3 3 4 4 4 4 ...
```

First, we need to develop our basic least squares regression model. We will do this with the “glm” function. This is because the “cv.glm” function (more on this later) only works when models are developed with the “glm” function. Below is the code.

`tax.glm<-glm(tax ~ mv+crim+zn+indus+chas+nox+rm+age+dis+rad+ptratio+blacks+lstat, data = Hedonic)`

We now need to calculate the MSE. To do this we will use the “cv.glm” function. Below is the code.

```
cv.error<-cv.glm(Hedonic,tax.glm)
cv.error$delta
```

`## [1] 4536.345 4536.075`

cv.error$delta contains two numbers. The first is the MSE for the training set and the second is the error for the LOOCV. As you can see the numbers are almost identical.

We will now repeat this process but with the inclusion of different polynomial models. The code for this is a little more complicated and is below.

```
cv.error=rep(0,5)
for (i in 1:5){
tax.loocv<-glm(tax ~ mv+poly(crim,i)+zn+indus+chas+nox+rm+poly(age,i)+dis+rad+ptratio+blacks+lstat, data = Hedonic)
cv.error[i]=cv.glm(Hedonic,tax.loocv)$delta[1]
}
cv.error
```

`## [1] 4536.345 4515.464 4710.878 7047.097 9814.748`

Here is what happen.

- First, we created an empty object called “cv.error” with five empty spots, which we will use to store information later.
- Next, we created a for loop that repeats 5 times
- Inside the for loop, we create the same regression model except we added the “poly” function in front of “age”” and also “crim”. These are the variables we want to try polynomials 1-5 one to see if it reduces the error.
- The results of the polynomial models are stored in the “cv.error” object and we specifically request the results of “delta” Finally, we printed “cv.error” to the console.

From the results, you can see that the error decreases at a second order polynomial but then increases after that. This means that high order polynomials are not beneficial generally.

**Conclusion**

LOOCV is another option in assessing different models and determining which is most appropriate. As such, this is a tool that is used by many data scientist.

Estimating error and looking for ways to reduce it is a key component of machine learning. In this post, we will look at a simple way of addressing this problem through the use of the validation set method.

The validation set method is a standard approach in model development. To put it simply, you divide your dataset into a training and a hold-out set. The model is developed on the training set and then the hold-out set is used for prediction purposes. The error rate of the hold-out set is assumed to be reflective of the test error rate.

In the example below, we will use the “Carseats” dataset from the “ISLR” package. Our goal is to predict the competitors’ price for a carseat based on the other available variables. Below is some initial code

```
library(ISLR)
data("Carseats")
str(Carseats)
```

```
## 'data.frame': 400 obs. of 11 variables:
## $ Sales : num 9.5 11.22 10.06 7.4 4.15 ...
## $ CompPrice : num 138 111 113 117 141 124 115 136 132 132 ...
## $ Income : num 73 48 35 100 64 113 105 81 110 113 ...
## $ Advertising: num 11 16 10 4 3 13 0 15 0 0 ...
## $ Population : num 276 260 269 466 340 501 45 425 108 131 ...
## $ Price : num 120 83 80 97 128 72 108 120 124 124 ...
## $ ShelveLoc : Factor w/ 3 levels "Bad","Good","Medium": 1 2 3 3 1 1 3 2 3 3 ...
## $ Age : num 42 65 59 55 38 78 71 67 76 76 ...
## $ Education : num 17 10 12 14 13 16 15 10 10 17 ...
## $ Urban : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 1 2 2 1 1 ...
## $ US : Factor w/ 2 levels "No","Yes": 2 2 2 2 1 2 1 2 1 2 ...
```

We need to divide our dataset into two part. One will be the training set and the other the hold-out set. Below is the code.

```
set.seed(7)
train<-sample(x=400,size=200)
```

Now, for those who are familiar with R you know that we haven’t actually made our training set. We are going to use the “train” object to index items from the “Carseat” dataset. What we did was set the seed so that the results can be replicated. Then we used the “sample” function using two arguments “x” and “size”. X represents the number of examples in the “Carseat” dataset. Size represents how big we want the sample to be. In other words, we want a sample size of 200 of the 400 examples to be in the training set. Therefore, R will randomly select 200 numbers from 400.

We will now fit our initial model

`car.lm<-lm(CompPrice ~ Income+Sales+Advertising+Population+Price+ShelveLoc+Age+Education+Urban, data = Carseats,subset = train)`

The code above should not be new. However, one unique twist is the use of the “subset” argument. What this argument does is tell R to only use rows that are in the “train” index. Next, we calculate the mean squared error.

`mean((Carseats$CompPrice-predict(car.lm,Carseats))[-train]^2)`

`## [1] 77.13932`

Here is what the code above means

- We took the “CompPrice” results and subtracted them from the prediction made by the “car.lm” model we developed.
- Used the test set which here is identified as “-train” minus means everything that is not in the “train”” index
- the results were squared.

The results here are the baseline comparison. We will now make two more models each with a polynomial in one of the variables. First, we will square the “income” variable

