Assignment 7, Chapter 8, Q 3,8,9

Question 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 ˆ pm1, 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`

p = seq(0, 1, 0.05)
gini = p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
class.err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini, entropy, class.err), pch = c(0, 15, 2))

Question 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.

library(ISLR)
library(caret)

## Loading required package: ggplot2

## Loading required package: lattice

library(tree)

## Warning: package 'tree' was built under R version 4.2.3

attach(Carseats)
set.seed(1)

(a) Split the data set into a training set and a test set.

inTrain <- createDataPartition(Carseats$Sales,p=0.5,list=FALSE)
train <- Carseats[inTrain,]
test <- Carseats[-inTrain,]
y.test <- test$Sales

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

carseat.tree <- tree(Sales~.,data=train)
summary(carseat.tree)

## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Income"      "Age"         "CompPrice"  
## [6] "Advertising"
## Number of terminal nodes:  17 
## Residual mean deviance:  2.272 = 418 / 184 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -4.76200 -0.95950  0.08568  0.00000  1.00600  3.55000

plot(carseat.tree)
text(carseat.tree, pretty=0)

yhat <- predict(carseat.tree, newdata=test)
carseat.MSE <- mean((yhat - y.test)^2)

The test MSE obtained was 4.5407647

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

cv.Carseats <- cv.tree(carseat.tree)
plot(cv.Carseats$size, cv.Carseats$dev, type='b')

prune.Carseats <- prune.tree(carseat.tree, best=14)
plot(prune.Carseats)
text(prune.Carseats, pretty=0)

yhat <- predict(prune.Carseats, test)
carseat.MSE2 <- mean((yhat-y.test)^2)

The test MSE is is 4.4820204, which is only 0.0587443 lower than the previous 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)

## Warning: package 'randomForest' was built under R version 4.2.3

## 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

bag.Carseats <- randomForest(Sales ~., data=train, mtry=10, importance=TRUE)

yhat.bag <- predict(bag.Carseats, newdata=test)
bag.MSE <- mean((yhat.bag-y.test)^2)
importance(bag.Carseats)

##                %IncMSE IncNodePurity
## CompPrice   20.0856500    160.997241
## Income      12.4774969     97.736143
## Advertising 15.5139101    121.667479
## Population  -0.7938376     50.286078
## Price       49.7490298    416.931988
## ShelveLoc   47.1510800    363.787790
## Age         16.9282480    140.719245
## Education   -0.5861685     53.488122
## Urban       -1.2541817      6.804677
## US           1.1445297      9.490164

The test MSE found is 2.5933931, which is lower than the previous MSE of 4.4820204

(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.

randforest.Carseats <- randomForest(Sales~., data=train, mtry=floor((ncol(Carseats)-1)/3),importance=TRUE)

yhat.randforest <- predict(randforest.Carseats, newdata=test)
randforest.MSE <- mean((yhat.randforest-y.test)^2)
importance(randforest.Carseats)

##                %IncMSE IncNodePurity
## CompPrice   11.5759305     146.19780
## Income       8.3447205     139.25753
## Advertising 12.8890094     131.73946
## Population   0.2337977     101.64228
## Price       35.7432757     333.97613
## ShelveLoc   33.6579308     274.06395
## Age         11.5117569     149.71564
## Education   -0.5650574      65.44572
## Urban       -3.2490911      12.95234
## US           4.7378338      19.12290

The MSE for random forest is 3.3579941

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

library(BART)

## Warning: package 'BART' was built under R version 4.2.3

## Loading required package: nlme

## Loading required package: nnet

## Loading required package: survival

## 
## Attaching package: 'survival'

## The following object is masked from 'package:caret':
## 
##     cluster

train =  sample(nrow(Carseats), nrow(Carseats) / 2)
test = Carseats[-train, "Sales"]
x <- Carseats[, 2:11]
y <- Carseats[, "Sales"]
xtrain <- x[train,]
ytrain <- y[train]
xtest <- x[test, ]
ytest <- y[test]
bartfit <- gbart(xtrain, ytrain, x.test=xtest)

## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 14, 197
## y1,yn: -3.000450, -1.940450
## x1,x[n*p]: 141.000000, 0.000000
## xp1,xp[np*p]: 132.000000, 0.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 64 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.284787,3,0.206647,7.15045
## *****sigma: 1.029983
## *****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: 3s
## trcnt,tecnt: 1000,1000

yhat.bart = bartfit$yhat.test.mean
BART.MSE <- mean((ytest - yhat.bart)^2)

The MSE with BART is 0.3068485

detach(Carseats)

Question 9

This problem involves the `OJ` data set which is part of the `ISLR2` package.

library(ISLR2)

## 
## Attaching package: 'ISLR2'

## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit

attach(OJ)
set.seed(1)

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

train  <-  sample(1:nrow(OJ), 800)
oj.train  <-  OJ[train,]
oj.test <-  OJ[-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~., 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 training error rate is 0.1588, and the tree is using 9 terminal nodes with 3 variables: LoyalCH, PriceDiff, and DiscMM.

(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 ) *

Node 7 identifies LoyalCH, a terminal node with a splitting value of 0.745, 260 observations below it, and a deviance of 84.77

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

plot(oj.tree)
text(oj.tree, pretty = 0, cex = 0.5)

This tree ends with the same amount of Terminal nodes on both sides (balanced?) with LoyalCH and PriceDiff being the two leading factors

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

oj.pred <-  predict(oj.tree, oj.test, type = "class")
table(oj.test$Purchase, oj.pred)

##     oj.pred
##       CH  MM
##   CH 160   8
##   MM  38  64

OJ.testError <- 1- mean(oj.test$Purchase == oj.pred)

The test error is 0.1703704

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

oj.cv <-  cv.tree(oj.tree, FUN=prune.misclass)
oj.cv

## $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"

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

plot(oj.cv$size, oj.cv$dev, xlab="Tree Size", ylab="Error Rate")

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

4 is the lowest

(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.pruned =  prune.tree(oj.tree, best=4)
oj.pruned

## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
## 1) root 800 1073.0 CH ( 0.60625 0.39375 )  
##   2) LoyalCH < 0.5036 365  441.6 MM ( 0.29315 0.70685 )  
##     4) LoyalCH < 0.280875 177  140.5 MM ( 0.13559 0.86441 ) *
##     5) LoyalCH > 0.280875 188  258.0 MM ( 0.44149 0.55851 ) *
##   3) LoyalCH > 0.5036 435  337.9 CH ( 0.86897 0.13103 )  
##     6) LoyalCH < 0.764572 174  201.0 CH ( 0.73563 0.26437 ) *
##     7) LoyalCH > 0.764572 261   91.2 CH ( 0.95785 0.04215 ) *

plot(oj.pruned)
text(oj.pruned, pretty = 0, cex = 0.5)

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

summary(oj.pruned)

## 
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 4:6)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes:  4 
## Residual mean deviance:  0.8678 = 690.7 / 796 
## Misclassification error rate: 0.205 = 164 / 800

The higher misclassification error rate belongs to oj.pruned at 0.1988; compared to oj.tree at 0.1588.

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

pred.oj.tree <-  predict(oj.tree, oj.test, type="class")
misclass.oj.tree <-  sum(oj.test$Purchase != pred.oj.tree)
misclass.oj.tree/length(pred.oj.tree)

## [1] 0.1703704

pred.oj.pruned <-  predict(oj.pruned, oj.test, type="class")
misclass.oj.pruned <-  sum(oj.test$Purchase != pred.oj.pruned)
misclass.oj.pruned/length(pred.oj.pruned)

## [1] 0.1851852

The higher test error rate belongs to the Pruned tree, at 0.1851852

Assignment 7, Chapter 8, Q 3,8,9

Garrett Garceau

2023-03-28

Question 3

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`

Question 8

(a) Split the data set into a training set and a test set.

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

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

(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.

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

Question 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.

(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?

(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.

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

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

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

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

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

(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.

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

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

Assignment 7, Chapter 8, Q 3,8,9

Garrett Garceau

2023-03-28

Question 3

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

Question 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.

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

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

(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.

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

Question 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.

(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?

(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.

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

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

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

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

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

(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.

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

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

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`

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.

(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.

(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.

This problem involves the `OJ` data set which is part of the `ISLR2` package.

(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?

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