Chapter 8
Applied Exercises
- In the lab, we applied random forests to the Boston data using mtry=6 and using ntree=25 and ntree=500. Create a plot displaying the test error resulting from random forests on this data set for a more comprehensive range of values for mtry and ntree. You can model your plot after Figure 8.10. Describe the results obtained.
library(MASS)
library(randomForest)## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
train <- sample(1:nrow(Boston), nrow(Boston) / 2)
Boston.train <- Boston[train, -14]
Boston.test <- Boston[-train, -14]
Y.train <- Boston[train, 14]
Y.test <- Boston[-train, 14]
rf.boston1 <- randomForest(Boston.train, y = Y.train, xtest = Boston.test, ytest = Y.test, mtry = ncol(Boston) - 1, ntree = 500)
rf.boston2 <- randomForest(Boston.train, y = Y.train, xtest = Boston.test, ytest = Y.test, mtry = (ncol(Boston) - 1) / 2, ntree = 500)
rf.boston3 <- randomForest(Boston.train, y = Y.train, xtest = Boston.test, ytest = Y.test, mtry = sqrt(ncol(Boston) - 1), ntree = 500)
plot(1:500, rf.boston1$test$mse, col = "green", type = "l", xlab = "Number of Trees", ylab = "Test MSE", ylim = c(10, 19))
lines(1:500, rf.boston2$test$mse, col = "red", type = "l")
lines(1:500, rf.boston3$test$mse, col = "blue", type = "l")
legend("topright", c("m = p", "m = p/2", "m = sqrt(p)"), col = c("green", "red", "blue"), cex = 1, lty = 1)
We may see that the Test MSE is very high for a single tree, it decreases as the number of trees increases. Also the Test MSE for all predictors is higher than for half the predictors or the square root of the number of predictors. ## 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)
set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats) / 2)
Car.train = Carseats[train, ]
Car.test = Carseats[-train,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)
reg.tree = tree(Sales~.,data = Carseats, subset=train)
reg.tree = tree(Sales~.,data = Car.train)
#Both above formulas outcome the same result.
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 =0)
yhat = predict(reg.tree,newdata = Car.test)
mean((yhat - Car.test$Sales)^2)## [1] 4.922039We conclude that the Test MSE is about 4.9. c. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.car = cv.tree(reg.tree)
plot(cv.car$size, cv.car$dev, type = "b")
In this case, the tree of size 8 is selected by cross-validation. We now prune the tree to obtain the 8-node tree.
prune.car = prune.tree(reg.tree, best = 8)
plot(prune.car)
text(prune.car,pretty=0)
yhat=predict(prune.car, newdata= Car.test)
mean((yhat-Car.test$Sales)^2)## [1] 5.113254We see that pruning the tree increases the Test MSE to 5.1. 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)
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$Sales)^2)## [1] 2.605253importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863varImpPlot(bag.car)
The most important variables are the price that company charges for car seats at each site and the quality of the shelving location for the car seats at each site. The test MSE associated with the bagegd regression tree iss 2.55, almost half that obtained using an optimally-pruned single tree. e. Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
library(randomForest)
set.seed(1)
rf.car = randomForest(Sales~.,data=Car.train,mtry = 3, importance = TRUE)
yhat.rf = predict(rf.car,newdata=Car.test)
mean((yhat.rf-Car.test$Sales)^2)## [1] 2.960559The test set MSE is 2.9; this indicates that random forests doesn’t yield an improvement over bagging in this case. ## 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.
