Assignment 7
Abel Garcia Chavez
2023-05-09
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 \(pˆm1\). 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_m1 <- seq(0, 1, 0.01)
gini.index <- 2 * p_m1 * (1 - p_m1)
class.error <- 1 - pmax(p_m1, 1 - p_m1)
cross.entropy <- - (p_m1 * log(p_m1) + (1 - p_m1) * log(1 - p_m1))
matplot(p_m1, cbind(gini.index, class.error, cross.entropy), col = c("black", "pink", "lightpink"))
Question 8
In the lab, a classification tree was applied to the Carseats data set af- ter 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)
set.seed(1)
train <- sample(1:nrow(Carseats), nrow(Carseats) / 2)
Carseats.train <- Carseats[train, ]
Carseats.test <- Carseats[-train, ](b) 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 = Carseats.train)
summary(tree.carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats.train)
## 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 = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 4.922039We get a test MSE of approx 4.92
(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
tree.min <- which.min(cv.carseats$dev)
points(tree.min, cv.carseats$dev[tree.min], col = "black", cex = 2, pch = 20)
prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)
yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)## [1] 5.113254Yes we can see that it improves to approx 5.11.
(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 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.bag.carseats <- randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500, importance = TRUE)
yhat.bag <- predict(bag.carseats, newdata = Carseats.test)
mean((yhat.bag - Carseats.test$Sales)^2)## [1] 2.657296importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 23.07909904 171.185734
## Income 2.82081527 94.079825
## Advertising 11.43295625 99.098941
## Population -3.92119532 59.818905
## Price 54.24314632 505.887016
## ShelveLoc 46.26912996 361.962753
## Age 14.24992212 159.740422
## Education -0.07662320 46.738585
## Urban 0.08530119 8.453749
## US 4.34349223 15.157608In this case it actually decreases to. approx 2.58 and after using the importance function we can see that ShelveLoc and Price are actually the two most important variables.
(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats <- randomForest(Sales ~ ., data = Carseats.train, mtry = 3, ntree = 500, importance = TRUE)
yhat.rf <- predict(rf.carseats, newdata = Carseats.test)
mean((yhat.rf - Carseats.test$Sales)^2)## [1] 3.049406importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 12.9489323 158.48521
## Income 2.2754686 129.59400
## Advertising 8.9977589 111.94374
## Population -2.2513981 102.84599
## Price 33.4226950 391.60804
## ShelveLoc 34.0233545 290.56502
## Age 12.2185108 171.83302
## Education 0.2592124 71.65413
## Urban 1.1382113 14.76798
## US 4.1925335 33.75554Here the test MSE is approx 3.02 and after using the importance function we can see its the same variables have importance.
Question 9
This problem involves the OJ data set which is part of the ISLR2 package. (a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
library(ISLR)
attach(OJ)
set.seed(1013)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = 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?
library(tree)
oj.tree = tree(Purchase ~ ., data = OJ.train)
summary(oj.tree)##
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "SalePriceMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7564 = 599.8 / 793
## Misclassification error rate: 0.1612 = 129 / 800LoyalCH PriceDiff ListPriceDiff SalePriceM It has 7 terminal nodes. Training error rate is 0.1612.
(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.
oj.tree## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1069.00 CH ( 0.61125 0.38875 )
## 2) LoyalCH < 0.5036 344 407.30 MM ( 0.27907 0.72093 )
## 4) LoyalCH < 0.276142 163 121.40 MM ( 0.12270 0.87730 ) *
## 5) LoyalCH > 0.276142 181 246.30 MM ( 0.41989 0.58011 )
## 10) PriceDiff < 0.065 75 75.06 MM ( 0.20000 0.80000 ) *
## 11) PriceDiff > 0.065 106 144.50 CH ( 0.57547 0.42453 ) *
## 3) LoyalCH > 0.5036 456 366.30 CH ( 0.86184 0.13816 )
## 6) LoyalCH < 0.753545 189 224.30 CH ( 0.71958 0.28042 )
## 12) ListPriceDiff < 0.235 79 109.40 MM ( 0.48101 0.51899 )
## 24) SalePriceMM < 1.64 22 20.86 MM ( 0.18182 0.81818 ) *
## 25) SalePriceMM > 1.64 57 76.88 CH ( 0.59649 0.40351 ) *
## 13) ListPriceDiff > 0.235 110 75.81 CH ( 0.89091 0.10909 ) *
## 7) LoyalCH > 0.753545 267 85.31 CH ( 0.96255 0.03745 ) *(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
LoyalCH is the most important variable of the tree for lower parts of
the tree PriceDiff is important.
(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?
oj.pred = predict(oj.tree, OJ.test, type = "class")
table(OJ.test$Purchase, oj.pred)## oj.pred
## CH MM
## CH 149 15
## MM 30 76the error rate is .16
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(oj.tree, FUN = prune.tree)(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
(h) Which tree size corresponds to the lowest cross-validated
classification error rate?
Size of 6 gives lowest cross-validation error.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned = prune.tree(oj.tree, best = 6)(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 12L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 6
## Residual mean deviance: 0.7701 = 611.5 / 794
## Misclassification error rate: 0.175 = 140 / 800Misclassification error of pruned tree is exactly same as that of original tree — 0.155.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1666667pred.pruned = predict(oj.pruned, OJ.test, type = "class")
misclass.pruned = sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.2The un-pruned one is better here.