Assignment 7
Jiwhan Yoo
7/31/2020
Question 6
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 ˆpm1. The xaxis should display ˆpm1, 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, ˆpm1 = 1− ˆpm2. 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= p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
classification_err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini_index, entropy, classification_err), col = c("red", "green", "blue"))
Question 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 3.6.3set.seed(1)
train <- sample(1:nrow(Carseats), nrow(Carseats) / 2)
Carseats.train <- Carseats[train, ]
Carseats.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 3.6.3tree.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" "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(tree.carseats)
text(tree.carseats, pretty = 0)
yhat <- predict(tree.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 4.922039The test MSE is 4.922
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.carseats = cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
Optimal level of tree complexity is 14
prune.carseats = prune.tree(tree.carseats, best = 14)
plot(prune.carseats)
text(prune.carseats,pretty=0)
yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 5.013738MSE has increased to 5.01 after pruning tree
(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 3.6.3## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.set.seed(1)
bag.carseats = randomForest(Sales~.,data=Carseats.train,mtry = 10, importance = TRUE)
yhat.bag = predict(bag.carseats,newdata=Carseats.test)
mean((yhat.bag-Carseats.test$Sales)^2)## [1] 2.605253importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863Bagging improves the MSE to 2.61. The Price, SelveLoc and CompPrice shows the highest prediction for sales
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables aremost important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
library(randomForest)
set.seed(1)
rf.Carseats = randomForest(Sales~.,data=Carseats.train,mtry = 3, importance = TRUE)
yhat.rf = predict(rf.Carseats,newdata=Carseats.test)
mean((yhat.rf-Carseats.test$Sales)^2)## [1] 2.960559importance(rf.Carseats)## %IncMSE IncNodePurity
## CompPrice 14.8840765 158.82956
## Income 4.3293950 125.64850
## Advertising 8.2215192 107.51700
## Population -0.9488134 97.06024
## Price 34.9793386 385.93142
## ShelveLoc 34.9248499 298.54210
## Age 14.3055912 178.42061
## Education 1.3117842 70.49202
## Urban -1.2680807 17.39986
## US 6.1139696 33.98963MSE improves slightly to 2.96. Price, ShelveLoc and CompPrice are still shows the highest prediction
Question 9
(a) Create a training set containing a random sample of 800 observations, 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 / 800The error rate is 0.1588. There are 9 terminal nodes in the tree
(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 ) *Node 9 shows a star next to it telling us that it denotes a terminal node. It shows that LoyalCH > 0.0356415 and the number of oberservations in the branch is 118. Deviance is 116.4 and we see that the 80% of the observations 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 LoyalCH variable is the most important variable in the tree. The decision also depends on values of ListPriceDiff,PriceDiff,PctDiscMM, and SpecialcCH but generally, LoyalCH < 0.28 predicts MM and the opposit predicts CH.
(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?
oj.pred = predict(oj.tree, OJ.test, type = "class")
table(OJ.test$Purchase, oj.pred)## oj.pred
## CH MM
## CH 160 8
## MM 38 64unpruned.test.err <- mean(oj.pred != OJ.test$Purchase)
print(unpruned.test.err)## [1] 0.1703704Test error rate is about 17.0%
(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.tree)
cv.oj## $size
## [1] 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 685.6493 698.8799 702.8083 702.8083 714.1093 725.4734 780.2099
## [8] 790.0301 1074.2062
##
## $k
## [1] -Inf 12.62207 13.94616 14.35384 26.21539 35.74964 43.07317
## [8] 45.67120 293.15784
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"
(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
tree size of 6 appears to be the lowest size
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.misclass(oj.tree, best = 6)
plot(prune.oj)
text(prune.oj, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(prune.oj)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800Pruned tree produces a lower misclassification error rate at 16.25%
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1703704pred.pruned = predict(prune.oj, OJ.test, type = "class")
misclass.pruned <- sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.162963Pruned tree shows a lower test error