Solutions of the exercises from Chapter 8
Conceptual
Q1. Draw an example (of your own invention) of a partition of two-dimensional feature space that could result from recursive binary splitting. Your example should contain at least six regions. Draw a decision tree corresponding to this partition. Be sure to label all aspects of your figures, including the regions \(R_1,R_2,...\), the cutpoints \(t_1,t_2,...\), and so forth.
par(xpd = NA)
plot(NA, NA, type = "n", xlim = c(0,100), ylim = c(0,100), xlab = "X", ylab = "Y")
# t1: x = 40; (40, 0) (40, 100)
lines(x = c(40,40), y = c(0,100))
text(x = 40, y = 108, labels = c("t1"), col = "red")
# t2: y = 75; (0, 75) (40, 75)
lines(x = c(0,40), y = c(75,75))
text(x = -8, y = 75, labels = c("t2"), col = "red")
# t3: x = 75; (75,0) (75, 100)
lines(x = c(75,75), y = c(0,100))
text(x = 75, y = 108, labels = c("t3"), col = "red")
# t4: x = 20; (20,0) (20, 75)
lines(x = c(20,20), y = c(0,75))
text(x = 20, y = 80, labels = c("t4"), col = "red")
# t5: y=25; (75,25) (100,25)
lines(x = c(75,100), y = c(25,25))
text(x = 70, y = 25, labels = c("t5"), col = "red")
text(x = (40+75)/2, y = 50, labels = c("R1"))
text(x = 20, y = (100+75)/2, labels = c("R2"))
text(x = (75+100)/2, y = (100+25)/2, labels = c("R3"))
text(x = (75+100)/2, y = 25/2, labels = c("R4"))
text(x = 30, y = 75/2, labels = c("R5"))
text(x = 10, y = 75/2, labels = c("R6"))
Q2. It is mentioned in Section 8.2.3 that boosting using depth-one trees (or stumps) leads to an additive model : that is, a model of the form \[f(X) = \sum_{j = 1}^p f_j(X_j).\] Explain why this is the case.
We begin by detailing the boost algorithm step by step. At first, we let \(\hat{f}(x) = 0\) and \(r_i = y_i\ \forall i\). At step one, we let \[\hat{f}^1(x) = c_1I(x_1 < t_1) + c_1' = \frac{1}{\lambda}f_1(x_1),\] then \(\hat{f}(x) = \lambda\hat{f}^1(x)\) and \(r_i = y_i - \lambda\hat{f}^1(x_i)\ \forall i\). At step two, we have \[\hat{f}^2(x) = c_2I(x_2 < t_2) + c_2' = \frac{1}{\lambda}f_2(x_2)\] as to maximize the fit to the residuals, another distinct stump must be fit. Then \(\hat{f}(x) = \lambda\hat{f}^1(x) + \lambda\hat{f}^2(x)\) and \(r_i = y_i - \lambda\hat{f}^1(x_i) - \lambda\hat{f}^2(x_i)\ \forall i\). Finally we have \[\hat{f}(x) = \sum_{j = 1}^p f_j(x_j).\]
Q3. Consider the Gini index, classification error, and cross-entropy in a simple setting with two classes. Create a single plot that displays each of these quantities as a function of \(\hat{p}_{m1}\). The \(x\)-axis should display \(\hat{p}_{m1}\), ranging from 0 to 1, and the \(y\)-axis should display the value of the Gini index, classification error, and entropy.
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"))
Q4. This question relates to the plots in Figure 8.12.
- Sketch the tree corresponding to the partition of the predictor space illustrated in the left-hand panel of Figure 8.12. The numbers inside the boxes indicate the mean of \(Y\) within each region.
If \(X_1 \ge 1\) then 5, else if \(X_2 \ge 1\) then 15, else if \(X_1 < 0\) then 3, else if \(X_2 <0\) then 10, else 0.
