Assignment 7
Obehi Ikpea
2023-04-20
library(tree)## Warning: package 'tree' was built under R version 4.1.3library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.1.3data("Carseats")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 ˆpm1. The x-axis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
pm1 <- seq(0,1, 0.001)
pm2 <- 1-pm1
E <- 1-pmax(pm1, pm2)
G <- pm1*(1-pm1) + pm2*(1-pm2)
D=-pm1*log(pm1)-pm2*log(pm2)
plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='f')
lines(pm1,G,type='l')
lines(pm1,E,col='blue')
lines(pm1,D,col='red')
legend(x='top',legend=c('gini','class error','cross entropy'),
col=c('black','blue','red'),lty=1,text.width = 0.22)
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.
set.seed(2)
train <- sample(1:nrow(Carseats), 200)
Carseats.test <- Carseats[-train, "Sales"]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.Carseats<- tree(Sales ~., Carseats, subset=train)
summary(tree.Carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats, subset = train)
## Variables actually used in tree construction:
## [1] "Price" "ShelveLoc" "CompPrice" "Age" "Advertising"
## [6] "Population"
## Number of terminal nodes: 14
## Residual mean deviance: 2.602 = 484 / 186
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.71700 -1.08700 -0.01026 0.00000 1.11300 4.06600plot(tree.Carseats)
text(tree.Carseats, pretty=0)
yhat <- predict(tree.Carseats, newdata = Carseats[-train, ])
mean((yhat-Carseats.test)^2)## [1] 4.471569test MSE is 4.47.
- Use cross-validation in order to determine the optimal level ofree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(tree.Carseats)
plot(cv.carseats$size, cv.carseats$dev, type="b")
# optimal level = 11
prune.carseats <- prune.tree(tree.Carseats, best=11)
plot(prune.carseats)
text(prune.carseats, pretty=0)
yhat <- predict(prune.carseats, newdata = Carseats[-train, ] )
mean((yhat-Carseats.test)^2)## [1] 4.644345Test error is 4.64. Pruning did not improve the test error rate.
library(randomForest)## Warning: package 'randomForest' was built under R version 4.1.3## randomForest 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.- 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.
set.seed(1)
bag.carseats <- randomForest(Sales~., data=Carseats, subset=train, mtry=10, importance=TRUE)
bag.carseats##
## Call:
## randomForest(formula = Sales ~ ., data = Carseats, mtry = 10, importance = TRUE, subset = train)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.953352
## % Var explained: 60.36yhat.bag <- predict(bag.carseats, newdata = Carseats[-train, ])
plot(yhat.bag, Carseats.test)
abline(0,1)
mean((yhat.bag-Carseats.test)^2)## [1] 2.555523importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 25.64599599 213.096278
## Income 5.67740403 76.385083
## Advertising 12.22093633 104.986108
## Population 0.59715138 61.294727
## Price 56.50749020 535.237246
## ShelveLoc 41.05022443 283.698935
## Age 9.11485795 118.103039
## Education -0.05758758 39.442533
## Urban 0.79459812 8.828001
## US 1.45384950 7.611851varImpPlot(bag.carseats)
Test mse: 2.56
- 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.
rf.carseats <- randomForest(Sales~., data=Carseats,subset=train, importance=T)
yhat.rf <- predict(rf.carseats, newdata=Carseats[-train, ])
mean((yhat.rf - Carseats.test)^2)## [1] 3.209712importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 10.8825891 157.96141
## Income 4.0892724 132.48555
## Advertising 9.8371786 126.18765
## Population -2.8000757 99.90701
## Price 36.4254627 403.76853
## ShelveLoc 30.4598617 238.61308
## Age 7.2107815 144.57156
## Education -0.0605914 69.02272
## Urban 1.3385164 14.16939
## US 1.6808343 15.63370varImpPlot(rf.carseats)
test error: 3.21 From the plot,we see that Price and Shelveloc are the
two most important variables.
- Now analyze the data using BART, and report your results.
library(BART)## Warning: package 'BART' was built under R version 4.1.3## Loading required package: nlme## Loading required package: nnet## Loading required package: survival## Warning: package 'survival' was built under R version 4.1.3x <- Carseats[, 1:10]
y <- Carseats[, "Sales"]
xtrain <- x[train,]
ytrain <- y[train]
xtest <- x[-train, ]
ytest <- y[-train]
set.seed(1)
bartfit <- gbart(xtrain, ytrain, x.test=xtest)## Warning in summary.lm(lm(y.train ~ ., data.frame(t(x.train), y.train))):
## essentially perfect fit: summary may be unreliable## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 13, 200
## y1,yn: 0.154000, -2.406000
## x1,x[n*p]: 7.500000, 1.000000
## xp1,xp[np*p]: 11.220000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 100 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.287616,3,9.47926e-32,7.346
## *****sigma: 0.000000
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,13,0
## *****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: 4s
## trcnt,tecnt: 1000,1000yhat.bart <- bartfit$yhat.test.mean
mean((ytest-yhat.bart)^2)## [1] 0.1158053The test error = 0.116 is lower than the test errors for the other techniques.