library(ISLR)
set.seed(1)
train = sample(dim(OJ)[1],800)
OJ.train = OJ[train,]
OJ.test = OJ[-train,]- 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 fitted tree has 8 terminal nodes and a training error rate of 0.1588. 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 ) *I pick the node labelled 8, which is a terminal node because of the asterisk. The split criterion is LoyalCH < 0.035, the number of observations in that branch is 59 with a deviance of 10.14 and an overall prediction for the branch of MM. Less than x% of the observations in that branch take the value of CH, and the remaining x% take the value of MM. d. Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree,pretty=TRUE)
The most important indicator of “Purchase” appears to be “LoyalCH”, since the first branch differentiates the intensity of customer brand loyalty to CH. In fact, the top three nodes contain “LoyalCH”. 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(160+64)/270## [1] 0.829629683% of the test observations are correctly classified so the test error rate is 17%. 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$dev,type='b', xlab = "Tree size", ylab = "Deviance")
h. Which tree size corresponds to the lowest cross-validated classification error rate? We might see that the 5-node tree is the smallest tree with the lowest classification error rate. i. Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ = prune.misclass(OJ.tree, best=5)
plot(prune.OJ)
text(prune.OJ,pretty=0)
j. Compare the training error rates between the pruned and unpruned trees. Which is higher?
tree.pred = predict(prune.OJ, newdata = OJ.test, type = "class")
table(tree.pred,OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 36
## MM 8 66(160+66)/270## [1] 0.837037In this case, the pruning has the same test error, but it produced a way more interpretable tree. ## 10. We now use boosting to predict Salary in the Hitters data set. a. Remove the observations for whom the salary information is unknown, and then log-transform the salaries.
Hitters = na.omit(Hitters)
Hitters$Salary = log(Hitters$Salary)- Create a training set consisting of the first 200 observations, and a test set consisting of the remaining observations.
train = 1:200
hitters.train = Hitters[train,]
hitters.test = Hitters[-train,]- Perform boosting on the training set with 1,000 trees for a range of values of the shrinkage parameter λ. Produce a plot with different shrinkage values on the x-axis and the corresponding training set MSE on the y-axis.
library(gbm)## Loaded gbm 2.1.8set.seed(1)
pows = seq(-10, -0.2, by = 0.1)
lambdas = 10^pows
train.err = rep(NA, length(lambdas))
for (i in 1:length(lambdas)) {
boost.hitters = gbm(Salary ~ ., data = hitters.train, distribution = "gaussian", n.trees = 1000, shrinkage = lambdas[i])
pred.train = predict(boost.hitters, hitters.train, n.trees = 1000)
train.err[i] = mean((pred.train - hitters.train$Salary)^2)
}
plot(lambdas, train.err, type = "b", xlab = "Shrinkage values", ylab = "Training MSE")
d. Produce a plot with different shrinkage values on the x-axis and the corresponding test set MSE on the y-axis.
set.seed(1)
test.err <- rep(NA, length(lambdas))
for (i in 1:length(lambdas)) {
boost.hitters = gbm(Salary ~ ., data = hitters.train, distribution = "gaussian", n.trees = 1000, shrinkage = lambdas[i])
yhat = predict(boost.hitters, hitters.test, n.trees = 1000)
test.err[i] = mean((yhat - hitters.test$Salary)^2)
}
plot(lambdas, test.err, type = "b", xlab = "Shrinkage values", ylab = "Test MSE")
min(test.err)## [1] 0.2540265lambdas[which.min(test.err)]## [1] 0.07943282e.Compare the test MSE of boosting to the test MSE that results from applying two of the regression approaches seen in Chapters 3 and 6.
library(glmnet)## Loading required package: Matrix## Loaded glmnet 4.0-2fit1 = lm(Salary ~ ., data = hitters.train)
pred1 = predict(fit1, hitters.test)
mean((pred1 - hitters.test$Salary)^2)## [1] 0.4917959x = model.matrix(Salary ~ ., data = hitters.train)
x.test = model.matrix(Salary ~ ., data = hitters.test)
y = hitters.train$Salary
fit2 = glmnet(x, y, alpha = 0)
pred2 = predict(fit2, s = 0.01, newx = x.test)
mean((pred2 - hitters.test$Salary)^2)## [1] 0.4570283The test MSE for boosting is lower than for linear regression and ridge regression. f. Which variables appear to be the most important predictors in the boosted model?