```
car.lm2<-lm(CompPrice ~ Income+Sales+Advertising+Population+I(Income^2)+Price+ShelveLoc+Age+Education+Urban, data = Carseats,subset = train)
mean((Carseats$CompPrice-predict(car.lm2,Carseats))[-train]^2)
```

`## [1] 75.68999`

You can see that there is a small decrease in the MSE. Also, notice the use of the “I” function which allows us to square “income”. Now, let’s try a cubic model

```
car.lm3<-lm(CompPrice ~ Income+Sales+Advertising+Population+I(Income^3)+Price+ShelveLoc+Age+Education+Urban, data = Carseats,subset = train)
mean((Carseats$CompPrice-predict(car.lm3,Carseats))[-train]^2)
```

`## [1] 75.84575`

This time there was an increase when compared to the second model. As such, higher order polynomials will probably not improve the model.

**Conclusion**

This post provided a simple example of assessing several different models use the validation approach. However, in practice, this approach is not used as frequently as there are so many more ways to do this now. Yet, it is still good to be familiar with a standard approach such as this.

Understanding boolean logic in r programming

Working with vectors in R

In regression, one of the assumptions is the additive assumption. This assumption states that the influence of a predictor variable on the dependent variable is independent of any other influence. However, in practice, it is common that this assumption does not hold.

In this post, we will look at how to address violations of the additive assumption through the use of interactions in a regression model.

An interaction effect is when you have two predictor variables whose effect on the dependent variable is not the same. As such, their effect must be considered simultaneously rather than separately. This is done through the use of an interaction term. An interaction term is the product of the two predictor variables.

Let’s begin by making a regular regression model with an interaction. To do this we will use the “Carseats” data from the “ISLR” package to predict “Sales”. Below is the code.

```
library(ISLR);library(ggplot2)
data(Carseats)
saleslm<-lm(Sales~.,Carseats)
summary(saleslm)
```

```
##
## Call:
## lm(formula = Sales ~ ., data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.8692 -0.6908 0.0211 0.6636 3.4115
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.6606231 0.6034487 9.380 < 2e-16 ***
## CompPrice 0.0928153 0.0041477 22.378 < 2e-16 ***
## Income 0.0158028 0.0018451 8.565 2.58e-16 ***
## Advertising 0.1230951 0.0111237 11.066 < 2e-16 ***
## Population 0.0002079 0.0003705 0.561 0.575
## Price -0.0953579 0.0026711 -35.700 < 2e-16 ***
## ShelveLocGood 4.8501827 0.1531100 31.678 < 2e-16 ***
## ShelveLocMedium 1.9567148 0.1261056 15.516 < 2e-16 ***
## Age -0.0460452 0.0031817 -14.472 < 2e-16 ***
## Education -0.0211018 0.0197205 -1.070 0.285
## UrbanYes 0.1228864 0.1129761 1.088 0.277
## USYes -0.1840928 0.1498423 -1.229 0.220
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.019 on 388 degrees of freedom
## Multiple R-squared: 0.8734, Adjusted R-squared: 0.8698
## F-statistic: 243.4 on 11 and 388 DF, p-value: < 2.2e-16
```

The results are rather excellent for the social sciences. The model explains 87.3% of the variance in “Sales”. The current results that we have are known as main effects. These are effects that directly influence the dependent variable. Most regression models only include main effects.

We will now examine an interaction effect between two continuous variables. Let’s see if there is an interaction between “Population” and “Income”.

```
saleslm1<-lm(Sales~CompPrice+Income+Advertising+Population+Price+Age+Education+US+
Urban+ShelveLoc+Population*Income, Carseats)
summary(saleslm1)
```

```
##
## Call:
## lm(formula = Sales ~ CompPrice + Income + Advertising + Population +
## Price + Age + Education + US + Urban + ShelveLoc + Population *
## Income, data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.8699 -0.7624 0.0139 0.6763 3.4344
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.195e+00 6.436e-01 9.625 <2e-16 ***
## CompPrice 9.262e-02 4.126e-03 22.449 <2e-16 ***
## Income 7.973e-03 3.869e-03 2.061 0.0400 *
## Advertising 1.237e-01 1.107e-02 11.181 <2e-16 ***
## Population -1.811e-03 9.524e-04 -1.901 0.0580 .
## Price -9.511e-02 2.659e-03 -35.773 <2e-16 ***
## Age -4.566e-02 3.169e-03 -14.409 <2e-16 ***
## Education -2.157e-02 1.961e-02 -1.100 0.2722
## USYes -2.160e-01 1.497e-01 -1.443 0.1498
## UrbanYes 1.330e-01 1.124e-01 1.183 0.2375
## ShelveLocGood 4.859e+00 1.523e-01 31.901 <2e-16 ***
## ShelveLocMedium 1.964e+00 1.255e-01 15.654 <2e-16 ***
## Income:Population 2.879e-05 1.253e-05 2.298 0.0221 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.013 on 387 degrees of freedom
## Multiple R-squared: 0.8751, Adjusted R-squared: 0.8712
## F-statistic: 226 on 12 and 387 DF, p-value: < 2.2e-16
```

The new contribution is at the bottom of the coefficient table and is the “Income:Population” coefficient. What this means is “the increase of Sales given a one unit increase in Income and Population simultaneously” In other words the “Income:Population” coefficient looks at their combined simultaneous effect on Sales rather than just their independent effect on Sales.

This makes practical sense as well. The larger the population the more available income and vice versa. However, for our current model, the improvement in the r-squared is relatively small. The actual effect is a small increase in sales. Below is a graph of income and population by sales. Notice how the lines cross. This is a visual of what an interaction looks like. The lines are not parallel by any means.

