Data Mining Homework 7
John Thomas
April 25, 2019
Homework 7
Problem 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 \(\hat{p}_{m1}\). The xaxis should display \(\hat{p}_{m1}\), 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, \(\hat{p}_{m1}\) = 1 - \(\hat{p}_{m2}\). You could make this plot by hand, but it will be much easier to make in R.
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("pink", "red", "purple"))
Problem 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.
Part 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, ]Part B
Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
library(tree)## Warning: package 'tree' was built under R version 3.5.3tree.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$Sales - pred.carseats)^2)## [1] 4.148897We do find the MSE is roughly 4.15.
Part C
Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$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] 4.993124Our MSE is now to be seen to be increased to 4.99.
Part 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 3.5.3## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500,
importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)## [1] 2.633915importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 16.9874366 126.852848
## Income 3.8985402 78.314126
## Advertising 16.5698586 123.702901
## Population 0.6487058 62.328851
## Price 55.3976775 514.654890
## ShelveLoc 42.7849818 319.133777
## Age 20.5135255 185.582077
## Education 3.4615211 42.253410
## Urban -2.5125087 8.700009
## US 7.3586645 18.180651Bagging improves the MSE we obtained to 2.58. We also see, using importance() that Price, Age and ShelveLoc are the three most important variables for Sale.
Part E
Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained
rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500,
importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)## [1] 2.816693importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 11.304167 126.68519
## Income 5.423214 102.44073
## Advertising 13.351166 137.98835
## Population 1.131119 82.24483
## Price 46.600559 451.70021
## ShelveLoc 37.352447 278.79756
## Age 19.992113 194.99430
## Education 1.945616 51.70741
## Urban -2.244558 10.87383
## US 6.261365 20.83998Random Forest actually worsens the MSE to 2.87. Changing m varies test MSE between 2.6 to 3. Once again though Price, ShelveLoc and Age are three most important predictors of Sale.
Problem 9
This problem involves the OJ data set which is part of the ISLR package.
Part 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, ]Part 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 / 800The tree only uses two variables: LoyalCH and PriceDiff. It has 7 terminal nodes. Training error rate (misclassification error) for the tree is 0.155.
Part 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 ) *Let’s pick terminal node labeled “10)”. The splitting variable at this node is PriceDiff. The splitting value of this node is 0.05. There are 79 points in the subtree below this node. The deviance for all points contained in region below this node is 80. A * in the line denotes that this is in fact a terminal node. The prediction at this node is Sales = MM. About 19% points in this node have CH as value of Sales. Remaining 81% points have MM as value of Sales.
Part 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, in fact top 3 nodes contain LoyalCH. If LoyalCH<0.27, the tree predicts MM. If LoyalCH>0.76, the tree predicts CH. For intermediate values of LoyalCH, the decision also depends on the value of PriceDiff.
Part 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$Purchase, oj.pred)## oj.pred
## CH MM
## CH 152 19
## MM 32 67Part F
Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj = cv.tree(oj.tree, FUN = prune.tree)Part G
Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")
Part H
Which tree size corresponds to the lowest cross-validated classification error rate?
Size of 6 gives lowest cross-validation error.
Part I
Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.pruned = prune.tree(oj.tree, best = 6)Part J
Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 13L)
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff"
## Number of terminal nodes: 6
## Residual mean deviance: 0.7689 = 610.5 / 794
## Misclassification error rate: 0.155 = 124 / 800Misclassification error of pruned tree is exactly same as that of original tree - 0.155.
Part K
Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$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] 0.1888889Pruned and unpruned trees have same test error rate of 0.189.