Question 9
- This problem involves the OJ data set which is part of the ISLR2 package.
data(OJ)- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
attach(OJ)
set.seed(2)
trainOJ <- sample(1:nrow(OJ), 800)
OJ.test <- OJ[-trainOJ, ]
Purchase.test <- Purchase[-trainOJ]- 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, subset=trainOJ)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ, subset = trainOJ)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7009 = 554.4 / 791
## Misclassification error rate: 0.1588 = 127 / 800The training error rate is 0.1588, and the tree has 9 terminal nodes.
- 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 1068.00 CH ( 0.61250 0.38750 )
## 2) LoyalCH < 0.5036 359 422.80 MM ( 0.27577 0.72423 )
## 4) LoyalCH < 0.280875 172 127.60 MM ( 0.12209 0.87791 )
## 8) LoyalCH < 0.035047 56 10.03 MM ( 0.01786 0.98214 ) *
## 9) LoyalCH > 0.035047 116 106.60 MM ( 0.17241 0.82759 ) *
## 5) LoyalCH > 0.280875 187 254.10 MM ( 0.41711 0.58289 )
## 10) PriceDiff < 0.05 73 71.36 MM ( 0.19178 0.80822 ) *
## 11) PriceDiff > 0.05 114 156.30 CH ( 0.56140 0.43860 ) *
## 3) LoyalCH > 0.5036 441 311.80 CH ( 0.88662 0.11338 )
## 6) LoyalCH < 0.737888 168 191.10 CH ( 0.74405 0.25595 )
## 12) PriceDiff < 0.265 93 125.00 CH ( 0.60215 0.39785 )
## 24) PriceDiff < -0.35 12 10.81 MM ( 0.16667 0.83333 ) *
## 25) PriceDiff > -0.35 81 103.10 CH ( 0.66667 0.33333 ) *
## 13) PriceDiff > 0.265 75 41.82 CH ( 0.92000 0.08000 ) *
## 7) LoyalCH > 0.737888 273 65.11 CH ( 0.97436 0.02564 )
## 14) PriceDiff < -0.39 11 12.89 CH ( 0.72727 0.27273 ) *
## 15) PriceDiff > -0.39 262 41.40 CH ( 0.98473 0.01527 ) *Node 1: The split criterion is loyalCH< 0.035047. There are 56 observations in the branch. The branch has a deviance of 10.03. The overall prediction for this branch is MM - Minute maid orange juice.
- Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty=0)
Interpretation: 1. Loyal CH is the most important factor in determing Purchase. 2.Customers with loyal CH<0.03 and having the lowest Price diff purchased minute maid. 3.Customers with a CH value higher than the previous and having Price Difference higher than the previous purchased Citrus Hill. 4. Customers having with loyal CH value greater than 0.5 but less than 0.7 and having Price Difference less than -0.35 purchased minute maid. 5. Customers having with loyal CH value greater than 0.5 but less than 0.7 and having Price Difference greater than -0.35 purchased citrus hill. 6. Customers having with loyal CH value greater than 0.7 and having Price Difference either less than or greater than -0.39 purchased citrus hill.
- 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 <- predict(oj.tree, OJ.test, type="class")
table(oj.pred, OJ.test$Purchase)##
## oj.pred CH MM
## CH 148 37
## MM 15 70(148+70)/270## [1] 0.8074074The test error rate is 0.807
- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(2)
cv.oj <- cv.tree(oj.tree, FUN=prune.misclass)
cv.carseats## $size
## [1] 14 13 12 11 10 9 8 7 6 4 3 2 1
##
## $dev
## [1] 944.4136 960.5082 975.5682 969.8304 1018.6218 1085.5631 1114.2876
## [8] 1116.6153 1156.7491 1160.1852 1142.0335 1320.0677 1574.5229
##
## $k
## [1] -Inf 16.92509 19.38585 23.44178 29.89370 36.28493 50.16562
## [8] 54.84825 65.75957 80.79945 90.11022 179.77305 277.78708
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
par(mfrow=c(1,2))
plot(cv.oj$size, cv.oj$dev, type="b")
- Which tree size corresponds to the lowest cross-validated classification error rate?
Ans: Tree size 4.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.prune <- prune.misclass(oj.tree,best=4)
prune.pred <- predict(oj.prune, OJ.test, type="class")
table(prune.pred, OJ.test$Purchase)##
## prune.pred CH MM
## CH 149 42
## MM 14 65summary(oj.prune)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 4:3)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 4
## Residual mean deviance: 0.8381 = 667.1 / 796
## Misclassification error rate: 0.1688 = 135 / 800(149+65)/270## [1] 0.7925926Test error is 0.793
Compare the training error rates between the pruned and unpruned trees. Which is higher? Ans: For unpruned trees: training error was 0.1588 For pruned trees: training error = 0.1688 Pruned tree had a higher training error rate.
Compare the test error rates between the pruned and unpruned trees. Which is higher?
Ans: Pruned: 0.793 Unpruned:0.807
Unpruned trees had a higher test error rate.