STA Assignment 7
Marina Allen
4/27/2021
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 ˆ pm1. Thexaxis should display ˆ pm1, 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, ˆ pm1 =1− ˆ pm2. 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 <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
par(bg = "lightcyan")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("gray67" , "lightpink2", "goldenrod2"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("gray67" , "lightpink2", "goldenrod2"), pch=c(15,17,19))
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.
Part A: Split the data set into a training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.3library(tree)## Warning: package 'tree' was built under R version 4.0.5library(dplyr)## Warning: package 'dplyr' was built under R version 4.0.3##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionset.seed(24)
c<-Carseats
train<-sample(1:nrow(c), nrow(c)/2)
c.tr<-c[train,]
c.te<-c[-train,]Part 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)
yhat.tree=predict(tree.c, c.te)
obs.sales<-c.te$Sales
mean((yhat.tree-obs.sales)^2)## [1] 4.541017Part 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)
cv.c## $size
## [1] 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 1012.989 1041.573 1020.262 1043.493 1057.997 1064.050 1131.822 1130.429
## [9] 1211.419 1182.640 1250.044 1239.531 1352.795 1438.036 1382.154 1686.631
##
## $k
## [1] -Inf 17.46067 18.21007 25.43324 29.77121 32.67272 44.19972
## [8] 46.09869 49.55016 50.87585 60.04473 63.14420 99.94859 132.15272
## [15] 143.22869 395.57046
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"plot(cv.c$size,cv.c$dev, xlab = 'size', ylab = 'cv error ')
abline(h = min(cv.c$dev) + .2*sd(cv.c$dev))
prune.c<-list()
yhat.prune<-list()
mse.prune<-list()
for (i in 2:16) {
prune.c[[i]]=prune.tree(tree.c, best = i)
yhat.prune[[i]]=predict(prune.c[[i]], c.te)
obs.sales<-c.te$Sales
mse.prune[[i]]<-mean((yhat.prune[[i]]-obs.sales)^2)
}
mse.mat<-as.matrix(mse.prune)
mse.df<-as.data.frame(mse.mat)
par(bg = "lavenderblush1" )
matplot(2:16, mse.df[-1,1], xlab = 'size', ylab = 'MSE', main = "What size tree \n gives the lowest MSE?")
lines(2:16, mse.df[-1,1], type = "o")
mse.df <- data.frame(size=2:16,lapply(mse.df, unlist) )
mse.df$mse.rank<-rank(mse.df$V1)
names(mse.df)[2]<-'mse'
mse.df[mse.df$mse.rank %in% 1:3,]## size mse mse.rank
## 13 14 4.727879 3
## 14 15 4.499964 1
## 15 16 4.541017 2Answer: It looks like 14 nodes provides the lowest MSE with 4.254
Part D: Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
library(randomForest)## Warning: package 'randomForest' was built under R version 4.0.5## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:dplyr':
##
## combineset.seed(24)
bag.c<-randomForest(Sales~., data = c.tr, mtry = 10, importance = TRUE)
yhat.bag<-predict(bag.c, newdata = c.te)
mean((yhat.bag - c.te$Sales)^2)## [1] 2.692938importance(bag.c)## %IncMSE IncNodePurity
## CompPrice 22.8321261 176.578526
## Income 4.1064747 77.183648
## Advertising 12.8743758 99.953925
## Population 0.3779757 76.359264
## Price 51.4435675 465.155918
## ShelveLoc 51.6714645 438.596460
## Age 21.2202209 201.116544
## Education 1.3558349 36.305982
## Urban -2.0566741 6.577210
## US 0.6451619 5.857904rf.c<- list()
yhat.rf<-list()
mse.rf<-list()
for ( i in 1:10 ) {
set.seed(24)
rf.c[[i]]<-randomForest(Sales ~ ., data = c.tr, mtry = i, importance = TRUE)
yhat.rf[[i]]<-predict(rf.c[[i]], newdata = c.te)
mse.rf[[i]]<- mean((yhat.rf[[i]] - c.te$Sales)^2)
}
par(bg = "rosybrown1" )
matplot(1:10, mse.rf, xlab = 'mtry', ylab = 'MSE', main = "What Number of predictors \n gives the lowest MSE?")
lines(1:10, mse.rf, type = "o")
Question 9:
###This problem involves the OJ data set which is part of the ISLR package.
