Assignment

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 ˆpm1. The xaxis should display ˆpm1, 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, ˆpm1 = 1− ˆpm2. You could make this plot by hand, but it will be much easier to make in R.

p=seq(0,1,0.01)

gini= 2*p*(1-p)
classerror= 1-pmax(p,1-p)
crossentropy= -(p*log(p)+(1-p)*log(1-p))

plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='f')

lines(p,gini,type='l')
lines(p,classerror,col='blue')
lines(p,crossentropy,col='red')

legend(x='top',legend=c('gini','class error','cross entropy'),
       col=c('black','blue','red'),lty=1,text.width = 0.22)

Problem 8

library(ISLR)
library(tree)
library(randomForest)
library(knitr)
library(kableExtra)

data("trees")

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.

set.seed(1)
train=sample(1:nrow(Carseats),nrow(Carseats)/2)

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

tree.carseats <- tree(formula=Sales~.,data=Carseats,subset = train)
tree.pred=predict(tree.carseats,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)

## [1] 4.922039

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

From the tree we can see that ShelveLoc and price are the two most important factors when predicting car seat sales. This is because they are at the top of the tree. The MSE we get from this chart 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)
tree.carseats.cv=cv.tree(tree.carseats) 
plot(tree.carseats.cv)

prune.carseats=prune.tree(tree.carseats,best=5)
plot(prune.carseats)
text(prune.carseats)

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

## [1] 5.186482

plot(tree.pred,Carseats[-train,'Sales'],xlab='prediction',ylab='actual')
abline(0,1)

From this test, you can see that the MSE does not improve.

(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.

d=ncol(Carseats)-1
set.seed(1)
carseats.rf=randomForest(Sales~.,data=Carseats,subset=train,mtry=d,importance=T,ntree=100)
tree.pred=predict(carseats.rf,Carseats[-train,])
mean((tree.pred-Carseats[-train,'Sales'])^2)

## [1] 2.616711

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

Here we can see that Price is the most important factor in predicting 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.

mse=c()

set.seed(1)
for(i in 3:10){
  carseats.rf=randomForest(Sales~.,data=Carseats,subset=train,mtry=5,importance=T,ntree=100)
  tree.pred=predict(carseats.rf,Carseats[-train,])
  mse=rbind(mse,mean((tree.pred-Carseats[-train,'Sales'])^2))}
plot(3:10,mse,type='b')

set.seed(42)
carseats.rf=randomForest(Sales~.,data=Carseats,subset=train,mtry=9,importance=T,ntree=100)
plot(carseats.rf)

kable(importance(carseats.rf))

	%IncMSE	IncNodePurity
CompPrice	11.9946846	164.902885
Income	1.3876551	96.815322
Advertising	5.9451314	98.013249
Population	-1.6967648	59.503288
Price	26.5838588	504.262677
ShelveLoc	22.1863369	369.834090
Age	6.5832931	164.935171
Education	0.8204575	44.799656
Urban	1.3786474	6.745175
US	-0.4019542	16.151430

Again the top two predictors are Price and Shelveloc. With only 9 predictors we can see that each tree gets lower training MSE and a higher test MSE.

Problem 9

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

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

data(OJ)
set.seed(1)
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~.,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 training error size is 15.88% and there are 9 terminal nodes. We have 5 predictors used in splits.

(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 < 0.196197 55   73.14 CH ( 0.61818 0.38182 ) *
##         25) PctDiscMM > 0.196197 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 ) *

Using node 11, PriceDiff > 0.05 109 147.00 CH ( 0.59633 0.40367 ) “” meaning that marks it as a terminal node. There were 109 observations.

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

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

LoyalCH is hands down the most important predictor variable. Citrus Hill(CH) and MinuteMaid(MM) are the lowest. The left shows that there is lower customer loyalty for CH vs MM. On the right we see that if MM was at a lower price, it would be chosen over CH.

(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?

OJ.pred.train=predict(OJ.tree,OJ.train,type = 'class')
kable(table(OJ.train[,'Purchase'],OJ.pred.train))

	CH	MM
CH	450	35
MM	92	223

kable(table(OJ.train[,'Purchase'],OJ.pred.train)/nrow(OJ.train))

	CH	MM
CH	0.5625	0.04375
MM	0.1150	0.27875

OJ.pred.test=predict(OJ.tree,OJ.test,type = 'class')
kable(table(OJ.test[,'Purchase'],OJ.pred.test))

	CH	MM
CH	160	8
MM	38	64

kable(table(OJ.test[,'Purchase'],OJ.pred.test)/nrow(OJ.test))

	CH	MM
CH	0.5925926	0.0296296
MM	0.1407407	0.2370370

The test error rate is around 23.70%

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

set.seed(1)
OJ.tree.cv=cv.tree(OJ.tree,K = 10,FUN = prune.misclass)
plot(OJ.tree.cv)

The most optimal size would be 2.

(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

Jesus Gonzalez

2022-05-02

Problem 3

Problem 8

Problem 9