```
ggplot(data=Carseats, aes(x=Income, y=Sales, group=1)) +geom_smooth(method=lm,se=F)+
geom_smooth(aes(Population,Sales), method=lm, se=F,color="black")+xlab("Income and Population")+labs(
title="Income in Blue Population in Black")
```

We will now repeat this process but this time using a categorical variable and a continuous variable. We will look at the interaction between “US” location (categorical) and “Advertising” (continuous).

```
saleslm2<-lm(Sales~CompPrice+Income+Advertising+Population+Price+Age+Education+US+
Urban+ShelveLoc+US*Advertising, Carseats)
summary(saleslm2)
```

```
##
## Call:
## lm(formula = Sales ~ CompPrice + Income + Advertising + Population +
## Price + Age + Education + US + Urban + ShelveLoc + US * Advertising,
## data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.8531 -0.7140 0.0266 0.6735 3.3773
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.6995305 0.6023074 9.463 < 2e-16 ***
## CompPrice 0.0926214 0.0041384 22.381 < 2e-16 ***
## Income 0.0159111 0.0018414 8.641 < 2e-16 ***
## Advertising 0.2130932 0.0530297 4.018 7.04e-05 ***
## Population 0.0001540 0.0003708 0.415 0.6782
## Price -0.0954623 0.0026649 -35.823 < 2e-16 ***
## Age -0.0463674 0.0031789 -14.586 < 2e-16 ***
## Education -0.0233500 0.0197122 -1.185 0.2369
## USYes -0.1057320 0.1561265 -0.677 0.4987
## UrbanYes 0.1191653 0.1127047 1.057 0.2910
## ShelveLocGood 4.8726025 0.1532599 31.793 < 2e-16 ***
## ShelveLocMedium 1.9665296 0.1259070 15.619 < 2e-16 ***
## Advertising:USYes -0.0933384 0.0537807 -1.736 0.0834 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.016 on 387 degrees of freedom
## Multiple R-squared: 0.8744, Adjusted R-squared: 0.8705
## F-statistic: 224.5 on 12 and 387 DF, p-value: < 2.2e-16
```

Again, you can see that when the store is in the US you have to also consider the advertising budget as well. When these two variables are considered there is a slight decline in sales. What this means in practice is that advertising in the US is not as beneficial as advertising outside the US.

Below you can again see a visual of the interaction effect when the lines for US yes and no cross each other in the plot below.

```
ggplot(data=Carseats, aes(x=Advertising, y=Sales, group = US, colour = US)) +
geom_smooth(method=lm,se=F)+scale_x_continuous(limits = c(0, 25))+scale_y_continuous(limits = c(0, 25))
```

Lastly, we will look at an interaction effect for two categorical variables.

```
saleslm3<-lm(Sales~CompPrice+Income+Advertising+Population+Price+Age+Education+US+
Urban+ShelveLoc+ShelveLoc*US, Carseats)
summary(saleslm3)
```

```
##
## Call:
## lm(formula = Sales ~ CompPrice + Income + Advertising + Population +
## Price + Age + Education + US + Urban + ShelveLoc + ShelveLoc *
## US, data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.8271 -0.6839 0.0213 0.6407 3.4537
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.8120748 0.6089695 9.544 <2e-16 ***
## CompPrice 0.0929370 0.0041283 22.512 <2e-16 ***
## Income 0.0158793 0.0018378 8.640 <2e-16 ***
## Advertising 0.1223281 0.0111143 11.006 <2e-16 ***
## Population 0.0001899 0.0003721 0.510 0.6100
## Price -0.0952439 0.0026585 -35.826 <2e-16 ***
## Age -0.0459380 0.0031830 -14.433 <2e-16 ***
## Education -0.0267021 0.0197807 -1.350 0.1778
## USYes -0.3683074 0.2379400 -1.548 0.1225
## UrbanYes 0.1438775 0.1128171 1.275 0.2030
## ShelveLocGood 4.3491643 0.2734344 15.906 <2e-16 ***
## ShelveLocMedium 1.8967193 0.2084496 9.099 <2e-16 ***
## USYes:ShelveLocGood 0.7184116 0.3320759 2.163 0.0311 *
## USYes:ShelveLocMedium 0.0907743 0.2631490 0.345 0.7303
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.014 on 386 degrees of freedom
## Multiple R-squared: 0.8753, Adjusted R-squared: 0.8711
## F-statistic: 208.4 on 13 and 386 DF, p-value: < 2.2e-16
```

In this case, we can see that when the store is in the US and the shelf location is good it has an effect on Sales when compared to a bad location. The plot below is a visual of this. However, it is harder to see this because the x-axis has only two categories