boost.hitters <- gbm(Salary ~ ., data = hitters.train, distribution = "gaussian", n.trees = 1000, shrinkage = lambdas[which.min(test.err)])
summary(boost.hitters)
## var rel.inf
## CAtBat CAtBat 20.8404970
## CRBI CRBI 12.3158959
## Walks Walks 7.4186037
## PutOuts PutOuts 7.1958539
## Years Years 6.3104535
## CWalks CWalks 6.0221656
## CHmRun CHmRun 5.7759763
## CHits CHits 4.8914360
## AtBat AtBat 4.2187460
## RBI RBI 4.0812410
## Hits Hits 4.0117255
## Assists Assists 3.8786634
## HmRun HmRun 3.6386178
## CRuns CRuns 3.3230296
## Errors Errors 2.6369128
## Runs Runs 2.2048386
## Division Division 0.5347342
## NewLeague NewLeague 0.4943540
## League League 0.2062551“CAtBat” is by far the most important variable. g. Now apply bagging to the training set. What is the test set MSE for this approach?
set.seed(1)
bag.hitters <- randomForest(Salary ~ ., data = hitters.train, mtry = 19, ntree = 500)
yhat.bag <- predict(bag.hitters, newdata = hitters.test)
mean((yhat.bag - hitters.test$Salary)^2)## [1] 0.2299324The test MSE for bagging is 0.23, which is slightly lower than the test MSE for boosting. ## 11. This question uses the Caravan data set. a. Create a training set consisting of the first 1,000 observations, and a test set consisting of the remaining observations.
train = 1:1000
Caravan$Purchase = ifelse(Caravan$Purchase == "Yes", 1, 0)
Caravan.train = Caravan[train,]
Caravan.test = Caravan[-train,]- Fit a boosting model to the training set with Purchase as the response and the other variables as predictors. Use 1,000 trees, and a shrinkage value of 0.01. Which predictors appear to be the most important?
set.seed(1)
boost.caravan = gbm(Purchase ~ ., data = Caravan.train, distribution = "gaussian", n.trees = 1000, shrinkage = 0.01)## Warning in gbm.fit(x = x, y = y, offset = offset, distribution = distribution, :
## variable 50: PVRAAUT has no variation.## Warning in gbm.fit(x = x, y = y, offset = offset, distribution = distribution, :
## variable 71: AVRAAUT has no variation.summary(boost.caravan)
## var rel.inf
## PPERSAUT PPERSAUT 13.51824557
## MKOOPKLA MKOOPKLA 10.24062778
## MOPLHOOG MOPLHOOG 7.32689780
## MBERMIDD MBERMIDD 6.35820558
## PBRAND PBRAND 4.98826360
## ABRAND ABRAND 4.54504653
## MGODGE MGODGE 4.26496875
## MINK3045 MINK3045 4.13253907
## PWAPART PWAPART 3.15612877
## MAUT1 MAUT1 2.76929763
## MOSTYPE MOSTYPE 2.56937935
## MAUT2 MAUT2 1.99879666
## MSKA MSKA 1.94618539
## MBERARBG MBERARBG 1.89917331
## PBYSTAND PBYSTAND 1.88591514
## MINKGEM MINKGEM 1.87131472
## MGODOV MGODOV 1.81673309
## MGODPR MGODPR 1.80814745
## MFWEKIND MFWEKIND 1.67884570
## MSKC MSKC 1.65075962
## MBERHOOG MBERHOOG 1.53559951
## MSKB1 MSKB1 1.43339514
## MOPLMIDD MOPLMIDD 1.10617074
## MHHUUR MHHUUR 1.09608784
## MRELGE MRELGE 1.09039794
## MINK7512 MINK7512 1.08772012
## MZFONDS MZFONDS 1.08427551
## MGODRK MGODRK 1.03126657
## MINK4575 MINK4575 1.02492795
## MZPART MZPART 0.98536712
## MRELOV MRELOV 0.80356854
## MFGEKIND