hw7
4/30/2021
Chapter 8, Questions 3,8,9
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 p̂m1. The xaxis should display p̂m1, 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, p̂m1 =1- p̂m2. You couldmake this plot by hand,but will be much easier with R
p = seq(0,1,0.01)
gini = p * (1-p) * 2
entropy = -(p* log(p) + (1-p) * log(1-p) )
class_error = 1 - pmax(p, 1 - p )
matplot(p, cbind(gini, entropy, class_error), col = c("red", "grey","orange"))
Question 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.
- Split the data set into training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.2attach(Carseats)
library(tree)## Warning: package 'tree' was built under R version 4.0.2set.seed(1)
train = sample(dim(Carseats)[1], dim(Carseats)[1]/2)
Carseats.train = Carseats[train, ]
Carseats.test = Carseats[-train,]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
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)
pred.carseats = predict( tree.carseats, Carseats.test)
mean(( Carseats.test$Sales - pred.carseats)^2)## [1] 4.922039We got the MSE is 4.922
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree )
par(mfrow = c(1,2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
pruned.carseats = prune.tree(tree.carseats, best = 9)
par(mfrow = c(1,1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)## [1] 4.918134MSE is 4.918
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 4.0.2## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10 , ntree = 500, imporance = T )
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)## [1] 2.622915importance(bag.carseats)## IncNodePurity
## CompPrice 169.829553
## Income 91.464308
## Advertising 102.156091
## Population 59.921849
## Price 495.642493
## ShelveLoc 381.170255
## Age 152.330816
## Education 45.600313
## Urban 9.229808
## US 15.470199Bagging imporves the MSE and we can see from above function output that Price, ShelveLoc and Age are the most important variables in Sales.
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained
rf.carseats = randomForest(Sales ~ . , data = Carseats.train, mtry = 5, ntree = 500 , importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)## [1] 2.732066importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 19.4239781 159.18237
## Income 4.8774271 110.16804
## Advertising 10.0913784 105.89067
## Population -2.4916697 80.15222
## Price 44.8465403 456.01518
## ShelveLoc 43.2983207 329.47794
## Age 13.4429336 169.72365
## Education 1.4438055 58.56204
## Urban -0.5453909 12.05602
## US 6.7127513 26.31279Random forest worsened the MSE. We see that again the Price, ShelveLoc and Age are three most important predictors of Sales.
Question 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.
attach(OJ)
set.seed(1013)
train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]- 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?
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 / 800- 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 ) *If we pick random node for example 25) SalePriceMM: the splittin criteria is 1.64 and we have 57 point in the subtree below this level of tree. The prediction at this node is CH. About 60% points in this node have CH value of Sales.
- Createa plot of the tree, and interpre the results
plot(oj.tree)
text(oj.tree, pretty = 0 )
LoyalCH is the root or lets say the most important variable of this tree. if LoyalCh < 0.27, MM is predict, if LoyalCH > 0.753, it predicts CH. PriceDiff gives the intermediate values of LoyalCH.
- 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$Purchase, oj.pred)## oj.pred
## CH MM
## CH 149 15
## MM 30 76- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(oj.tree, FUN = prune.tree)- Produce a plot with tree size on the x-axis and cross validation classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type= "b", xlab = "Tree Size", ylab = "Deviance")
h. Which tree size corresponds to the lowest cross-validated classification error rate? Size 5 gives lowest cross-validation error.
- produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminals nodes.
oj.pruned = prune.tree(oj.tree, best = 6)- Compare the training error rates between the prunes and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 12L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 6
## Residual mean deviance: 0.7701 = 611.5 / 794
## Misclassification error rate: 0.175 = 140 / 800For original tree, Misclassification error rate: 0.1612. After pruning, it changed to 0.175
- Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned / length(pred.unpruned)## [1] 0.1666667pred.pruned = predict(oj.tree, OJ.test, type = "class")
misclass.pruned = sum(OJ.test$Purchase != pred.pruned)
misclass.pruned / length(pred.pruned)## [1] 0.1666667Pruned and unpruned trees have same test error rate of 0.1666