```
ggplot(data=Carseats, aes(x=US, y=Sales, group = ShelveLoc, colour = ShelveLoc)) +
geom_smooth(method=lm,se=F)
```

**Conclusion**

Interactions effects are a great way to fine-tune a model, especially for explanatory purposes. Often, the change in r-square is not strong enough for prediction but can be used for nuanced understanding of the relationships among the variables.

Vectors and functions in R

Understanding the use of vectors in Rstudio

In this post, we are going to look at Bayesian regression. In particular, we will compare the results of ordinary least squares regression with Bayesian regression.

**Bayesian Statistics**

Bayesian statistics involves the use of probabilities rather than frequencies when addressing uncertainty. This allows you to determine the distribution of the model parameters and not only the values. This is done through averaging over the model parameters through marginalizing the joint probability distribution.

**Linear Regression**

We will now develop our two models. The first model will be a normal regression and the second a Bayesian model. We will be looking at factors that affect the tax rate of homes in the “Hedonic” dataset in the “Ecdat” package. We will load our packages and partition our data. Below is some initial code

`library(ISLR);library(caret);library(arm);library(Ecdat);library(gridExtra)`

```
data("Hedonic")
inTrain<-createDataPartition(y=Hedonic$tax,p=0.7, list=FALSE)
trainingset <- Hedonic[inTrain, ]
testingset <- Hedonic[-inTrain, ]
str(Hedonic)
```

```
## 'data.frame': 506 obs. of 15 variables:
## $ mv : num 10.09 9.98 10.45 10.42 10.5 ...
## $ crim : num 0.00632 0.02731 0.0273 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 ...
## $ chas : Factor w/ 2 levels "no","yes": 1 1 1 1 1 1 1 1 1 1 ...
## $ nox : num 28.9 22 22 21 21 ...
## $ rm : num 43.2 41.2 51.6 49 51.1 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 ...
## $ dis : num 1.41 1.6 1.6 1.8 1.8 ...
## $ rad : num 0 0.693 0.693 1.099 1.099 ...
## $ tax : int 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 ...
## $ blacks : num 0.397 0.397 0.393 0.395 0.397 ...
## $ lstat : num -3 -2.39 -3.21 -3.53 -2.93 ...
## $ townid : int 1 2 2 3 3 3 4 4 4 4 ...
```

We will now create our regression model

```
ols.reg<-lm(tax~.,trainingset)
summary(ols.reg)
```

```
##
## Call:
## lm(formula = tax ~ ., data = trainingset)
##
## Residuals:
## Min 1Q Median 3Q Max
## -180.898 -35.276 2.731 33.574 200.308
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 305.1928 192.3024 1.587 0.11343
## mv -41.8746 18.8490 -2.222 0.02697 *
## crim 0.3068 0.6068 0.506 0.61339
## zn 1.3278 0.2006 6.618 1.42e-10 ***
## indus 7.0685 0.8786 8.045 1.44e-14 ***
## chasyes -17.0506 15.1883 -1.123 0.26239
## nox 0.7005 0.4797 1.460 0.14518
## rm -0.1840 0.5875 -0.313 0.75431
## age 0.3054 0.2265 1.349 0.17831
## dis -7.4484 14.4654 -0.515 0.60695
## rad 98.9580 6.0964 16.232 < 2e-16 ***
## ptratio 6.8961 2.1657 3.184 0.00158 **
## blacks -29.6389 45.0043 -0.659 0.51061
## lstat -18.6637 12.4674 -1.497 0.13532
## townid 1.1142 0.1649 6.758 6.07e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 63.72 on 341 degrees of freedom
## Multiple R-squared: 0.8653, Adjusted R-squared: 0.8597
## F-statistic: 156.4 on 14 and 341 DF, p-value: < 2.2e-16
```

The model does a reasonable job. Next, we will do our prediction and compare the results with the test set using correlation, summary statistics, and the mean absolute error. In the code below, we use the “predict.lm” function and include the arguments “interval” for the prediction as well as “se.fit” for the standard error

`ols.regTest<-predict.lm(ols.reg,testingset,interval = 'prediction',se.fit = T)`

Below is the code for the correlation, summary stats, and mean absolute error. For MAE, we need to create a function.

`cor(testingset$tax,ols.regTest$fit[,1])`

`## [1] 0.9313795`

`summary(ols.regTest$fit[,1])`

```
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 144.7 288.3 347.6 399.4 518.4 684.1
```

`summary(trainingset$tax)`

```
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 188.0 279.0 330.0 410.4 666.0 711.0
```

```
MAE<-function(actual, predicted){
mean(abs(actual-predicted))
}
MAE(ols.regTest$fit[,1], testingset$tax)
```

`## [1] 41.07212`

The correlation is excellent. The summary stats are similar and the error is not unreasonable. Below is a plot of the actual and predicted values

We now need to combine some data into one dataframe. In particular, we need the following actual dependent variable results predicted dependent variable results The upper confidence value of the prediction THe lower confidence value of the prediction

The code is below

