Assignment7
Karthik Munnam
2024-04-19
library(ISLR)
library(caret)## Loading required package: ggplot2## Loading required package: latticelibrary(tree)
library(rattle)## Loading required package: tibble## Loading required package: bitops## Rattle: A free graphical interface for data science with R.
## Version 5.5.1 Copyright (c) 2006-2021 Togaware Pty Ltd.
## Type 'rattle()' to shake, rattle, and roll your data.library(randomForest)## randomForest 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:rattle':
##
## importance## The following object is masked from 'package:ggplot2':
##
## marginlibrary(gbm)## Loaded gbm 2.1.9## This version of gbm is no longer under development. Consider transitioning to gbm3, https://github.com/gbm-developers/gbm3library(rsample)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 pˆm1. The x- axis should display pˆm1, 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.
p = seq(0, 1, 0.001)
gini.index = 2 * p * (1 - p)
class.error = 1 - pmax(p, 1 - p)
cross.entropy = - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), ylab = "gini.index, class.error, cross.entropy", col = c("red", "green", "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.
(a) Split the data set into a training set and a test
set.
set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats)/2)
strain = Carseats[train, ]
stest = Carseats[-train, ](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
tree.seats = tree(Sales ~ ., data = strain)
summary(tree.seats)##
## Regression tree:
## tree(formula = Sales ~ ., data = strain)
## 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.seats)
text(tree.seats, pretty = 0)
treeseat.pred = predict(tree.seats, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)## [1] 4.9220394.922039 is the test MSE.
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(1)
cv.seats = cv.tree(tree.seats)
plot(cv.seats$size, cv.seats$dev, type = "b")
prune.car = prune.tree(tree.seats, best = 10)
plot(prune.car)
text(prune.car,pretty=0)
treeseat.pred = predict(prune.car, newdata = stest)
mean((treeseat.pred - stest$Sales)^2)## [1] 4.918134Pruning the tree improved the test MSE
(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.seats = randomForest(Sales~., data = strain, mtry = 10, ntree = 551, importance = TRUE)
bagseat.pred = predict(bag.seats, newdata = stest)
mean((bagseat.pred - stest$Sales)^2)## [1] 2.599099importance(bag.seats)## %IncMSE IncNodePurity
## CompPrice 26.18616309 170.781666
## Income 5.25063979 90.717958
## Advertising 13.25673204 97.498810
## Population -2.14346969 58.289311
## Price 60.58241525 503.478806
## ShelveLoc 50.77308639 380.258594
## Age 19.03720001 158.282846
## Education 1.24264920 44.834257
## Urban -0.08461165 9.883299
## US 4.71515903 17.907727There is a 2.60 test MSE. Price and ShelveLoc are the two most significant variables.
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
set.seed(1)
rando.seats = randomForest(Sales~., data = strain, mtry = 10, importance = TRUE)
randseat.pred = predict(rando.seats, newdata = stest)
mean((randseat.pred - stest$Sales)^2)## [1] 2.605253importance(rando.seats)## %IncMSE IncNodePurity
## CompPrice 24.8888481 170.182937
## Income 4.7121131 91.264880
## Advertising 12.7692401 97.164338
## Population -1.8074075 58.244596
## Price 56.3326252 502.903407
## ShelveLoc 48.8886689 380.032715
## Age 17.7275460 157.846774
## Education 0.5962186 44.598731
## Urban 0.1728373 9.822082
## US 4.2172102 18.073863There is a 2.61 test MSE. Price and ShelveLoc remain the most important variables.
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)
OJtrain = OJ[train, ]
OJtest = 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?
tree.OJ = tree(Purchase ~ ., data = OJtrain)
summary(tree.OJ)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## 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 / 800plot(tree.OJ)
text(tree.OJ, pretty = 0)
The development of the tree actually made use of 5 variables. There is a 0.1588 training error rate. The tree has 9 terminal nodes.
(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 ) *Line 9 (LoyalCH) contains 118 observations. It also indicates that the LoyalCH value of branch 9 is less than 0.0356415. In branch 9, MM is the value assigned to more than 80% of the observations, and CH is the value assigned to less than 20% of the observations.
(d) Create a plot of the tree, and interpret the results.
plot(tree.OJ)
text(tree.OJ, pretty = 0)
The most important variables are ListPriceDiff, PriceDiff, PctDiscMM, LoyalCH, and SpecialCH.
(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?
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)##
## treeOJ.pred CH MM
## CH 160 38
## MM 8 64(38 + 8) / 270## [1] 0.1703704test error rate: 17%
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
OJcv = cv.tree(tree.OJ, FUN = prune.misclass)
OJcv## $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
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJcv$size, OJcv$dev, type = "b", xlab = "Tree Size", ylab = "classification error rate")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
The cross-validated classification error rate that is lowest corresponds with a tree size of 7.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune.OJ=prune.tree(tree.OJ,best=7)
plot(prune.OJ)
text(prune.OJ, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(tree.OJ)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJtrain)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800summary(prune.OJ)##
## Classification tree:
## snip.tree(tree = tree.OJ, nodes = c(10L, 4L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800Compared to the unpruned tree (0.1588), the pruned tree has a higher training error rate (0.1625).
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
treeOJ.pred = predict(tree.OJ, newdata = OJtest, type = "class")
table(treeOJ.pred, OJtest$Purchase)##
## treeOJ.pred CH MM
## CH 160 38
## MM 8 64(38 + 8) / 270## [1] 0.1703704pruneOJ.pred = predict(prune.OJ, newdata = OJtest, type = "class")
table(pruneOJ.pred, OJtest$Purchase)##
## pruneOJ.pred CH MM
## CH 160 36
## MM 8 66(36 + 8) / 270## [1] 0.162963Compared to the unpruned tree (0.1703704), the pruned tree has a lower test error rate (0.162963). The test error rate is higher in the unpruned tree.