STA4133 Chapter 8 (3,8,9)
Tu Thai
2023-04-19
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, 0.001)
gini = p * (1 - p) * 2
entropy = -(p * log(p) + (1 - p) * log(1 - p))
class.err = 1 - pmax(p, 1 - p)
matplot(p, cbind(gini, entropy, class.err), col = c("blue", "red", "green"))
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.
library(ISLR)(a) Split the data set into a training set and a test set.
tr.ind <- sample(1:nrow(Carseats),size = 0.8*nrow(Carseats))
train <- Carseats[tr.ind,]
test <- Carseats[-tr.ind,](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)
library(rattle)## Warning: package 'rattle' was built under R version 4.2.3## 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(rpart)
tree <- rpart(Sales ~ ., data = train)
rattle::fancyRpartPlot(tree, sub = "")
From the decision tree, we know that ‘Shelve’ is the most important
indicator for ‘Sales’ since it’s at the beginning of the split. Then
‘price’ and ‘age’
tree.pred = predict(tree, newdata=test)
mean((tree.pred - test$Sales)^2)## [1] 4.519835The test MSE obtained is 5.203
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
The optimal level is at 11 based on the plot below
plotcp(tree)
opt_cp <- tree$cptable[which.min(tree$cptable[,"xerror"]),"CP"]
# Prune the tree to the optimal complexity parameter
pruned_tree <- prune(tree, cp = opt_cp)
rattle::fancyRpartPlot(tree, sub = "")
# Find the MSE
tree.pred = predict(pruned_tree, newdata=test)
mean((tree.pred - test$Sales)^2)## [1] 4.52347The new MSE is 5.097. This is lower compared to before pruning
(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 4.2.3## 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':
##
## importancelibrary(caret)## Loading required package: ggplot2##
## Attaching package: 'ggplot2'## The following object is masked from 'package:randomForest':
##
## margin## Loading required package: lattice#Bagging approach
set.seed(1)
bagging <- randomForest(Sales ~ ., data = train, ntree = 1000, mtry = 10, importance = TRUE)
#Test MSE
bag.pred <- predict(bagging, newdata = test)
mean((bag.pred - test$Sales)^2)## [1] 2.417814#Important variable
varImp(bagging)## Overall
## CompPrice 51.422800
## Income 17.763775
## Advertising 37.613176
## Population -1.294712
## Price 109.929671
## ShelveLoc 109.035701
## Age 29.174163
## Education 3.177931
## Urban -2.352881
## US 2.416496Test error is 2.68 if we use bagging approach. From the importance function, we can see that Price, ShelveLocation is the most important variables
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
#Random forest approach
set.seed(1)
rf <- randomForest(Sales ~ ., data = train, importance = TRUE)
rf##
## Call:
## randomForest(formula = Sales ~ ., data = train, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 3
##
## Mean of squared residuals: 2.864861
## % Var explained: 63.98#Test MSE
rf.pred <- predict(rf, newdata = test)
mean((rf.pred - test$Sales)^2)## [1] 2.659032#Important
varImp(rf)## Overall
## CompPrice 17.334141
## Income 5.204413
## Advertising 17.647821
## Population -2.380962
## Price 50.004776
## ShelveLoc 46.724213
## Age 18.244700
## Education 2.004686
## Urban -1.584318
## US 2.736328The test MSE obtained is 2.966, which is worse than bagging. Random Forest functions build 500 trees, with 3 variables at each split, and the independent variables can explained 65.03% of Sales variable. The most important variables are ShelveLoc and Price, similar to boosting.
9. This problem involves the OJ data set which is part of the ISLR2 package.
library(ISLR)(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
tr.ind <- sample(1:nrow(OJ), size = 800)
train <- OJ[tr.ind,]
test <- OJ[-tr.ind,](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 <- tree(Purchase ~., data = train)
summary(tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = 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 / 800There are 9 nodes in tree. The error rate is 15.88. Most important variables are LoyalCH, PriceDiff, and SpecialCH
(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## 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 ) *If we choose the second node, then LoyalCh is splitted when it’s smaller than 0.5036. It’s deviance is 441.60 and overall prediction is MM (minute maid orange juice). Fraction of observations in this branch that takes on values of “CH” and “MM” = (0.29315 0.70685)
(d) Create a plot of the tree, and interpret the results.
plot(tree)
text(tree, pretty = 0)
The tree determine that LoyalCH is the most important indicator. If LoyalCH is between 0.280 and 0.503, PricedDiff and SpecialCH will determine whether the customer will buy MM or CH orange juice. Most from this branch results in a buying of MM.
If LoyalCH is between 0.503 and 0.764, then ListPriceDiff and PctDiscMM will determine the outcome. Most from this branch results in a buying of CH
(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, newdata=test, type="class")
table(tree.pred,test$Purchase)##
## tree.pred CH MM
## CH 160 38
## MM 8 64(8+38)/270## [1] 0.1703704The test error rate is 17.03.
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
OJcv = cv.tree(tree, FUN = prune.misclass)(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJcv$size, OJcv$dev, type = "b")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
From the plot, tree size of 8 produce the lowest error rate. However, since there is not much different from size 4 to 8, i would choose tree size of 4 for my optimal point for easier of interpretation and the tree would be more efficient.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
OJprune=prune.tree(tree,best=4)
plot(OJprune)
text(OJprune, pretty = 0)
(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
#Initial tree training error rates
summary(tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800#Prune tree training error rates
summary(OJprune)##
## Classification tree:
## snip.tree(tree = tree, nodes = 4:6)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes: 4
## Residual mean deviance: 0.8678 = 690.7 / 796
## Misclassification error rate: 0.205 = 164 / 800As we can see, the training error rates of pruned tree is at 20.5, which is a little higher than original tree at 15.88
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
#Initial tree test error rates: 17.03 (from part e)
#Prune tree test error rates:
prune.pred <- predict(OJprune, newdata=test, type="class")
table(prune.pred,test$Purchase)##
## prune.pred CH MM
## CH 142 24
## MM 26 78(26+24)/270## [1] 0.1851852From this, we can see that prune tree also has a higher testing error rates of 18.52, when comparing test error rate of unpruned ones (17.03)