```
yout.ols <- as.data.frame(cbind(testingset$tax,ols.regTest$fit))
ols.upr <- yout.ols$upr
ols.lwr <- yout.ols$lwr
```

We can now plot this

```
p.ols <- ggplot(data = yout.ols, aes(x = testingset$tax, y = ols.regTest$fit[,1])) + geom_point() + ggtitle("Ordinary Regression") + labs(x = "Actual", y = "Predicted")
p.ols + geom_errorbar(ymin = ols.lwr, ymax = ols.upr)
```

You can see the strong linear relationship. However, the confidence intervals are rather wide. Let’s see how Bayes does.

**Bayes Regression**

Bayes regression uses the “bayesglm” function from the “arm” package. We need to set the family to “gaussian” and the link to “identity”. In addition, we have to set the “prior.df” (prior degrees of freedom) to infinity as this indicates we want a normal distribution

`bayes.reg<-bayesglm(tax~.,family=gaussian(link=identity),trainingset,prior.df = Inf)`

```
bayes.regTest<-predict.glm(bayes.reg,newdata = testingset,se.fit = T)
cor(testingset$tax,bayes.regTest$fit)
```

`## [1] 0.9313793`

`summary(bayes.regTest$fit)`

```
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 144.7 288.3 347.5 399.4 518.4 684.1
```

`summary(trainingset$tax)`

```
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 188.0 279.0 330.0 410.4 666.0 711.0
```

`MAE(bayes.regTest$fit, testingset$tax)`

`## [1] 41.07352`

The numbers are essentially the same. This leads to the question of what is the benefit of Bayesian regression? The answer is in the confidence intervals. Next, we will calculate the confidence intervals for the Bayesian model.

```
yout.bayes <- as.data.frame(cbind(testingset$tax,bayes.regTest$fit))
names(yout.bayes) <- c("tax", "fit")
critval <- 1.96 #approx for 95% CI
bayes.upr <- bayes.regTest$fit + critval * bayes.regTest$se.fit
bayes.lwr <- bayes.regTest$fit - critval * bayes.regTest$se.fit
```

We now create our Bayesian regression plot.

`p.bayes <- ggplot(data = yout.bayes, aes(x = yout.bayes$tax, y = yout.bayes$fit)) + geom_point() + ggtitle("Bayesian Regression Prediction") + labs(x = "Actual", y = "Predicted")`

Lastly, we display both plots as a comparison.

```
ols.plot <- p.ols + geom_errorbar(ymin = ols.lwr, ymax = ols.upr)
bayes.plot <- p.bayes + geom_errorbar(ymin = bayes.lwr, ymax = bayes.upr)
grid.arrange(ols.plot,bayes.plot,ncol=2)
```

As you can see, the Bayesian approach gives much more compact confidence intervals. This is because the Bayesian approach a distribution of parameters is calculated from a posterior distribution. These values are then averaged to get the final prediction that appears on the plot. This reduces the variance and strengthens the confidence we can have in each individual example.

Data preparation aka data munging is what most data scientist spend the majority of their time doing. Extracting and transforming data is difficult, to say the least. Every dataset is different with unique problems. This makes it hard to generalize best practices for transforming data so that it is suitable for analysis.

In this post, we will look at how to use the various functions in the “dplyr”” package. This package provides numerous ways to develop features as well as explore the data. We will use the “attitude” dataset from base r for our analysis. Below is some initial code.

`library(dplyr)`

```
data("attitude")
str(attitude)
```

```
## 'data.frame': 30 obs. of 7 variables:
## $ rating : num 43 63 71 61 81 43 58 71 72 67 ...
## $ complaints: num 51 64 70 63 78 55 67 75 82 61 ...
## $ privileges: num 30 51 68 45 56 49 42 50 72 45 ...
## $ learning : num 39 54 69 47 66 44 56 55 67 47 ...
## $ raises : num 61 63 76 54 71 54 66 70 71 62 ...
## $ critical : num 92 73 86 84 83 49 68 66 83 80 ...
## $ advance : num 45 47 48 35 47 34 35 41 31 41 ...
```

You can see we have seven variables and only 30 observations. Our first function that we will learn to use is the “select” function. This function allows you to select columns of data you want to use. In order to use this feature, you need to know the names of the columns you want. Therefore, we will first use the “names” function to determine the names of the columns and then use the “select”” function.

`names(attitude)[1:3]`

`## [1] "rating" "complaints" "privileges"`

```
smallset<-select(attitude,rating:privileges)
head(smallset)
```

```
## rating complaints privileges
## 1 43 51 30
## 2 63 64 51
## 3 71 70 68
## 4 61 63 45
## 5 81 78 56
## 6 43 55 49
```

The difference is probably obvious. Using the “select” function we have 3 instead of 7 variables. We can also exclude columns we do not want by placing a negative in front of the names of the columns. Below is the code

`head(select(attitude,-(rating:privileges)))`

```
## learning raises critical advance
## 1 39 61 92 45
## 2 54 63 73 47
## 3 69 76 86 48
## 4 47 54 84 35
## 5 66 71 83 47
## 6 44 54 49 34
```

We can also use the “rename” function to change the names of columns. In our example below, we will change the name of the “rating” to “rates.” The code is below. Keep in mind that the new name for the column is to the left of the equal sign and the old name is to the right

```
attitude<-rename(attitude,rates=rating)
head(attitude)
```

```
## rates complaints privileges learning raises critical advance
## 1 43 51 30 39 61 92 45
## 2 63 64 51 54 63 73 47
## 3 71 70 68 69 76 86 48
## 4 61 63 45 47 54 84 35
## 5 81 78 56 66 71 83 47
## 6 43 55 49 44 54 49 34
```

The “select”” function can be used in combination with other functions to find specific columns in the dataset. For example, we will use the “ends_with” function inside the “select” function to find all columns that end with the letter s.

