Homework 7
Ashley Mata
April 26, 2019
3. Consider the Gini index, classification error, and entropy in asimple classification setting with two classes. Create a single plot that displays each of these quantities as a function of ^pm1. The xaxis should display ^pm1, 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, 0.01)
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("red", "yellow", "pink"))
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)
attach(Carseats)
set.seed(1)
train = sample(dim(Carseats)[1], dim(Carseats)[1]/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" "Income"
## [6] "CompPrice"
## Number of terminal nodes: 18
## Residual mean deviance: 2.36 = 429.5 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.2570 -1.0360 0.1024 0.0000 0.9301 3.9130plot(tree.carseats)
text(tree.carseats, pretty = 0)
pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)## [1] 4.148897(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
# Best size = 9
pruned.carseats = prune.tree(tree.carseats, best = 9)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)## [1] 4.993124(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.
(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 worsens the MSE on test set to 2.87. Changing m varies test MSE between 2.6 to 3.The Price, ShelveLoc and Age are three most important predictors of Sale.
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.
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"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7517 = 596.1 / 793
## Misclassification error rate: 0.155 = 124 / 800(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 1075.00 CH ( 0.60250 0.39750 )
## 2) LoyalCH < 0.5036 359 422.80 MM ( 0.27577 0.72423 )
## 4) LoyalCH < 0.276142 170 119.10 MM ( 0.11176 0.88824 ) *
## 5) LoyalCH > 0.276142 189 257.50 MM ( 0.42328 0.57672 )
## 10) PriceDiff < 0.05 79 76.79 MM ( 0.18987 0.81013 ) *
## 11) PriceDiff > 0.05 110 148.80 CH ( 0.59091 0.40909 ) *
## 3) LoyalCH > 0.5036 441 343.30 CH ( 0.86848 0.13152 )
## 6) LoyalCH < 0.764572 186 210.30 CH ( 0.74731 0.25269 )
## 12) PriceDiff < -0.165 29 34.16 MM ( 0.27586 0.72414 ) *
## 13) PriceDiff > -0.165 157 140.90 CH ( 0.83439 0.16561 )
## 26) PriceDiff < 0.265 82 95.37 CH ( 0.73171 0.26829 ) *
## 27) PriceDiff > 0.265 75 31.23 CH ( 0.94667 0.05333 ) *
## 7) LoyalCH > 0.764572 255 90.67 CH ( 0.95686 0.04314 ) *(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, 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?
oj.pred = predict(oj.tree, OJ.test, type = "class")
table(OJ.test$Purchase, oj.pred)## oj.pred
## CH MM
## CH 152 19
## MM 32 67(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(oj.tree, FUN = prune.tree)(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
Size of 6 gives lowest cross-validation error.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned = prune.tree(oj.tree, best = 6)(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 13L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 6
## Residual mean deviance: 0.7689 = 610.5 / 794
## Misclassification error rate: 0.155 = 124 / 800(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)## [1] 0.1888889pred.pruned = predict(oj.pruned, OJ.test, type = "class")
misclass.pruned = sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)## [1] 0.1888889