Assignment #7
Raegan Gutierrez
4/30/2021
Question 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 \(ˆpm1\). The x-axis should display \(ˆpm1\), 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 <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.ent <- - (p * log(p) + (1 - p) * log(1 - p))
plot(NA,NA,xlim=c(0,1),ylim=c(0,1),xlab='p',ylab='f')
lines(p,gini,type='l')
lines(p,class.error,col='blue')
lines(p,cross.ent,col='red')
legend(x='top',legend=c('gini','class error','cross entropy'),
col=c('black','blue','red'),lty=1,text.width = 0.22)
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)
set.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 4.0.5tree.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 4.09900yhat <- predict(tree.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 4.922039I obtained a MSE of 4.922.
c. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 1)
yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 5.113254In this case, the pruning did not improve the test MSE. The test MSE increased to 5.113.
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 4.0.5## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.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] 2.623527importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 24.6476262 168.57576
## Income 5.3355144 90.64911
## Advertising 11.7975748 100.81807
## Population -2.4347577 58.94510
## Price 55.1387362 500.32765
## ShelveLoc 46.3295341 376.78200
## Age 17.9245890 162.68705
## Education 0.8706811 43.04402
## Urban 0.6850649 8.77639
## US 4.3005762 17.96535I obtained a test MSE of 2.624, and conclude that Price and ShelveLoc are the most important variables.
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.
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] 3.001375importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 15.0614295 155.38762
## Income 2.8372504 125.35841
## Advertising 8.5912531 108.52715
## Population -2.2524534 101.40095
## Price 37.9562323 398.55509
## ShelveLoc 36.9148773 289.37326
## Age 11.0749075 173.58313
## Education 0.9820296 70.20401
## Urban 0.7161522 15.45718
## US 6.1094256 33.75805I obtained an MSE of 3.001 and see that the same two variables, Priceand ShelveLoc, are the most important. Changing \(m\) has made the mean decrease accuracy lower, from 55% and 46% tov 38% and 37%, respectively. This means the second model’s accuracy will decrease less if the two most important variables are removed compared to the first model.
Question 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)
train <- sample(1:nrow(OJ), 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?
library(tree)
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"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800The training error rate = 0.1588 with 9 terminal nodes on 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.
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 < 0.196197 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 ) *Observing the terminal node labeled 8, the split criterion is LoyalCH < 0.0356. The number of observations in that branch is 59 with a deviance of 10.14 and an overall prediction for the branch of MM. 1.695% of the observations in that branch take the value of CH, and the remaining 98.305% take the value of MM.
d. Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 1)
All of the top 3 nodes use the factor LoyalCH, making it the most important factor to determine a purchase. The next row of nodes contain Price, meaning the price of the carseat is the second determining factor in a purchase.
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(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 641-(147+62) /270 ## [1] 0.2259259The test error rate = 22.59%
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)
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$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
h. Which tree size corresponds to the lowest cross-validated classification error rate?
a tree with 7 terminal nodes will produce the lowest classification error rate.
i. Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.misclass(tree.oj, best = 7)
plot(prune.oj)
text(prune.oj, pretty = 0)
j. Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The training error rate for the pruned tree is slightly higher at 0.1625.
k. Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, OJ.test, type = "class")
table(prune.pred, OJ.test$Purchase)##
## prune.pred CH MM
## CH 160 36
## MM 8 661 - (160 + 66) / 270## [1] 0.162963tree.pred <- predict(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 641 - (160 + 64) / 270## [1] 0.1703704The test error rate for the unpruned tree is higher at 17.037%