ISLR Q8.9 - Regression Trees/OJ
Question
p334
This problem involves the OJ data set which is part of the ISLR package.
Create a training set containing a random sample of 800 obser- vations, and a test set containing the remaining observations.
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?
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.
Create a plot of the tree, and interpret the results.
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?
Apply the cv.tree() function to the training set in order to determine the optimal tree size.
Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
Which tree size corresponds to the lowest cross-validated classi- fication error rate?
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.
Compare the training error rates between the pruned and un- pruned trees. Which is higher?
Compare the test error rates between the pruned and unpruned trees. Which is higher?
library(ISLR)
library(tree)Similar to Lab: 8.3.1 Fitting Classification Trees
target = OJ$Purchase
Purchase: A factor with levels CH and MM indicating whether the customer purchased Citrus Hill or Minute Maid Orange Juice
Overview
- Build Tree with Training Data: tree.oj
- Predict Training Data Error on unpruned tree
- Prune Tree: prune.oj
- Predict Training Data Error on pruned tree
- Predict Test Data Error on unpruned tree
- Predict Test Data Error on pruned tree
9a
Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
dim(OJ)## [1] 1070 18set.seed(1)
train = sample(1:nrow(OJ), 800)
# Don't actually use these ???
oj.train = OJ[train,]
oj.test = OJ[-train,]
oj.train.y = OJ[train,"Purchase"]
oj.test.y = OJ[-train,"Purchase"]9b Fit Tree to Training
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?
tree.oj = tree(Purchase ~ ., OJ, subset=train)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ, subset = 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 / 800Uses only 5 predictors to split the tree.
Training error rate
Training error rate: Misclassification rate: 15.88% (p324)
How many terminal nodes does the tree have?
terminal nodes: 9
9c
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.
- denotes terminal node
tree.oj## 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 ) *Interpret one terminal node
"8) LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ) *"
"9) LoyalCH > 0.0356415 118 116.40 MM ( 0.19492 0.80508 ) *"Branch 8
59 - Number observations in that branch
10.14 - Deviance
MM - Predicted class
( 0.01695 0.98305 ) - (Prob CH, Prob MM)This branching (8) looks redundant because MM is always chosen.
Branch 9
118 - Number observations in that branch
9d Plot Unpruned Tree
Create a plot of the tree, and interpret the results.
{plot(tree.oj)
text(tree.oj, pretty=0)
}
Interpretation of Results
9 terminal nodes. At least least 1 redundant node.
9e Predict Test Data from Unpruned
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?
Predict Test Response SKIP
tree.pred = predict(tree.oj, oj.test, type="class")Calculate Error Rate of Training Data
yhat = predict(tree.oj, newdata = OJ[-train ,], type = "class")
oj.test.y = OJ[-train, "Purchase"] # Y target vectorConfusion Matrix
table(yhat, oj.test.y)## oj.test.y
## yhat CH MM
## CH 160 38
## MM 8 64Calculate Test Error Rate for unpruned tree
(160+64)/270## [1] 0.8296296summary(OJ$Purchase)## CH MM
## 653 4179f Find Optimal Prune Size
Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj=cv.tree(tree.oj, FUN=prune.misclass)
summary(cv.oj)## Length Class Mode
## size 6 -none- numeric
## dev 6 -none- numeric
## k 6 -none- numeric
## method 1 -none- characternames(cv.oj)## [1] "size" "dev" "k" "method"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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"7 is the optimal number of terminal nodes. Small misclassification error with 149 ($dev=149)
9g Plot Tree Size vs Classification Error Rate
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="Classification Error")
Plot the error rate as a function of both size and k. (p326) Type=“b” means plot both “p” points, “l” lines
par(mfrow=c(1,2))
plot(cv.oj$size, cv.oj$dev, type="b")
plot(cv.oj$k,cv.oj$dev, type="b")
9h
Which tree size corresponds to the lowest cross-validated classification error rate?
7 is the optimal.
9i Prune Training Tree
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.tree(tree.oj, best=7)
{plot(prune.oj)
text(prune.oj, pretty=0)
}
9j Training Error
Compare the training error rates between the pruned and unpruned trees. Which is higher?
train.predict = predict(tree.oj, newdata = oj.train, type="class")
table(oj.train$Purchase, train.predict)## train.predict
## CH MM
## CH 450 35
## MM 92 223(450+223)/800## [1] 0.84125Predict on Prune
train.pruned.predict = predict(prune.oj, newdata = oj.train, type="class")
table(oj.train$Purchase, train.pruned.predict)## train.pruned.predict
## CH MM
## CH 441 44
## MM 86 229table(oj.train$Purchase, train.predict)
(441+229)/800## [1] 0.8375Unpruned is overfitting so it gives a better result.
9k Test Error
Compare the test error rates between the pruned and unpruned trees. Which is higher?
Unpruned error from 9e: 0.8296296
test.pruned.predict = predict(prune.oj, newdata = oj.test, type="class")
table(oj.test$Purchase, test.pruned.predict)## test.pruned.predict
## CH MM
## CH 160 8
## MM 36 66(160 + 66)/270## [1] 0.837037Pruned gives a better test error rate.