- Create a diagram similar to the left-hand panel of Figure 8.12, using the tree illustrated in the right-hand panel of the same figure. You should divide up the predictor space into the correct regions, and indicate the mean for each region.
par(xpd = NA)
plot(NA, NA, type = "n", xlim = c(-2, 2), ylim = c(-3, 3), xlab = "X1", ylab = "X2")
# X2 < 1
lines(x = c(-2, 2), y = c(1, 1))
# X1 < 1 with X2 < 1
lines(x = c(1, 1), y = c(-3, 1))
text(x = (-2 + 1)/2, y = -1, labels = c(-1.8))
text(x = 1.5, y = -1, labels = c(0.63))
# X2 < 2 with X2 >= 1
lines(x = c(-2, 2), y = c(2, 2))
text(x = 0, y = 2.5, labels = c(2.49))
# X1 < 0 with X2<2 and X2>=1
lines(x = c(0, 0), y = c(1, 2))
text(x = -1, y = 1.5, labels = c(-1.06))
text(x = 1, y = 1.5, labels = c(0.21))
Q5. Suppose we produce ten bootstrapped samples from a data set containing red and green classes. We then apply a classification tree to each bootstrapped sample and, for a specific value of \(X\), produce 10 estimates of \(P(\text{Class is Red}|X)\) : \[0.1,0.15,0.2,0.2,,0.55,0.6,0.6,0.65,0.7,0.75.\] There are two common ways to combine these results together into a single class prediction. One is the majority vote approach discussed in this chapter. The second approach is to classify based on the average probability. In this example, what is the final classification under each of these two approaches ?
With the majority vote approach, we classify \(X\) as Red as it is the most commonly occurring class among the 10 predictions (6 for Red vs 4 for Green). With the average probability approach, we classify \(X\) as Green as the average of the 10 probabilities is 0.45.
Q6. Provide a detailed explanation of the algorithm that is used to fit a regression tree.
First we do recursive binary splitting on the data. This is a top-down approach where recursively and greedily we find the best single partitioning of the data such that the reduction of RSS is the greatest. This process is applied to each of the split parts seperately until some minimal number of observations is present on each of the leaves.
Then we apply cost complexity pruning of this larger tree formed in step 1 to obtain a sequence of best subtrees as a function of a parameter, \(\alpha\). Each value of \(\alpha\) corresponds to a different subtree which minimizes the equation \[\sum_{m=i}^{|T|}\sum_{i:x_i\in R_m}(y_i - \hat y_{R_m})^2 + \alpha |T|.\] Here \(|T|\) is the number of terminal nodes on the tree. When \(\alpha=0\) we have the original tree, and as \(\alpha\) increases we get a more pruned version of the tree.
Next, using K-fold CV, choose \(\alpha\). For each fold, repeat steps 1 and 2, and then evaluate the MSE as a function of \(\alpha\) on the held out fold. Chose an \(\alpha\) that minimizes the average error.
Given the \(\alpha\) chosen in previous step, return the tree calculated using the formula laid out in step 2 on the entire dataset with that chosen value of \(\alpha\).
Applied
Q7. 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”. Describe the results obtained.
library(MASS)
library(randomForest)## randomForest 4.6-10
## 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.
Q8. 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.
- 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)
Carseats.train <- Carseats[train, ]
Carseats.test <- Carseats[-train, ]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test error rate do you obtain ?
library(tree)
tree.carseats <- tree(Sales ~ ., data = Carseats.train)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Advertising" "Income"
## [6] "CompPrice"
## Number of terminal nodes: 18
## Residual mean deviance: 2.36 = 429.5 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.2570 -1.0360 0.1024 0.0000 0.9301 3.9130plot(tree.carseats)
text(tree.carseats, pretty = 0)
yhat <- predict(tree.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 4.148897We may conclude that the Test MSE is about 4.15.