```
s_set<-head(select(attitude,ends_with("s")))
s_set
```

```
## rates complaints privileges raises
## 1 43 51 30 61
## 2 63 64 51 63
## 3 71 70 68 76
## 4 61 63 45 54
## 5 81 78 56 71
## 6 43 55 49 54
```

The “filter” function allows you to select rows from a dataset based on criteria. In the code below we will select only rows that have a 75 or higher in the “raises” variable.

```
bigraise<-filter(attitude,raises>75)
bigraise
```

```
## rates complaints privileges learning raises critical advance
## 1 71 70 68 69 76 86 48
## 2 77 77 54 72 79 77 46
## 3 74 85 64 69 79 79 63
## 4 66 77 66 63 88 76 72
## 5 78 75 58 74 80 78 49
## 6 85 85 71 71 77 74 55
```

If you look closely all values in the “raise” column are greater than 75. Of course, you can have more than one criteria. IN the code below there are two.

`filter(attitude, raises>70 & learning<67)`

```
## rates complaints privileges learning raises critical advance
## 1 81 78 56 66 71 83 47
## 2 65 70 46 57 75 85 46
## 3 66 77 66 63 88 76 72
```

The “arrange” function allows you to sort the order of the rows. In the code below we first sort the data ascending by the “critical” variable. Then we sort it descendingly by adding the “desc” function.

```
ascCritical<-arrange(attitude, critical)
head(ascCritical)
```

```
## rates complaints privileges learning raises critical advance
## 1 43 55 49 44 54 49 34
## 2 81 90 50 72 60 54 36
## 3 40 37 42 58 50 57 49
## 4 69 62 57 42 55 63 25
## 5 50 40 33 34 43 64 33
## 6 71 75 50 55 70 66 41
```

```
descCritical<-arrange(attitude, desc(critical))
head(descCritical)
```

```
## rates complaints privileges learning raises critical advance
## 1 43 51 30 39 61 92 45
## 2 71 70 68 69 76 86 48
## 3 65 70 46 57 75 85 46
## 4 61 63 45 47 54 84 35
## 5 81 78 56 66 71 83 47
## 6 72 82 72 67 71 83 31
```

The “mutate” function is useful for engineering features. In the code below we will transform the “learning” variable by subtracting its mean from its self

```
attitude<-mutate(attitude,learningtrend=learning-mean(learning))
head(attitude)
```

```
## rates complaints privileges learning raises critical advance
## 1 43 51 30 39 61 92 45
## 2 63 64 51 54 63 73 47
## 3 71 70 68 69 76 86 48
## 4 61 63 45 47 54 84 35
## 5 81 78 56 66 71 83 47
## 6 43 55 49 44 54 49 34
## learningtrend
## 1 -17.366667
## 2 -2.366667
## 3 12.633333
## 4 -9.366667
## 5 9.633333
## 6 -12.366667
```

You can also create logical variables with the “mutate” function.In the code below, we create a logical variable that is true when the “critical” variable” is higher than 80 and false when “critical”” is less than 80. The new variable is called “highCritical”

```
attitude<-mutate(attitude,highCritical=critical>=80)
head(attitude)
```

```
## rates complaints privileges learning raises critical advance
## 1 43 51 30 39 61 92 45
## 2 63 64 51 54 63 73 47
## 3 71 70 68 69 76 86 48
## 4 61 63 45 47 54 84 35
## 5 81 78 56 66 71 83 47
## 6 43 55 49 44 54 49 34
## learningtrend highCritical
## 1 -17.366667 TRUE
## 2 -2.366667 FALSE
## 3 12.633333 TRUE
## 4 -9.366667 TRUE
## 5 9.633333 TRUE
## 6 -12.366667 FALSE
```

The “group_by” function is used for creating summary statistics based on a specific variable. It is similar to the “aggregate” function in R. This function works in combination with the “summarize” function for our purposes here. We will group our data by the “highCritical” variable. This means our data will be viewed as either TRUE for “highCritical” or FALSE. The results of this function will be saved in an object called “hcgroups”

