Exercise 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 pˆm1.The x-axis should display pˆm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.
p1 = seq(0, 1, 0.001)
p2 = 1 - p1
gini.index = 2 * p1 * p2
class.error = 1 - pmax(p1, p2)
cross.entropy = - (p1 * log(p1) + p2 * log(p2))
matplot(p1, cbind(gini.index, class.error, cross.entropy), col = c("lightcoral", "gray52", "lightsteelblue"))
Exercise 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.
attach(Carseats)
set.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 inter-pret the results. What test MSE do you obtain?
reg.tree = tree(Sales~., data = Car.train)
plot(reg.tree)
text(reg.tree ,pretty =0, cex = .50)
##
## 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 4.09900
test_pred = predict(reg.tree, Car.test)
tree_mse = mean((test_pred - Car.test$Sales)^2)
phrase = "We conclude that the test MSE is approximately: "
paste(phrase, tree_mse)
## [1] "We conclude that the test MSE is approximately: 4.92203909112416"
- (c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.car = cv.tree(reg.tree)
plot(cv.car$size, cv.car$dev, type = "b")
best_tree_size = cv.car$size[which.min(cv.car$dev)]
phrase = "The best tree size is: "
paste(phrase, best_tree_size)
## [1] "The best tree size is: 18"
- We now prune the tree to obtain the 18-node tree. We see that pruning did not improve the test MSE. 4.92203909112416 vs 4.92203909112416
prune.car = prune.tree(reg.tree, best = 18)
plot(prune.car)
text(prune.car,pretty=0, cex = .50)
prune_pred = predict(prune.car, Car.test)
prune_mse = mean((prune_pred - Car.test$Sales)^2)
phrase = "We conclude that the test MSE is approximately: "
paste(phrase, prune_mse)
## [1] "We conclude that the test MSE is approximately: 4.92203909112416"
- (d) Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the
importance() function to de-termine which variables are most important.
set.seed(1)
bag.car = randomForest(Sales~.,data = Car.train, mtry = 10, importance = T)
test_pred = predict(bag.car, Car.test)
bag_MSE = mean((test_pred - Car.test$Sales)^2)
phrase = "We conclude that the test MSE is approximately: "
paste(phrase, bag_MSE)
## [1] "We conclude that the test MSE is approximately: 2.60525306410819"
## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863
- (e) Use random forests to analyze this data. What test MSE do you obtain? Use the
importance() function to determine which vari-ables are most important. Describe the effect of m,the numberof variables considered at each split, on the error rate obtained.
rf.car = randomForest(Sales~.,data=Car.train,mtry = 3, importance = TRUE)
test_pred = predict(rf.car, Car.test)
rf_MSE = mean((test_pred-Car.test$Sales)^2)
phrase = "We conclude that the test MSE is approximately: "
paste(phrase, rf_MSE)
## [1] "We conclude that the test MSE is approximately: 3.05430594288339"
## %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 33.91622
Exercise 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 obser-vations, and a test set containing the remaining observations.
set.seed(1)
train = sample(dim(OJ)[1],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 / 800
- The fitted tree has 9 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.
## 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 ) *
- I select node #8. The split criterion is LoyalCH < 0.0356415, 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 2% of the observations in that branch take the value of CH, and the remaining 98% take the value of MM.
- (d) Create a plot of the tree, and interpret the results.
plot(OJ.tree)
text(OJ.tree,pretty=0, cex = .50)
- The most important indicator appears to be
LoyalCH. The top three nodes contain this indicator.
- (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
test_error = 1 - ((160 + 64) / 270)
phrase = "The test error rate is: "
paste(phrase, test_error)
## [1] "The test error rate is: 0.17037037037037"
- (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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"
- (g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.OJ$size,cv.OJ$dev,type='b', xlab = "Tree size", ylab = "Deviance")
- (h) Which tree size corresponds to the lowest cross-validated classi-fication error rate?
- We see that 7-node tree correspond to the lowest classification 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=7)
plot(prune.OJ)
text(prune.OJ,pretty=0, cex = 0.50)