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test error rate ?
cv.carseats <- cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
tree.min <- which.min(cv.carseats$dev)
points(tree.min, cv.carseats$dev[tree.min], col = "red", cex = 2, pch = 20)
In this case, the tree of size 8 is selected by cross-validation. We now prune the tree to obtain the 8-node tree.
prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 5.09085We may see that pruning the tree increases the Test MSE to 5.1.
- Use the bagging approach in order to analyze this data. What test error rate do you obtain ? Use the “importance()” function to determine which variables are most important.
bag.carseats <- randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500, importance = TRUE)
yhat.bag <- predict(bag.carseats, newdata = Carseats.test)
mean((yhat.bag - Carseats.test$Sales)^2)## [1] 2.604369We may see that bagging decreases the Test MSE to 2.6.
importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 14.4124562 133.731797
## Income 6.5147532 74.346961
## Advertising 15.7607104 117.822651
## Population 0.6031237 60.227867
## Price 57.8206926 514.802084
## ShelveLoc 43.0486065 319.117972
## Age 19.8789659 192.880596
## Education 2.9319161 39.490093
## Urban -3.1300102 8.695529
## US 7.6298722 15.723975We may conclude that “Price” and “ShelveLoc” are the two most important variables.
- 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 = Carseats.train, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(rf.carseats, newdata = Carseats.test)
mean((yhat.rf - Carseats.test$Sales)^2)## [1] 3.296078In this case, with \(m = \sqrt{p}\), we have a Test MSE of 3.3.
importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 7.5233429 127.36625
## Income 4.3612064 119.19152
## Advertising 12.5799388 138.13567
## Population -0.2974474 100.28836
## Price 37.1612032 383.12126
## ShelveLoc 30.3751253 246.19930
## Age 16.0261885 197.44865
## Education 1.7855151 63.87939
## Urban -1.3928225 16.01173
## US 5.6393475 32.85850We may conclude that, in this case also, “Price” and “ShelveLoc” are the two most important variables.
Q9. This problem involves the “OJ” data set which is part of the “ISLR” package.
- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
train <- sample(1:nrow(OJ), 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 except for “Buy” 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 ~ ., data = OJ.train)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7305 = 578.6 / 792
## Misclassification error rate: 0.165 = 132 / 800The fitted tree has 8 terminal nodes and a training error rate of 0.165.
- 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.
tree.oj## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1064.00 CH ( 0.61750 0.38250 )
## 2) LoyalCH < 0.508643 350 409.30 MM ( 0.27143 0.72857 )
## 4) LoyalCH < 0.264232 166 122.10 MM ( 0.12048 0.87952 )
## 8) LoyalCH < 0.0356415 57 10.07 MM ( 0.01754 0.98246 ) *
## 9) LoyalCH > 0.0356415 109 100.90 MM ( 0.17431 0.82569 ) *
## 5) LoyalCH > 0.264232 184 248.80 MM ( 0.40761 0.59239 )
## 10) PriceDiff < 0.195 83 91.66 MM ( 0.24096 0.75904 )
## 20) SpecialCH < 0.5 70 60.89 MM ( 0.15714 0.84286 ) *
## 21) SpecialCH > 0.5 13 16.05 CH ( 0.69231 0.30769 ) *
## 11) PriceDiff > 0.195 101 139.20 CH ( 0.54455 0.45545 ) *
## 3) LoyalCH > 0.508643 450 318.10 CH ( 0.88667 0.11333 )
## 6) LoyalCH < 0.764572 172 188.90 CH ( 0.76163 0.23837 )
## 12) ListPriceDiff < 0.235 70 95.61 CH ( 0.57143 0.42857 ) *
## 13) ListPriceDiff > 0.235 102 69.76 CH ( 0.89216 0.10784 ) *
## 7) LoyalCH > 0.764572 278 86.14 CH ( 0.96403 0.03597 ) *We 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 57 with a deviance of 10.07 and an overall prediction for the branch of MM. Less than 2% of the observations in that branch take the value of CH, and the remaining 98% take the value of MM.
- Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)
We may see that 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”.