```
hcgroups<-group_by(attitude,highCritical)
head(hcgroups)
```

```
## # A tibble: 6 x 9
## # Groups: highCritical [2]
## rates complaints privileges learning raises critical advance
##
## 1 43 51 30 39 61 92 45
## 2 63 64 51 54 63 73 47
## 3 71 70 68 69 76 86 48
## 4 61 63 45 47 54 84 35
## 5 81 78 56 66 71 83 47
## 6 43 55 49 44 54 49 34
## # ... with 2 more variables: learningtrend , highCritical
```

Looking at the data you probably saw no difference. This is because we are not done yet. We need to summarize the data in order to see the results for our two groups in the “highCritical” variable.

We will now generate the summary statistics by using the “summarize” function. We specifically want to know the mean of the “complaint” variable based on the variable “highCritical.” Below is the code

`summarize(hcgroups,complaintsAve=mean(complaints))`

```
## # A tibble: 2 x 2
## highCritical complaintsAve
##
## 1 FALSE 67.31579
## 2 TRUE 65.36364
```

Of course, you could have learned this through doing a t.test but this is another approach.

**Conclusion**

The “dplyr” package is one powerful tool for wrestling with data. There is nothing new in this package. Instead, the coding is simpler than what you can excute using base r.

In this post, we will look at analyzing tweets from Twitter using R. Before beginning, if you plan to replicate this on your own, you will need to set up a developer account with Twitter. Below are the steps

**Twitter Setup**

- Go to https://dev.twitter.com/apps
- Create a twitter account if you do not already have one
- Next, you want to click “create new app”
- After entering the requested information be sure to keep the following information for R; consumer key, consumer secret, request token URL, authorize URL, access token URL

The instruction here are primarily for users of Linux. If you are using a windows machine you need to download a cecert.pem file below is the code

download.file(url=‘http://curl.haxx.se/ca/cacert.pem’,destfile=‘/YOUR_LOCATION/cacert.pem’)

You need to save this file where it is appropriate. Below we will begin the analysis by loading the appropriate libraries.

**R Setup**

`library(twitteR);library(ROAuth);library(RCurl);library(tm);library(wordcloud)`

Next, we need to use all of the application information we generate when we created the developer account at twitter. We will save the information in objects to use in R. In the example code below “XXXX” is used where you should provide your own information. Sharing this kind of information would allow others to use my twitter developer account. Below is the code

```
my.key<-"XXXX" #consumer key
my.secret<-"XXXX" #consumer secret
my.accesstoken<-'XXXX' #access token
my.accesssecret<-'XXXX' ##access secret
```

Some of the information we just stored now needs to be passed to the “OAuthFactory” function of the “ROAuth” package. We will be passing the “my.key” and “my.secret”. We also need to add the request URL, access URL, and auth URL. Below is the code for all this.

```
cred<-OAuthFactory$new(consumerKey=my.key,consumerSecret=my.secret,requestURL='https://api.twitter/oauth/request_token',
accessURL='https://api.twitter/oauth/access_token',authURL='https://api.twitter/oauth/authorize')
```

If you are a windows user you need to code below for the cacert.pem. You need to use the “cred$handshake(cainfo=”LOCATION OF CACERT.PEM” to complete the setup process. make sure to save your authentication and then use the “registerTwitterOAuth(cred)” to finish this. For Linux users, the code is below.

`setup_twitter_oauth(my.key, my.secret, my.accesstoken, my.accesssecret)`

**Data Preparation**

We can now begin the analysis. We are going to search twitter for the term “Data Science.” We will look for 1,500 of the most recent tweets that contain this term. To do this we will use the “searchTwitter” function. The code is as follows

`ds_tweets<-searchTwitter("data science",n=1500)`

We know need to some things that are a little more complicated. First, we need to convert our “ds_tweets” object to a dataframe. This is just to save our search so we don’t have to research again. The code below performs this.

`ds_tweets.df<-do.call(rbind,lapply(ds_tweets,as.data.frame))`

Second, we need to find all the text in our “ds_tweets” object and convert this into a list. We will use the “sapply” function along with a “getText” Below is the code

`ds_tweets.list<-sapply(ds_tweets,function(x) x$getText())`

Third, we need to turn our “ds_tweets.list” into a corpus.

`ds_tweets.corpus<-Corpus(VectorSource(ds_tweets.list)) `

Now we need to do a lot of cleaning of the text. In particular, we need to *make all words lower case *remove punctuation *Get rid of funny characters (i.e. #,/, etc) *remove stopwords (words that lack meaning such as “the”)

To do this we need to use a combination of functions in the “tm” package as well as some personally made functions

```
ds_tweets.corpus<-tm_map(ds_tweets.corpus,removePunctuation)
removeSpecialChars <- function(x) gsub("[^a-zA-Z0-9 ]","",x)#remove garbage terms
ds_tweets.corpus<-tm_map(ds_tweets.corpus,removeSpecialChars) #application of custom function
ds_tweets.corpus<-tm_map(ds_tweets.corpus,function(x) removeWords(x,stopwords())) #removes stop words
ds_tweets.corpus<-tm_map(ds_tweets.corpus,tolower)
```

**Data Analysis**

We can make a word cloud for fun now

```
wordcloud(ds_tweets.corpus)
```

We now need to convert our corpus to a matrix for further analysis. In addition, we need to remove sparse terms as this reduces the size of the matrix without losing much information. The value to set it to is at the discretion of the researcher. Below is the code