Part A: Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(10)
library(ISLR)
library(tree)
samp<-sample(1:nrow(OJ), 800)
oj.tr<-OJ[samp,]
oj.te<-OJ[-samp,]Part 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" "DiscMM" "PriceDiff"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7983 = 633 / 793
## Misclassification error rate: 0.1775 = 142 / 800Answer: Misclassification error rate: 16.25%, Number of terminal nodes: 7, Uses only 3 of the 18 variables: “LoyalCH” “SpecialCH” “PriceDiff”
Part 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 1067.000 CH ( 0.61375 0.38625 )
## 2) LoyalCH < 0.48285 290 315.900 MM ( 0.23448 0.76552 )
## 4) LoyalCH < 0.035047 51 9.844 MM ( 0.01961 0.98039 ) *
## 5) LoyalCH > 0.035047 239 283.600 MM ( 0.28033 0.71967 )
## 10) DiscMM < 0.47 220 270.500 MM ( 0.30455 0.69545 ) *
## 11) DiscMM > 0.47 19 0.000 MM ( 0.00000 1.00000 ) *
## 3) LoyalCH > 0.48285 510 466.000 CH ( 0.82941 0.17059 )
## 6) LoyalCH < 0.764572 245 300.200 CH ( 0.69796 0.30204 )
## 12) PriceDiff < 0.145 99 137.000 MM ( 0.47475 0.52525 )
## 24) DiscMM < 0.47 82 112.900 CH ( 0.54878 0.45122 ) *
## 25) DiscMM > 0.47 17 12.320 MM ( 0.11765 0.88235 ) *
## 13) PriceDiff > 0.145 146 123.800 CH ( 0.84932 0.15068 ) *
## 7) LoyalCH > 0.764572 265 103.700 CH ( 0.95094 0.04906 ) *Part D: Create a plot of the tree, and interpret the results.
par(bg = "lightpink")
plot(tree.oj)
text(tree.oj, pretty = 0)
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 135 20
## MM 27 88
##
## Accuracy : 0.8259
## 95% CI : (0.7753, 0.8692)
## No Information Rate : 0.6
## P-Value [Acc > NIR] : 9.992e-16
##
## Kappa : 0.6412
##
## Mcnemar's Test P-Value : 0.3815
##
## Sensitivity : 0.8333
## Specificity : 0.8148
## Pos Pred Value : 0.8710
## Neg Pred Value : 0.7652
## Prevalence : 0.6000
## Detection Rate : 0.5000
## Detection Prevalence : 0.5741
## Balanced Accuracy : 0.8241
##
## 'Positive' Class : CH
## Part 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)Part 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/800, type ="b", xlab = "tree size", ylab = 'CV Error Rate', main = 'What size tree has\n the lowest CV Error Rate?')
Part H: Which tree size corresponds to the lowest cross-validated classification error rate?
best.trees<-data.frame(tree_size = cv.oj$size, CvErrors = cv.oj$dev, Rate = paste0(cv.oj$dev/8,"%"))
best.trees[order(best.trees$Rate),]## tree_size CvErrors Rate
## 1 7 157 19.625%
## 2 5 157 19.625%
## 3 2 161 20.125%
## 4 1 309 38.625%Part 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 = 5)Part J: Compare the training error rates between the pruned and unpruned trees. Which is higher?
prune.predz<-predict(prune.oj, oj.te, type = "class")
caret::confusionMatrix(prune.predz, obs.purch)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 135 20
## MM 27 88
##
## Accuracy : 0.8259
## 95% CI : (0.7753, 0.8692)
## No Information Rate : 0.6
## P-Value [Acc > NIR] : 9.992e-16
##
## Kappa : 0.6412
##
## Mcnemar's Test P-Value : 0.3815
##
## Sensitivity : 0.8333
## Specificity : 0.8148
## Pos Pred Value : 0.8710
## Neg Pred Value : 0.7652
## Prevalence : 0.6000
## Detection Rate : 0.5000
## Detection Prevalence : 0.5741
## Balanced Accuracy : 0.8241
##
## 'Positive' Class : CH
##