HW 7 CH8 Trees and Forrests
Manuel S. H. Valles
April 18, 2019
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 = "papayawhip")
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19) ,ylab = "gini.index, class.error, cross.entropy",col = c("darkolivegreen4" , "wheat", "tomato2"), type = 'b')
legend('bottom', inset=.01, legend = c('gini.index', 'class.error', 'cross.entropy'), col = c("darkolivegreen4" , "wheat", "tomato2"), pch=c(15,17,19))
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 3.5.2library(tree)## Warning: package 'tree' was built under R version 3.5.3library(dplyr)##
## 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,]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] 4.380944C.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)
#Were Looking for lowest $dev value
cv.c## $size
## [1] 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 1095.379 1190.538 1189.565 1198.805 1208.513 1190.543 1215.122
## [8] 1249.185 1296.663 1305.450 1287.363 1287.135 1326.605 1309.057
## [15] 1441.186 1847.726
##
## $k
## [1] -Inf 21.14139 22.25737 25.47605 26.96703 28.51798 37.51636
## [8] 43.64950 50.73523 58.36588 66.92904 75.29555 102.60990 103.46708
## [15] 262.82101 442.72190
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"plot(cv.c$size,cv.c$dev, xlab = 'size', ylab = 'cv error ')
abline(h = min(cv.c$dev) + .2*sd(cv.c$dev))
While the size with all 16 variables has the lowest cv error, there a couple close that have less variables. Lets try all these bad boys out
prune.c<-list()
yhat.prune<-list()
mse.prune<-list()
for (i in 2:16) {
prune.c[[i]]=prune.tree(tree.c, best = i)
yhat.prune[[i]]=predict(prune.c[[i]], c.te)
obs.sales<-c.te$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 = "cornsilk3" )
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$mse.rank %in% 1:3,]## size mse mse.rank
## 8 9 4.278954 3
## 13 14 4.254269 1
## 14 15 4.272504 2It looks like 14 nodes provides the lowest MSE with 4.254
D.Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
library(randomForest)## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:dplyr':
##
## combineset.seed(24)
bag.c<-randomForest(Sales~., data = c.tr, mtry = 10, importance = TRUE)
yhat.bag<-predict(bag.c, newdata = c.te)
mean((yhat.bag - c.te$Sales)^2)## [1] 2.484656importance(bag.c)## %IncMSE IncNodePurity
## CompPrice 24.5184459 181.565931
## Income 6.5892916 83.524935
## Advertising 14.5588651 119.608270
## Population 1.0922213 79.001601
## Price 56.6984008 634.062290
## ShelveLoc 55.0596062 486.097209
## Age 14.9663083 161.743021
## Education -0.2642658 41.939529
## Urban -0.9996057 7.871980
## US 0.5995533 3.998345I know this is not always applicable in real life, but for this data lets try mtry = 1:10
rf.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$Sales)^2)
}
par(bg = "cornflowerblue" )
matplot(1:10, mse.rf, xlab = 'mtry', ylab = 'MSE', main = "What # of predictors \n gives the lowest MSE?")
lines(1:10, mse.rf, type = "o")
mseRF<-t(as.data.frame(lapply(mse.rf, unlist)))
mse.rf<-data.frame(size=1:10,mseRF)
mse.rf.rownames<-paste0("mtry", 1:10)
rownames(mse.rf)<-mse.rf.rownames
mse.rf$mse.rank<-rank(mse.rf$mseRF)