```
ds_tweets.tdm<-TermDocumentMatrix(ds_tweets.corpus)
ds_tweets.tdm<-removeSparseTerms(ds_tweets.tdm,sparse = .8)#remove sparse terms
```

We’ve looked at how to find the most frequent terms in another post. Below is the code for the 15 most common words

`findFreqTerms(ds_tweets.tdm,15)`

```
## [1] "datasto" "demonstrates" "download" "executed"
## [5] "hard" "key" "leaka" "locally"
## [9] "memory" "mitchellvii" "now" "portable"
## [13] "science" "similarly" "data"
```

Below are words that are highly correlated with the term “key”.

`findAssocs(ds_tweets.tdm,'key',.95)`

```
## $key
## demonstrates download executed leaka locally
## 0.99 0.99 0.99 0.99 0.99
## memory datasto hard mitchellvii portable
## 0.99 0.98 0.98 0.98 0.98
## similarly
## 0.98
```

For the final trick, we will make a hierarchical agglomerative cluster. This will clump words that are more similar next to each other. We first need to convert our current “ds_tweets.tdm” into a regular matrix. Then we need to scale it because the distances need to be standardized. Below is the code.

```
ds_tweets.mat<-as.matrix(ds_tweets.tdm)
ds_tweets.mat.scale<-scale(ds_tweets.mat)
```

Now, we need to calculate the distance statistically

`ds_tweets.dist<-dist(ds_tweets.mat.scale,method = 'euclidean')`

At last, we can make the clusters,

`ds_tweets.fit<-hclust(ds_tweets.dist,method = 'ward')`

`plot(ds_tweets.fit)`

Looking at the chart, it appears we have six main clusters we can highlight them using the code below

```
plot(ds_tweets.fit)
groups<-cutree(ds_tweets.fit,k=6)
rect.hclust(ds_tweets.fit,k=6)
```

**Conclusion**

This post provided an example of how to pull data from twitter for text analysis. There are many steps but also some useful insights can be gained from this sort of research.

In this post, we will learn how to conduct a diversity and lexical dispersion analysis in R. Diversity analysis is a measure of the breadth of an author’s vocabulary in a text. Are provides several calculations of this in their output

Lexical dispersion is used for knowing where or when certain words are used in a text. This is useful for identifying patterns if this is a goal of your data exploration.

We will conduct our two analysis by comparing two famous philosophical texts

- Analects
- The Prince

These books are available at the Gutenberg Project. You can go to the site type in the titles and download them to your computer.

We will use the “qdap” package in order to complete the sentiment analysis. Below is some initial code.

library(qdap)

**Data Preparation**

Below are the steps we need to take to prepare the data

- Paste the text files into R
- Convert the text files to ASCII format
- Convert the ASCII format to data frames
- Split the sentences in the data frame
- Add a variable that indicates the book name
- Combine the two books into one dataframe

We now need to prepare the three text. First, we move them into R using the “paste” function.

analects<-paste(scan(file ="C:/Users/darrin/Documents/R/R working directory/blog/blog/Text/Analects.txt",what='character'),collapse=" ") prince<-paste(scan(file ="C:/Users/darrin/Documents/R/R working directory/blog/blog/Text/Prince.txt",what='character'),collapse=" ")

We must convert the text files to ASCII format see that R is able to interpret them.

analects<-iconv(analects,"latin1","ASCII","") prince<-iconv(prince,"latin1","ASCII","")

For each book, we need to make a dataframe. The argument “texts” gives our dataframe one variable called “texts” which contains all the words in each book. Below is the code

data frame

analects<-data.frame(texts=analects) prince<-data.frame(texts=prince)

With the dataframes completed. We can now split the variable “texts” in each dataframe by sentence. The “sentSplit” function will do this.

analects<-sentSplit(analects,'texts') prince<-sentSplit(prince,'texts')

Next, we add the variable “book” to each dataframe. What this does is that for each row or sentence in the dataframe the “book” variable will tell you which book the sentence came from. This will be valuable for comparative purposes.

analects$book<-"analects" prince$book<-"prince"

Now we combine the two books into one dataframe. The data preparation is now complete.

twobooks<-rbind(analects,prince)

**Data Analysis**

We will begin with the diversity analysis

div<-diversity(twobooks$texts,twobooks$book) div book wc simpson shannon collision berger_parker brillouin 1 analects 30995 0.989 6.106 4.480 0.067 5.944 2 prince 52105 0.989 6.324 4.531 0.059 6.177

For most of the metrics, the diversity in the use of vocabulary is the same despite being different books from different eras in history. How these numbers are calculated is beyond the scope of this post.

Next, we will calculate the lexical dispersion of the two books. Will look at three common themes in history money, war, and marriage.

dispersion_plot(twobooks$texts,grouping.var=twobooks$book,c("money","war",'marriage'))

The tick marks show when each word appears. For example, money appears at the beginning of Analects only but is more spread out in tThe PRince. War is evenly dispersed in both books and marriage only appears in The Prince

**Conclusion**

This analysis showed additional tools that can be used to analyze text in R.

%d bloggers like this: