Chapter 8
Brittany Parffrey
4/28/2021
Chapter 08 (page 332): 3, 8, 9
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 ˆ pm1. Thexaxis 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, ˆ pm1 =1− ˆ pm2. You could make this plot by hand, but it will be much easier to make in R.
p <- seq(0, 1, 0.01)
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), col = c("red", "green", "blue"))
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)## Warning: package 'ISLR' was built under R version 4.0.3set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats) / 2)
Car.train = Carseats[train, ]
Car.test = Carseats[-train,]B.Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)## Warning: package 'tree' was built under R version 4.0.5reg.tree = tree(Sales~.,data = Carseats, subset=train)
reg.tree = tree(Sales~.,data = Car.train)
summary(reg.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = Car.train)
## 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 4.09900plot(reg.tree)
text(reg.tree, pretty = .01)
yhat = predict(reg.tree, newdata = Car.test)
mean((yhat - Car.test$Sales)^2)## [1] 4.922039MSE = 4.148897
C.Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.car = cv.tree(reg.tree)
plot(cv.car$size, cv.car$dev, type = "b")
The optimal level of Tree Complexity appears to be 8.
prune.car = prune.tree(reg.tree, best = 8)
plot(prune.car)
text(prune.car,pretty=0)
yhat=predict(prune.car, newdata= Car.test)
mean((yhat-Car.test$Sales)^2)## [1] 5.113254MSE = 5.09085 which increased due to pruning
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 4.0.5## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
bag.car = randomForest(Sales~.,data=Car.train,mtry = 10, importance = TRUE)
yhat.bag = predict(bag.car,newdata=Car.test)
mean((yhat.bag-Car.test$Sales)^2)## [1] 2.605253MSE = 2.614642
importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863varImpPlot(bag.car)
Price and ShelveLoc are the most important variables shown.
E. Use random forests to analyze this data. What test error rate do you obtain ? Use the “importance()” function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats <- randomForest(Sales ~ ., data = Car.train, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(rf.carseats, newdata = Car.test)
mean((yhat.rf - Car.test$Sales)^2)## [1] 3.054306MSE = 3.267744
importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 12.9540442 157.53376
## Income 2.1683293 129.18612
## Advertising 8.7289900 111.38250
## Population -2.5290493 102.78681
## Price 33.9482500 393.61313
## ShelveLoc 34.1358807 289.28756
## Age 12.0804387 172.03776
## Education 0.2213600 72.02479
## Urban 0.9793293 14.73763
## US 4.1072742 33.91622varImpPlot(rf.carseats)
It appears that Price and ShelveLoc are also the most important variables here as well.
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.
attach(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 / 800The training error rate is 0.165 and There are 8 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 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 ) *The split criterion is given 1st with its threshold value. The observations within the branch is followed along with its deviance and the overall prediction of the branch.In the (__) at the end shows the percent of observations within the branch that fall under each of the classifiers. For example, Node 7 : LoyalCH > 0.764572 is the split critia 278 Observations within node 7 w/ a deviance of 86.14 CH is the predicted of node 7 96% for CH and 03% for MM
D. Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree)
The 1st split we see is LoyalCH < 0.508643 The next split goes further with LoyalCH. LoyalCH is of high importance when predicting Purchase.
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?
tree.pred = predict(OJ.tree, newdata = OJ.test, type = "class")
table(tree.pred,OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 64(147+62)/270## [1] 0.774074177% of the test data are correct resulting in a test erro rate of 23%.
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 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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.OJ$size,cv.OJ$dev,type='b', xlab = "Tree size", ylab = "Deviance")
H. Which tree size corresponds to the lowest cross-validated classification error rate? 5 is the lowest tree size
I. Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ = prune.misclass(OJ.tree, best=5)
plot(prune.OJ)
text(prune.OJ,pretty=0)
J. Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(OJ.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.OJ)##
## Classification tree:
## snip.tree(tree = OJ.tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The Error Rate is the same however the simpler model is more desirable showing that the prune.OJ model is just as good as the 8 node containing model.
K. Compare the test error rates between the pruned and unpruned trees. Which is higher?
tree.pred = predict(prune.OJ, newdata = OJ.test, type = "class")
table(tree.pred,OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 36
## MM 8 66(147+62)/270## [1] 0.774074177% Error rate, same as the 8 node tree but the prune tree is simpler and more interpertable