MFGEKIND 0.80335689
## MBERARBO MBERARBO 0.60909852
## APERSAUT APERSAUT 0.56707821
## MGEMOMV MGEMOMV 0.55589456
## MOSHOOFD MOSHOOFD 0.55498375
## MAUT0 MAUT0 0.54748481
## PMOTSCO PMOTSCO 0.43362597
## MSKB2 MSKB2 0.43075446
## MSKD MSKD 0.42751490
## MINK123M MINK123M 0.40920707
## MINKM30 MINKM30 0.36996576
## MHKOOP MHKOOP 0.34941518
## MBERBOER MBERBOER 0.28967068
## MFALLEEN MFALLEEN 0.28877552
## MGEMLEEF MGEMLEEF 0.20084195
## MOPLLAAG MOPLLAAG 0.15750616
## MBERZELF MBERZELF 0.11203381
## PLEVEN PLEVEN 0.11030994
## MRELSA MRELSA 0.04500507
## MAANTHUI MAANTHUI 0.03322830
## PWABEDR PWABEDR 0.00000000
## PWALAND PWALAND 0.00000000
## PBESAUT PBESAUT 0.00000000
## PVRAAUT PVRAAUT 0.00000000
## PAANHANG PAANHANG 0.00000000
## PTRACTOR PTRACTOR 0.00000000
## PWERKT PWERKT 0.00000000
## PBROM PBROM 0.00000000
## PPERSONG PPERSONG 0.00000000
## PGEZONG PGEZONG 0.00000000
## PWAOREG PWAOREG 0.00000000
## PZEILPL PZEILPL 0.00000000
## PPLEZIER PPLEZIER 0.00000000
## PFIETS PFIETS 0.00000000
## PINBOED PINBOED 0.00000000
## AWAPART AWAPART 0.00000000
## AWABEDR AWABEDR 0.00000000
## AWALAND AWALAND 0.00000000
## ABESAUT ABESAUT 0.00000000
## AMOTSCO AMOTSCO 0.00000000
## AVRAAUT AVRAAUT 0.00000000
## AAANHANG AAANHANG 0.00000000
## ATRACTOR ATRACTOR 0.00000000
## AWERKT AWERKT 0.00000000
## ABROM ABROM 0.00000000
## ALEVEN ALEVEN 0.00000000
## APERSONG APERSONG 0.00000000
## AGEZONG AGEZONG 0.00000000
## AWAOREG AWAOREG 0.00000000
## AZEILPL AZEILPL 0.00000000
## APLEZIER APLEZIER 0.00000000
## AFIETS AFIETS 0.00000000
## AINBOED AINBOED 0.00000000
## ABYSTAND ABYSTAND 0.00000000The variables “PPERSAUT” and “MKOOPKLA” are the two most important variables. c. Use the boosting model to predict the response on the test data. Predict that a person will make a purchase if the estimated prob- ability of purchase is greater than 20 %. Form a confusion ma- trix. What fraction of the people predicted to make a purchase do in fact make one? How does this compare with the results obtained from applying KNN or logistic regression to this data set?
probs.test <- predict(boost.caravan, Caravan.test, n.trees = 1000, type = "response")
pred.test <- ifelse(probs.test > 0.2, 1, 0)
table(Caravan.test$Purchase, pred.test)## pred.test
## 0 1
## 0 4493 40
## 1 278 11For boosting, the fraction of people predicted to make a purchase that in fact make one is 0.2156863.
logit.caravan <- glm(Purchase ~ ., data = Caravan.train, family = "binomial")## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs.test2 <- predict(logit.caravan, Caravan.test, type = "response")## Warning in predict.lm(object, newdata, se.fit, scale = 1, type = if (type == :
## prediction from a rank-deficient fit may be misleadingpred.test2 <- ifelse(probs.test > 0.2, 1, 0)
table(Caravan.test$Purchase, pred.test2)## pred.test2
## 0 1
## 0 4493 40
## 1 278 11For logistic regression, the fraction of people predicted to make a purchase that in fact make one is again 0.2156863.