Problem 3

Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of \(\hat{p}_{m1}\) . The x-axis should display \(\hat{p}_{m1}\) , ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
Hint: In a setting with two classes, \(\hat{p}_{m1} = 1 − \hat{p}_{m2}\) . You could make this plot by hand, but it will be much easier to make in R.

library(tidyverse)
library(plotly)
gini<-as_tibble(list(P=seq(0, 1, 0.001))) %>% mutate(Value=P*(1-P)*2, Measure="Gini")
ent<-as_tibble(list(P=seq(0, 1, 0.001))) %>% mutate(Value=-(P*log(P)+(1-P)*log(1-P)), Measure="Entropy")
error=as_tibble(list(P=seq(0, 1, 0.001))) %>% mutate(Value=1-pmax(P,1-P), Measure="Classification Error")
df<-bind_rows(gini, ent, error) %>% arrange(Measure,P)
c<-ggplot(df, aes(x=P,y=Value,col=Measure))+geom_line()+scale_color_manual(values=c("#377eb8","#e41a1c","#4daf4a"))+labs(x="P",y="Value for Split")+theme_minimal() 
fig<-ggplotly(c,width=600,height=300)
fig

Problem 8

In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.
(a) Split the data set into a training set and a test set.

library(ISLR2)
attach(Carseats)
set.seed(1)

train=sample(dim(Carseats)[1], dim(Carseats)[1]/2)
test=-train

(b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

library(tree)
tree.carseats=tree(formula=Sales~.,data=Carseats[train,])
tree.pred=predict(tree.carseats,Carseats[-train,])

plot(tree.carseats)
text(tree.carseats)

mean((tree.pred-Carseats[-train,'Sales'])^2)
## [1] 4.922039

(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?

tree.carseats.cv=cv.tree(tree.carseats) 
plot(tree.carseats.cv)

prune.carseats=prune.tree(tree.carseats,best=10)

plot(prune.carseats)
text(prune.carseats) 

tree.pred=predict(prune.carseats,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)
## [1] 4.918134

This is a very slight improvement to the MSE.

(d) Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.

library(randomForest)
set.seed(1)

carseats.rf=randomForest(Sales~.,data=Carseats[train,],mtry=10,importance=T,ntree=100)
tree.pred=predict(carseats.rf,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)
## [1] 2.616711
knitr::kable(importance(carseats.rf))
%IncMSE IncNodePurity
CompPrice 11.4033686 169.11991
Income 1.4751526 92.34927
Advertising 5.3836500 96.15208
Population -2.0667575 62.75983
Price 27.2224905 492.32337
ShelveLoc 22.7210914 363.27721
Age 9.1141420 154.10317
Education -0.6365854 46.57603
Urban -0.6581541 10.55949
US 2.1808079 16.20339

The MSE has improved to 2.616711. Price, ShelveLoc, and CompPrice are the most important predictors for Sales.

(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.

carseats.rf=randomForest(Sales~.,data=Carseats[train,],mtry=7,importance=T,ntree=100)
tree.pred=predict(carseats.rf,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)
## [1] 2.637936

In this case, random forests did not result in an improvement over bagging. The same variables are still the most important predictors.

(f) Now analyze the data using BART, and report your results.

library(BART)
x=Carseats[,2:11]
y=Carseats[,"Sales"]
xtrain=x[train,]
ytrain=y[train]
xtest=x[test,]
ytest=y[test]
set.seed(1)
bartfit=gbart(xtrain,ytrain,x.test=xtest)
## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 200
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 107.000000, 1.000000
## xp1,xp[np*p]: 111.000000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 63 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,0.23074,7.57815
## *****sigma: 1.088371
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
## *****printevery: 100
## 
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 2s
## trcnt,tecnt: 1000,1000
yhat.bart=bartfit$yhat.test.mean
mean((ytest-yhat.bart)^2)
## [1] 1.450842
detach(Carseats)

Problem 9

This problem involves the OJ data set which is part of the ISLR2 package.
(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

attach(OJ)
set.seed(1)

train=sample(nrow(OJ),800)
test=-train

(b) Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?

