Chapter 08: 3, 8, 9

library(ISLR)
attach(Carseats)
library(tree)
library(rpart)
library(caret)
## Loading required package: ggplot2
## Warning: package 'ggplot2' was built under R version 4.3.2
## Loading required package: lattice
library(randomForest)
## randomForest 4.7-1.1
## Type rfNews() to see new features/changes/bug fixes.
## 
## Attaching package: 'randomForest'
## The following object is masked from 'package:ggplot2':
## 
##     margin
library(car)
## Loading required package: carData
library(tree)
  1. 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 pˆm1.
p = seq(0, 1, 0.001)
gini.index = 2 * p * (1 - p)
class.error = 1 - pmax(p, 1 - p)
cross.entropy = - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), ylab = "gini.index, class.error, cross.entropy", col = c("yellow", "blue", "red"))

  1. In the lab, a classification tree was applied to the Carseats data set af- ter 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.
  1. Split the data set into a training set and a test set.
set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats)/2)
strain = Carseats[train, ]
stest = Carseats[-train, ]
  1. Fit a regression tree to the training set. Plot the tree, and inter- pret the results. What test MSE do you obtain?
tree.seats = tree(Sales ~ ., data = strain)
summary(tree.seats)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = strain)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Advertising" "CompPrice"  
## [6] "US"         
## Number of terminal nodes:  18 
## Residual mean deviance:  2.167 = 394.3 / 182 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -3.88200 -0.88200 -0.08712  0.00000  0.89590  4.09900
plot(tree.seats)
text(tree.seats, pretty = 0)

treeseat.pred = predict(tree.seats, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)
## [1] 4.922039

The test MSE is 4.922039. (c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?

set.seed(1)
cv.seats = cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b")

prune.car = prune.tree(tree.seats, best = 10)
plot(prune.car)
text(prune.car,pretty=0)

treeseat.pred = predict(prune.car, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)
## [1] 4.918134

Pruning the tree does improve the MSE as the new MSE is 4.918134, as opposed to the previous MSE of 4.922039.

  1. Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to de- termine which variables are most important.
set.seed(1)
bag.seats = randomForest(Sales~., data = strain, mtry = 10, ntree = 551, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = stest)
mean((bagseat.pred - stest$Sales)^2)
## [1] 2.599099

The test MSE is 2.599099.

importance(bag.seats)
##                 %IncMSE IncNodePurity
## CompPrice   26.18616309    170.781666
## Income       5.25063979     90.717958
## Advertising 13.25673204     97.498810
## Population  -2.14346969     58.289311
## Price       60.58241525    503.478806
## ShelveLoc   50.77308639    380.258594
## Age         19.03720001    158.282846
## Education    1.24264920     44.834257
## Urban       -0.08461165      9.883299
## US           4.71515903     17.907727
  1. 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.
set.seed(1)
rando.seats = randomForest(Sales~., data = strain, mtry = 10, importance = TRUE)
randseat.pred = predict(rando.seats, newdata = stest)
mean((randseat.pred - stest$Sales)^2)
## [1] 2.605253

The test MSE is 2.605253.

importance(rando.seats)
##                %IncMSE IncNodePurity
## CompPrice   24.8888481    170.182937
## Income       4.7121131     91.264880
## Advertising 12.7692401     97.164338
## Population  -1.8074075     58.244596
## Price       56.3326252    502.903407
## ShelveLoc   48.8886689    380.032715
## Age         17.7275460    157.846774
## Education    0.5962186     44.598731
## Urban        0.1728373      9.822082
## US           4.2172102     18.073863

It looks like Random Forest increases the MSE. Price, ShelveLoc, and Age are the three most important predictors of Sale.

  1. Now analyze the data using BART, and report your results.
library(BART)
## Warning: package 'BART' was built under R version 4.3.2
## Loading required package: nlme
## Loading required package: nnet
## Loading required package: survival
## 
## Attaching package: 'survival'
## The following object is masked from 'package:caret':
## 
##     cluster
xtrain <- Carseats[train, 2:11]
ytrain <- Carseats[train, "Sales"]
xtest <- Carseats[-train, 2:11]

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: 7s
## trcnt,tecnt: 1000,1000
  1. This problem involves the OJ data set which is part of the ISLR2 package.
  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train = sample(1:nrow(OJ), 800)
OJtrain = OJ[train, ]
OJtest = OJ[-train, ]
  1. 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?
tree.OJ = tree(Purchase ~ ., data = OJtrain)
summary(tree.OJ)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## 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
plot(tree.OJ)
text(tree.OJ, pretty = 0)

5 variables are used here, the training error rate is 0.1588, and there are 9 terminal nodes.

  1. 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.
tree.OJ
## 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 ) *

Line 9 (LoyalCH) has 118 observations. It also shows that branch 9 has a value of LoyalCH < 0.0356415. over 80% of the observations in branch 9 take the value of MM and just under 20% of the observations take the value of CH.

  1. Create a plot of the tree, and interpret the results.
plot(tree.OJ)
text(tree.OJ, pretty = 0)

LoyalCH, SpecialCH, PriceDiff, PctDiscMM, and ListPriceDiff are the most important variables.

  1. 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?
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)
##            
## treeOJ.pred  CH  MM
##          CH 160  38
##          MM   8  64
(38 + 8) / 270
## [1] 0.1703704

The test error rate is 0.1703704.

  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
OJcv = cv.tree(tree.OJ, FUN = prune.misclass)
OJcv
## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 150 150 149 158 172 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJcv$size, OJcv$dev, type = "b", xlab = "Tree Size", ylab = "cv classification error rate")

  1. Which tree size corresponds to the lowest cross-validated classi- fication error rate? The tree size of 7 corresponds to the lowest cross-validated classification error rate.

  2. 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.

prune.OJ=prune.tree(tree.OJ,best=7)
  1. Compare the training error rates between the pruned and un- pruned trees. Which is higher?
summary(tree.OJ)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## 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
summary(prune.OJ)
## 
## Classification tree:
## snip.tree(tree = tree.OJ, nodes = c(10L, 4L))
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "ListPriceDiff" "PctDiscMM"    
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7748 = 614.4 / 793 
## Misclassification error rate: 0.1625 = 130 / 800

The pruned tree has a higher training error rate (0.1625) than the un-pruned tree (0.1588).

  1. Compare the test error rates between the pruned and unpruned trees. Which is higher? The pruned tree has a lower test error rate (0.162963) than the un-pruned tree (0.1703704). The un-pruned tree has a higher test error rate.
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)
##            
## treeOJ.pred  CH  MM
##          CH 160  38
##          MM   8  64
unprunedOJvalerr = (38 + 8) / 270
unprunedOJvalerr
## [1] 0.1703704
pruneOJ.pred = predict(prune.OJ, newdata = OJtest, type = "class")
table(pruneOJ.pred, OJtest$Purchase)
##             
## pruneOJ.pred  CH  MM
##           CH 160  36
##           MM   8  66
prunedOJvalerr = (36 + 8) / 270
prunedOJvalerr
## [1] 0.162963