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

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.

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 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.

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.