OJ.tree=tree(Purchase~.,data=OJ[train,])
summary(OJ.tree)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ[train, ])
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "SpecialCH"     "ListPriceDiff"
## [5] "PctDiscMM"    
## Number of terminal nodes:  9 
## Residual mean deviance:  0.7432 = 587.8 / 791 
## Misclassification error rate: 0.1588 = 127 / 800

The tree uses 5 variables and has 9 terminal nodes. We have a training error rate of 0.1588.

(c) Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed.

OJ.tree
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 800 1073.00 CH ( 0.60625 0.39375 )  
##    2) LoyalCH < 0.5036 365  441.60 MM ( 0.29315 0.70685 )  
##      4) LoyalCH < 0.280875 177  140.50 MM ( 0.13559 0.86441 )  
##        8) LoyalCH < 0.0356415 59   10.14 MM ( 0.01695 0.98305 ) *
##        9) LoyalCH > 0.0356415 118  116.40 MM ( 0.19492 0.80508 ) *
##      5) LoyalCH > 0.280875 188  258.00 MM ( 0.44149 0.55851 )  
##       10) PriceDiff < 0.05 79   84.79 MM ( 0.22785 0.77215 )  
##         20) SpecialCH < 0.5 64   51.98 MM ( 0.14062 0.85938 ) *
##         21) SpecialCH > 0.5 15   20.19 CH ( 0.60000 0.40000 ) *
##       11) PriceDiff > 0.05 109  147.00 CH ( 0.59633 0.40367 ) *
##    3) LoyalCH > 0.5036 435  337.90 CH ( 0.86897 0.13103 )  
##      6) LoyalCH < 0.764572 174  201.00 CH ( 0.73563 0.26437 )  
##       12) ListPriceDiff < 0.235 72   99.81 MM ( 0.50000 0.50000 )  
##         24) PctDiscMM < 0.196196 55   73.14 CH ( 0.61818 0.38182 ) *
##         25) PctDiscMM > 0.196196 17   12.32 MM ( 0.11765 0.88235 ) *
##       13) ListPriceDiff > 0.235 102   65.43 CH ( 0.90196 0.09804 ) *
##      7) LoyalCH > 0.764572 261   91.20 CH ( 0.95785 0.04215 ) *

For terminal node 8) the splitting variable is LoyalCH. The splitting value is 0.0356415. There are 59 points in the subtree below this node. The deviance for all points in this region is 10.14. The prediction at this node is Sales = MM.

(d) Create a plot of the tree, and interpret the results.

plot(OJ.tree)
text(OJ.tree)

(e) Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate?

library(caret)
OJ.pred=predict(OJ.tree,OJ[test,],type='class')
confusionMatrix(OJ[test,'Purchase'],OJ.pred)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160   8
##         MM  38  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.7333          
##     P-Value [Acc > NIR] : 0.0001259       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.8081          
##             Specificity : 0.8889          
##          Pos Pred Value : 0.9524          
##          Neg Pred Value : 0.6275          
##              Prevalence : 0.7333          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.6222          
##       Balanced Accuracy : 0.8485          
##                                           
##        'Positive' Class : CH              
##

(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.

OJ.tree.cv=cv.tree(OJ.tree,FUN=prune.tree)

(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

plot(OJ.tree.cv)

(h) Which tree size corresponds to the lowest cross-validated classification error rate?

Size 8 produces the lowest cross-validation error.

(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.

OJ.tree.prune=prune.tree(OJ.tree,best=8)

(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?

summary(OJ.tree.prune)
## 
## Classification tree:
## snip.tree(tree = OJ.tree, nodes = 10L)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "ListPriceDiff" "PctDiscMM"    
## Number of terminal nodes:  8 
## Residual mean deviance:  0.7582 = 600.5 / 792 
## Misclassification error rate: 0.1625 = 130 / 800

(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?

unp.err= sum(OJ[test,'Purchase'] != OJ.pred)
unp.err/length(OJ.pred)
## [1] 0.1703704
OJ.pred.prune=predict(OJ.tree.prune,OJ[test,],type='class')
p.err= sum(OJ[test,'Purchase'] != OJ.pred.prune)
p.err/length(OJ.pred.prune)
## [1] 0.162963

The unpruned tree has a higher test error rate.