Assignment #7
Jared Chappelle
2022-10-30
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. Hint: In a setting with two classes, \(pˆm1 = 1 − pˆm2\). You could make this plot by hand, but it will be much easier to make in R.
p=seq(0,1,0.01)
gini=p*(1-p)*2
class.error=1-pmax(p,1-p)
entropy=-(p*log(p)+(1-p)*log(1-p))
matplot(p,cbind(gini,class.error,entropy),col=c("magenta","navy", "cyan"))
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(ISLR)
library(tree)## Warning: package 'tree' was built under R version 4.2.2attach(Carseats)
set.seed(5)
train=sample(1:nrow(Carseats),nrow(Carseats)/2)
train.carseats=Carseats[train,]
test.carseats=Carseats[-train,](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
#Making our tree
tree.carseats=tree(Sales~.,data=train.carseats)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = train.carseats)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "CompPrice" "Income"
## [6] "Advertising"
## Number of terminal nodes: 20
## Residual mean deviance: 2.184 = 393.1 / 180
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.79500 -1.06000 0.08106 0.00000 0.91510 4.47500#Plotting the tree
plot(tree.carseats)
text(tree.carseats,pretty=0)
#Test MSE
yhat=predict(tree.carseats,newdata=test.carseats)
mean((yhat-test.carseats$Sales)^2)## [1] 5.041825The test MSE for our tree is 5.04!
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
#Pruning
set.seed(5)
cv.carseats=cv.tree(tree.carseats)
plot(cv.carseats$size,cv.carseats$dev,type='b')
prune.carseats=prune.tree(tree.carseats,best=15)
plot(prune.carseats)
text(prune.carseats,pretty=0)
It looks like the best size is 15.
#Test MSE
yhat.prune=predict(prune.carseats,newdata=test.carseats)
mean((yhat.prune-test.carseats$Sales)^2)## [1] 4.805951Now the test MSE is 4.81 so yes test MSE has improved!
(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 4.2.2## randomForest 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.bag.carseats=randomForest(Sales~.,data=train.carseats,mtry=10,ntree=500,importance=TRUE)
yhat.bag=predict(bag.carseats,newdata=test.carseats)
mean((yhat.bag-test.carseats$Sales)^2)## [1] 2.581029importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 24.2063791 193.847782
## Income 7.2214771 144.584129
## Advertising 13.5699320 107.769020
## Population 0.6714347 55.113182
## Price 43.8908481 357.153136
## ShelveLoc 59.1547061 521.377367
## Age 15.2755356 140.369728
## Education 3.2737774 49.289733
## Urban -0.4309196 8.233032
## US 2.5729516 8.930978The test MSE we got was 2.58. The most important variables seem to be ShelveLoc, Price, and CompPrice, with Income and Age getting honorable mentions.
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats=randomForest(Sales~.,data=train.carseats,mtry=5,importance=TRUE)
yhat.rf=predict(rf.carseats,newdata=test.carseats)
mean((yhat.rf-test.carseats$Sales)^2)## [1] 2.760039importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 18.2539268 188.39380
## Income 8.4858735 161.81450
## Advertising 10.7619350 125.93143
## Population -0.6079235 71.10633
## Price 34.9889921 330.31882
## ShelveLoc 50.2360133 459.48512
## Age 13.1500191 152.07561
## Education 1.0709442 58.79071
## Urban -0.9298438 10.64918
## US 3.1920582 13.64481varImpPlot(rf.carseats)
Our test MSE this time is 2.76, so a little worse than it was for Bagging. The most important variables here are once again ShelveLoc, Price, and CompPrice, with the same honorable mentions of Age and Income.
(f) Now analyze the data using BART, and report your results.
#BART
library(BART)## Warning: package 'BART' was built under R version 4.2.2## Loading required package: nlme## Loading required package: nnet## Loading required package: survivalx=Carseats[,1:11]
y=Carseats[,"Sales"]
xtrain=x[train,]
ytrain=y[train]
xtest=x[-train,]
ytest=y[-train]
set.seed(5)
bart.fit=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, 15, 200
## y1,yn: 0.138150, 0.058150
## x1,x[n*p]: 7.520000, 0.000000
## xp1,xp[np*p]: 10.810000, 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: 2,0.95,0.284787,3,7.42266e-31,7.38185
## *****sigma: 0.000000
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,15,0
## *****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: 1000,1000yhat.bart=bart.fit$yhat.test.mean
mean((ytest-yhat.bart)^2)## [1] 0.1274798It seems that using BART, our test MSE was .1275.