- 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(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)##
## tree.pred CH MM
## CH 147 49
## MM 12 621 - (147 + 62) / 270## [1] 0.2259259We may conclude that the test error rate is about 22%.
- Apply the “cv.tree()” function to the training set in order to determine the optimal size tree.
cv.oj <- cv.tree(tree.oj, FUN = prune.misclass)
cv.oj## $size
## [1] 8 5 2 1
##
## $dev
## [1] 156 156 156 306
##
## $k
## [1] -Inf 0.000000 4.666667 160.000000
##
## $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")
- Which tree size corresponds to the lowest cross-validated classification error rate ?
We may see that the 2-node tree is the smallest tree with the lowest classification 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.
prune.oj <- prune.misclass(tree.oj, best = 2)
plot(prune.oj)
text(prune.oj, pretty = 0)
- Compare the training error rates between the pruned and unpruned trees. Which is higher ?
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## Number of terminal nodes: 8
## Residual mean deviance: 0.7305 = 578.6 / 792
## Misclassification error rate: 0.165 = 132 / 800summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(3L, 2L))
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 2
## Residual mean deviance: 0.9115 = 727.4 / 798
## Misclassification error rate: 0.1825 = 146 / 800The misclassification error rate is slightly higher for the pruned tree (0.1825 vs 0.165).
- Compare the test error rates between the pruned and unpruned trees. Which is higher ?
prune.pred <- predict(prune.oj, OJ.test, type = "class")
table(prune.pred, OJ.test$Purchase)##
## prune.pred CH MM
## CH 119 30
## MM 40 811 - (119 + 81) / 270## [1] 0.2592593In this case, the pruning process increased the test error rate to about 26%, but it produced a way more interpretable tree.
Q10. We now use boosting to predict “Salary” in the “Hitters” data set.
- Remove the observations for whom the salary information is unknown, and then log-transform the salaries.
Hitters <- na.omit(Hitters)
Hitters$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 1000 trees for a range of values of the shrinkage parameter \(\lambda\). Produce a plot with different shrinkage values on the \(x\)-axis and the corresponding training set MSE on the \(y\)-axis.
library(gbm)
set.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")
- 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] 0.07943282The minimum test MSE is 0.25, and is obtained for \(\lambda = 0.079\).
- 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 1.9-8fit1 <- 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$Salary)^2)## [1] 0.4570283The test MSE for boosting is lower than for linear regression and ridge regression.
- Which variables appear to be the most important predictors in the boosted model ?
library(gbm)## Loading required package: survival
## Loading required package: lattice
## Loading required package: splines
## Loading required package: parallel
## Loaded gbm 2.1boost.hitters <- gbm(Salary ~ ., data = Hitters.train, distribution = "gaussian", n.trees = 1000, shrinkage = lambdas[which.min(test.err)])
summary(boost.hitters)
## var rel.inf
## CAtBat CAtBat 23.6278225
## Walks Walks 7.4858538
## PutOuts PutOuts 7.4438741
## CWalks CWalks 7.3437919
## CRBI CRBI 6.7581127
## Years Years 6.0435181
## CRuns CRuns 5.8389531
## Assists Assists 5.2910978
## CHmRun CHmRun 4.9321654
## Hits Hits 4.4338073
## CHits CHits 4.4008947
## RBI RBI 4.2852503
## HmRun HmRun 3.2708823
## AtBat AtBat 3.1569920
## Errors Errors 2.6144780
## Runs Runs 1.8095817
## Division Division 0.6302155
## NewLeague NewLeague 0.5079414
## League League 0.1247675We may see that “CAtBat” is by far the most important variable.
- Now apply bagging to the training set. What is the test set MSE for this approach ?
set.seed(1)
bag.hitters <- randomForest(Salary ~ ., data = Hitters.train, mtry = 19, ntree = 500)
yhat.bag <- predict(bag.hitters, newdata = Hitters.test)
mean((yhat.bag - Hitters.test$Salary)^2)## [1] 0.2313593The test MSE for bagging is 0.23, which is slightly lower than the test MSE for boosting.
