Assignment7
Lily He
4/21/2022
library(ISLR)
library(tree)
library(dplyr)
library(randomForest)Problem 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 p̂ m1. The xaxis should display p̂ m1, 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, p̂ m1 = 1 - p̂ m2. You could make this plot by hand, but it will be much easier to make in R.
p = seq(0, 1, 0.01)
gini = p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
class.err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini, entropy, class.err), col = c("pink", "red", "purple"))
# Problem 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.
set.seed(24)
c<-Carseats
train<-sample(1:nrow(c), nrow(c)/2)
c.tr<-c[train,]
c.te<-c[-train,]B.Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.c=tree(Sales~., data=c.tr)
plot(tree.c)
text(tree.c, pretty=0)
pred.tree=predict(tree.c, c.te)
obs.sales<-c.te$Sales
mean((pred.tree-obs.sales)^2)## [1] 4.541017- We can see the MSE is roughly 4.54.
C.Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.c = cv.tree(tree.c, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.c$size, cv.c$dev, type = "b")
plot(cv.c$k, cv.c$dev, type = "b")
# Best size = 7
pruned.c = prune.tree(tree.c, best = 7)
par(mfrow = c(1, 1))
plot(pruned.c)
text(pruned.c, pretty = 0)
pred.pruned = predict(pruned.c, c.te)
mean((c.te$Sales - pred.pruned)^2)## [1] 4.886626- Pruning the tree improved the test MSE from 4.54 to 4.89.
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.
set.seed(1)
bag.c<-randomForest(Sales~., data = c.tr, mtry = 10, importance = TRUE)
yhat.bag<-predict(bag.c, newdata = c.te)
mean((yhat.bag - c.te$Sales)^2)## [1] 2.730285importance(bag.c)## %IncMSE IncNodePurity
## CompPrice 24.469745 169.299552
## Income 3.243752 84.351895
## Advertising 13.544325 99.588388
## Population 2.880161 78.093002
## Price 54.280757 466.183454
## ShelveLoc 53.819809 439.484997
## Age 21.119729 201.544314
## Education 3.882171 34.191473
## Urban -3.192860 5.634045
## US 1.813198 5.186044- Bagging improves the MSE from 4.89 to 2.73. We also see, using importance() that Price, ShelveLoc and CompPrice are the three most important variables for Sale.
Problem 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.
set.seed(1)
samp<-sample(1:nrow(OJ), 800)
oj.tr<-OJ[samp,]
oj.te<-OJ[-samp,]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?
tree.oj<-tree(Purchase~., data = oj.tr)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = oj.tr)
## 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- Training error rate: 15.88%
- Number of terminal nodes: 9
- Uses 5 of the 18 variables:“LoyalCH”,“PriceDiff”,“SpecialCH”,“ListPriceDiff”, “PctDiscMM”
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.
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 ) *- Let’s pick terminal node labeled “10)”. The splitting variable at this node is PriceDiff. The splitting value of this node is 0.05. There are 79 points in the subtree below this node. The deviance for all points contained in region below this node is 80. A * in the line denotes that this is in fact a terminal node. The prediction at this node is Sales = MM. About 23% points in this node have CH as value of Sales. Remaining 77% points have MM as value of Sales.
D. Create a plot of the tree, and interpret the results
plot(tree.oj)
text(tree.oj, pretty = 0)
* LoyalCH is the most important variable of the tree, in fact top 5 nodes contain LoyalCH. If LoyalCH<0.28, the tree predicts MM. If LoyalCH>0.76, the tree predicts CH. For intermediate values of LoyalCH, the decision also depends on the value of PriceDiff.
Part 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.predz<-predict(tree.oj, oj.te, type = 'class')
obs.purch<-oj.te$Purchase
caret::confusionMatrix(tree.predz, obs.purch)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 160 38
## MM 8 64
##
## Accuracy : 0.8296
## 95% CI : (0.7794, 0.8725)
## No Information Rate : 0.6222
## P-Value [Acc > NIR] : 8.077e-14
##
## Kappa : 0.6154
##
## Mcnemar's Test P-Value : 1.904e-05
##
## Sensitivity : 0.9524
## Specificity : 0.6275
## Pos Pred Value : 0.8081
## Neg Pred Value : 0.8889
## Prevalence : 0.6222
## Detection Rate : 0.5926
## Detection Prevalence : 0.7333
## Balanced Accuracy : 0.7899
##
## 'Positive' Class : CH
## - We get an error rate of 0.1704
F. 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)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', main = 'CV Error Rate')
H. Which tree size corresponds to the lowest cross-validated classification error rate?
- Size of 7 gives lowest cross-validation error.
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(tree.oj, best = 7)J. Compare the test error rates between the pruned and unpruned trees. Which is higher?
summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, 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 / 800- Misclassification error of pruned tree is 0.1625, unpruned tree is 0.1704, so the test error rates for unpruned tree is higher.