Assignment 7
Diana Cisneros
2022-04-28
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 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, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.
gini=function(m1){
return(2*(m1*(1-m1)))
}
ent=function(m1){
m2=1-m1
return(-((m1*log(m1))+(m2*log(m2))))
}
classerr=function(m1){
m2=1-m1
return(1-max(m1,m2))
}
err=seq(0,1,by=0.01)
c.err=sapply(err,classerr)
g=sapply(err,gini)
e=sapply(err,ent)
d=data.frame(Gini.Index=g,Cross.Entropy=e)
plot(err,c.err,type='l',col="green",xlab="m1",ylim=c(0,0.8),ylab="value")
matlines(err,d,col=c("orange","blue"))
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.
- Split the data set into a training set and a test set.
library(ISLR)
library(rpart)
set.seed(1)
training = sample(1:nrow(Carseats), nrow(Carseats) / 2)
Car.training = Carseats[training, ]
Car.testing = Carseats[-training,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)
tree.carseats <- tree(Sales ~ ., data = Car.training)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = Car.training)
## 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 = Car.testing)
mean((yhat - Car.testing$Sales)^2)## [1] 4.922039The test MSE is 4.92
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.car = cv.tree(tree.carseats)
plot(cv.car$size, cv.car$dev, type = "b")
prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
yhat <- predict(prune.carseats, newdata = Car.testing)
mean((yhat - Car.testing$Sales)^2)## [1] 5.113254The test MSE is higher now (5.11)
- 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)## randomForest 4.7-1## Type rfNews() to see new features/changes/bug fixes.bag.carseats <- randomForest(Sales ~ ., data = Car.training, mtry = 10, ntree = 500, importance = TRUE)
yhat.bag <- predict(bag.carseats, newdata = Car.testing)
mean((yhat.bag - Car.testing$Sales)^2)## [1] 2.588149The method bagging lowered the Test MSE to 2.597
importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 25.6142224 170.664874
## Income 4.5915046 91.047400
## Advertising 12.9597786 98.653166
## Population -0.7823257 57.653647
## Price 55.2302897 510.311641
## ShelveLoc 48.5480002 381.630885
## Age 16.7267043 158.261715
## Education 0.9105144 43.460189
## Urban 0.2260420 9.347446
## US 5.8455044 18.261630the higher variables are Price and ShelveLoc.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
randfor.carseats <- randomForest(Sales ~ ., data = Car.training, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(randfor.carseats, newdata = Car.testing)
mean((yhat.rf - Car.testing$Sales)^2)## [1] 3.054306The test MSE is 3.09
importance(randfor.carseats)## %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 33.91622The higher variables are Price and ShelveLoc.
9. This problem involves the OJ data set which is part of the ISLR package.
library(ISLR)
attach(OJ)
View(OJ)- 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.training <- OJ[train, ]
OJ.testing <- OJ[-train, ]- 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.training)
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.training)
## 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 MSE is 0.1588 with 8 nodes.
- 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 ) *The split criterion is LoyalCH < 0.035 (Node 8), 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)
The top three nodes contain LoyalCH, so this variable is the most important indicator of Purchase.
- 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.testing, type = "class")
table(tree.pred, OJ.testing$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 641 - (160 + 64) / 270## [1] 0.1703704The error rate is 17%
- 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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
- Which tree size corresponds to the lowest cross-validated classification error rate?
The 7-node tree is the smallest tree with the lowest classification error rate.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.oj <- prune.misclass(tree.oj, best = 7)
plot(prune.oj)
text(prune.oj, pretty = 0)
- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(tree.oj)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.training)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.oj)##
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(4L, 10L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800The MSE is higher for pruned tree.
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
For the pruned tree:
prune.pred <- predict(prune.oj, OJ.testing, type = "class")
table(prune.pred, OJ.testing$Purchase)##
## prune.pred CH MM
## CH 160 36
## MM 8 661 - (160 + 66) / 270## [1] 0.162963For the unpruned tree:
prune.pred <- predict(tree.oj, OJ.testing, type = "class")
table(prune.pred, OJ.testing$Purchase)##
## prune.pred CH MM
## CH 160 38
## MM 8 641 - (160 + 64) / 270## [1] 0.1703704The pruned tree had an error of 17%, this is higher than the unpruned tree.