Q11. This question uses the “Caravan” data set.
- Create a training set consisting of the first 1000 observations, and a test set consisting of the remaining observations.
set.seed(1)
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 1000 trees, and a shrinkage value of 0.01. Which predictors appear to be 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, y, offset = offset, distribution = distribution, w =
## w, : variable 50: PVRAAUT has no variation.## Warning in gbm.fit(x, y, offset = offset, distribution = distribution, w =
## w, : 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.
- Use the boosting model to predict the response on the test data. Predict that a person will make a purchase if the estimated probability of purchase is greater than 20%. Form a confusion matrix. 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$Purchase, pred.test)## pred.test
## 0 1
## 0 4493 40
## 1 278 11For boosting, the fraction of people predicted to make a purchase that in fact make one is 0.2156863.
logit.caravan <- glm(Purchase ~ ., data = Caravan.train, family = "binomial")## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs.test2 <- predict(logit.caravan, Caravan.test, type = "response")## Warning in predict.lm(object, newdata, se.fit, scale = 1, type =
## ifelse(type == : prediction from a rank-deficient fit may be misleadingpred.test2 <- ifelse(probs.test > 0.2, 1, 0)
table(Caravan.test$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.
Q12. Apply boosting, bagging, and random forests to a data set of your choice. Be sure to fit models on a training set and to evaluate their performance on a test set. How accurate are the results compared to simple methods like linear or logistic regression ? Which of these approaches yields the best performance ?
We will use the “Weekly” data set from the “ISLR” package to predict the “Direction” variable.
set.seed(1)
train <- sample(nrow(Weekly), nrow(Weekly) / 2)
Weekly$Direction <- ifelse(Weekly$Direction == "Up", 1, 0)
Weekly.train <- Weekly[train, ]
Weekly.test <- Weekly[-train, ]We begin with logistic regression.
logit.fit <- glm(Direction ~ . - Year - Today, data = Weekly.train, family = "binomial")
logit.probs <- predict(logit.fit, newdata = Weekly.test, type = "response")
logit.pred <- ifelse(logit.probs > 0.5, 1, 0)
table(Weekly.test$Direction, logit.pred)## logit.pred
## 0 1
## 0 11 244
## 1 8 282We have a classification error of 0.4623853. We continue with boosting.
boost.fit <- gbm(Direction ~ . - Year - Today, data = Weekly.train, distribution = "bernoulli", n.trees = 5000)
boost.probs <- predict(boost.fit, newdata = Weekly.test, n.trees = 5000)
boost.pred <- ifelse(boost.probs > 0.5, 1, 0)
table(Weekly.test$Direction, boost.pred)## boost.pred
## 0 1
## 0 166 89
## 1 181 109We have a classification error of 0.4954128. We continue with bagging.
bag.fit <- randomForest(Direction ~ . - Year - Today, data = Weekly.train, mtry = 6)## Warning in randomForest.default(m, y, ...): The response has five or fewer
## unique values. Are you sure you want to do regression?bag.probs <- predict(bag.fit, newdata = Weekly.test)
bag.pred <- ifelse(bag.probs > 0.5, 1, 0)
table(Weekly.test$Direction, bag.pred)## bag.pred
## 0 1
## 0 85 170
## 1 71 219We have a classification error of 0.7137615. We end with random forests.
rf.fit <- randomForest(Direction ~ . - Year - Today, data = Weekly.train, mtry = 2)## Warning in randomForest.default(m, y, ...): The response has five or fewer
## unique values. Are you sure you want to do regression?rf.probs <- predict(rf.fit, newdata = Weekly.test)
rf.pred <- ifelse(rf.probs > 0.5, 1, 0)
table(Weekly.test$Direction, rf.pred)## rf.pred
## 0 1
## 0 69 186
## 1 62 228We have a classification error of 0.4550459.
We may conclude that random forests gave the lowest classification error.