mse.rf[order(mse.rf$mse.rank),]## size mseRF mse.rank
## mtry5 5 2.369562 1
## mtry6 6 2.412489 2
## mtry8 8 2.419069 3
## mtry7 7 2.425788 4
## mtry4 4 2.431394 5
## mtry9 9 2.458708 6
## mtry10 10 2.484656 7
## mtry3 3 2.529006 8
## mtry2 2 2.823754 9
## mtry1 1 3.932894 10Wow. mtry = 10, has the 7th lowest mse, while mtry = 5 has the best with an mse of 2.369562.
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(10)
library(ISLR)
library(tree)
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" "SpecialCH" "PriceDiff"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7936 = 629.3 / 793
## Misclassification error rate: 0.1625 = 130 / 800Misclassification error rate: 16.25%
Number of terminal nodes: 7
Uses only 3 of the 18 variables: “LoyalCH” “SpecialCH” “PriceDiff”
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 1080.000 CH ( 0.59500 0.40500 )
## 2) LoyalCH < 0.450956 290 295.700 MM ( 0.20690 0.79310 )
## 4) LoyalCH < 0.276142 172 119.500 MM ( 0.11047 0.88953 ) *
## 5) LoyalCH > 0.276142 118 152.400 MM ( 0.34746 0.65254 )
## 10) SpecialCH < 0.5 104 125.000 MM ( 0.28846 0.71154 ) *
## 11) SpecialCH > 0.5 14 14.550 CH ( 0.78571 0.21429 ) *
## 3) LoyalCH > 0.450956 510 487.400 CH ( 0.81569 0.18431 )
## 6) LoyalCH < 0.764572 258 321.100 CH ( 0.68605 0.31395 )
## 12) PriceDiff < 0.05 87 119.200 MM ( 0.43678 0.56322 )
## 24) PriceDiff < -0.34 18 7.724 MM ( 0.05556 0.94444 ) *
## 25) PriceDiff > -0.34 69 95.290 CH ( 0.53623 0.46377 ) *
## 13) PriceDiff > 0.05 171 164.900 CH ( 0.81287 0.18713 ) *
## 7) LoyalCH > 0.764572 252 102.400 CH ( 0.94841 0.05159 ) *4) LoyalCH < 0.276142 172 119.500 MM ( 0.11047 0.88953 ) *
If LoyalCH is less than 0.450956 AND less than 0.276142, the Purchase is MM
D. Create a plot of the tree, and interpret the results
par(bg = "darkorange")
plot(tree.oj)
text(tree.oj, pretty = 0)
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$Purchase
caret::confusionMatrix(tree.predz, obs.purch)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 155 27
## MM 22 66
##
## Accuracy : 0.8185
## 95% CI : (0.7673, 0.8626)
## No Information Rate : 0.6556
## P-Value [Acc > NIR] : 2.277e-09
##
## Kappa : 0.5929
## Mcnemar's Test P-Value : 0.5677
##
## Sensitivity : 0.8757
## Specificity : 0.7097
## Pos Pred Value : 0.8516
## Neg Pred Value : 0.7500
## Prevalence : 0.6556
## Detection Rate : 0.5741
## Detection Prevalence : 0.6741
## Balanced Accuracy : 0.7927
##
## 'Positive' Class : CH
## We get an error rate of 18.15%
F. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj<-cv.tree(tree.oj, FUN = prune.misclass)G. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size,cv.oj$dev/800, type ="b", xlab = "tree size", ylab = 'CV Error Rate', main = 'What size tree has\n the lowest CV Error Rate?')
H. Which tree size corresponds to the lowest cross-validated classification error rate?
best.trees<-data.frame(tree_size = cv.oj$size, CvErrors = cv.oj$dev, Rate = paste0(cv.oj$dev/8,"%"))
best.trees[order(best.trees$Rate),]## tree_size CvErrors Rate
## 2 5 149 18.625%
## 1 7 151 18.875%
## 3 4 159 19.875%
## 4 2 159 19.875%
## 5 1 324 40.5%It looks like a tree size of 5 gives us the lowest CV error rate of 18.625%
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)J. Compare the test 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 151 25
## MM 26 68
##
## Accuracy : 0.8111
## 95% CI : (0.7592, 0.856)
## No Information Rate : 0.6556
## P-Value [Acc > NIR] : 1.242e-08
##
## Kappa : 0.5828
## Mcnemar's Test P-Value : 1
##
## Sensitivity : 0.8531
## Specificity : 0.7312
## Pos Pred Value : 0.8580
## Neg Pred Value : 0.7234
## Prevalence : 0.6556
## Detection Rate : 0.5593
## Detection Prevalence : 0.6519
## Balanced Accuracy : 0.7921
##
## 'Positive' Class : CH
##