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.
detach(Carseats)
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.2.2##
## Attaching package: 'ISLR2'## The following objects are masked from 'package:ISLR':
##
## Auto, Creditattach(OJ)
set.seed(5)
train=sample(dim(OJ)[1],800)
train.OJ=OJ[train,]
test.OJ=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=train.OJ)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train.OJ)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7347 = 581.1 / 791
## Misclassification error rate: 0.1662 = 133 / 800The error rate we got with this tree is .1662 and the number of terminal nodes the tree has is 9. The variables used in this tree were LoyalCH, PriceDiff, and ListPriceDiff.
(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 1068.00 CH ( 0.61250 0.38750 )
## 2) LoyalCH < 0.5036 346 412.40 MM ( 0.28324 0.71676 )
## 4) LoyalCH < 0.280875 164 125.50 MM ( 0.12805 0.87195 )
## 8) LoyalCH < 0.0356415 56 10.03 MM ( 0.01786 0.98214 ) *
## 9) LoyalCH > 0.0356415 108 103.50 MM ( 0.18519 0.81481 ) *
## 5) LoyalCH > 0.280875 182 248.00 MM ( 0.42308 0.57692 )
## 10) PriceDiff < 0.05 71 67.60 MM ( 0.18310 0.81690 ) *
## 11) PriceDiff > 0.05 111 151.30 CH ( 0.57658 0.42342 ) *
## 3) LoyalCH > 0.5036 454 362.00 CH ( 0.86344 0.13656 )
## 6) PriceDiff < -0.39 31 40.32 MM ( 0.35484 0.64516 )
## 12) LoyalCH < 0.638841 10 0.00 MM ( 0.00000 1.00000 ) *
## 13) LoyalCH > 0.638841 21 29.06 CH ( 0.52381 0.47619 ) *
## 7) PriceDiff > -0.39 423 273.70 CH ( 0.90071 0.09929 )
## 14) LoyalCH < 0.705326 135 143.00 CH ( 0.77778 0.22222 )
## 28) ListPriceDiff < 0.255 67 89.49 CH ( 0.61194 0.38806 ) *
## 29) ListPriceDiff > 0.255 68 30.43 CH ( 0.94118 0.05882 ) *
## 15) LoyalCH > 0.705326 288 99.77 CH ( 0.95833 0.04167 ) *I picked terminal node 28, and this node has 67 observations in it. Its splitting point is .255. The branch has a deviance of 89.43, and this branch predicts that Citrus Hill is being purchased. However, this is only correct on 61.19% of the observations.
(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree,pretty=0)
The splitting variable at the top is LoyalCH, which is the loyalty of CH’s customers to the brand. Therefore, it seems that LoyalCH is the most important variable, with PriceDiff and ListPriceDiff being the next most important variables here. The less loyal customers are to Citrus Hill, the more likely they are to purchase Minute Maid. With the more loyal Citrus Hill Customers, it takes a bigger price difference to get them to jump ship.
(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=predict(oj.tree,newdata=test.OJ,type="class")
table(test.OJ$Purchase,oj.pred)## oj.pred
## CH MM
## CH 148 15
## MM 32 75mean(oj.pred!=test.OJ$Purchase)## [1] 0.1740741The test error rate here is 17.41%.
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj=cv.tree(oj.tree,FUN=prune.misclass)
cv.oj## $size
## [1] 9 6 5 3 2 1
##
## $dev
## [1] 155 156 156 172 170 310
##
## $k
## [1] -Inf 0.0 1.0 8.5 9.0 150.0
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size,cv.oj$dev,type="b",xlab="Tree Size",ylab="Deviance")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
The tree size with the lowest CV error rate is a tree that has 9 nodes.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned=prune.misclass(oj.tree,best=9)
plot(oj.pruned)
text(oj.pruned,pretty=0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train.OJ)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7347 = 581.1 / 791
## Misclassification error rate: 0.1662 = 133 / 800The test error rate is the same between the pruned and unpruned trees because the pruned tree is not actually pruned. Using CV, it was determined that the best size was with 9 terminal nodes.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
ojprune.pred=predict(oj.pruned, newdata=test.OJ, type="class")
mean(ojprune.pred != test.OJ$Purchase)## [1] 0.1740741Again, we get the same test error rate because CV found that the unpruned tree was best.