hmw6
Diego Ramos
May 4, 2020
- 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 ^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. Hint: In a setting with two classes, ^pm1 = 1??? ^pm2. 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("red", "green", "blue"))
- 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 a training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 3.6.2attach(Carseats)
set.seed(420)
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?
library(tree)## Warning: package 'tree' was built under R version 3.6.3tree.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" "Advertising" "CompPrice" "Income"
## [6] "Age" "US"
## Number of terminal nodes: 18
## Residual mean deviance: 1.989 = 362 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -2.87900 -0.97120 -0.01688 0.00000 0.98090 4.00600plot(tree.carseats)
text(tree.carseats, pretty = 0)
pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)## [1] 5.804668test mse is 5.14
- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")
# Best size = 9
pruned.carseats = prune.tree(tree.carseats, best = 7)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)
pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)## [1] 5.590505pruning the test gives us 5.34 test MSE
- 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 3.6.3## 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$Sales - bag.pred)^2)## [1] 3.203475importance(bag.carseats)## %IncMSE IncNodePurity
## CompPrice 19.3293891 142.375289
## Income 9.7172257 101.151280
## Advertising 14.3091177 121.277258
## Population 1.7268708 58.159152
## Price 53.2910661 484.272326
## ShelveLoc 49.8205619 316.790453
## Age 7.2600657 99.165252
## Education -0.2298514 34.633861
## Urban -1.1816569 5.797032
## US 2.6532566 10.184712our MSE is 3.0, with price, shelveloc, and compprice being the most important predictors
- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables aremost important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500,
importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)## [1] 3.344854importance(rf.carseats)## %IncMSE IncNodePurity
## CompPrice 13.1862220 135.396201
## Income 5.8183410 115.426670
## Advertising 10.8474857 136.779370
## Population 0.2648359 78.451301
## Price 44.1353013 420.848684
## ShelveLoc 39.5808402 279.267052
## Age 5.3569496 113.405360
## Education 1.4016062 47.632886
## Urban -1.0096678 7.777727
## US 3.6910967 14.432180our test mse got slightly better with a 2.91 mse, our predictors have changed to shelvloc, price, and advertising.
- This problem involves the OJ data set which is part of the ISLR package.
- 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, ]- 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 / 800only uses LoyalCH and PriceDiff, 7 terminal nodes and missclassification error of 0.155
- 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 ) *LoyalCH < 0.764572 186 210.30 CH ( 0.74731 0.25269 ) LoyalCH has a splititng point of 0.76, there are 186 nodes below this subtree, the deviance is 210.30
(d) Create a plot of the tree, and interpret the results.
plot(oj.tree)
text(oj.tree, pretty = 0)
LoyalCH appears to be the most important variable of the tree, the 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 depends on PriceDiff.
- 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$Purchase, oj.pred)## oj.pred
## CH MM
## CH 149 15
## MM 30 76- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
table(OJ.test$Purchase, oj.pred)## oj.pred
## CH MM
## CH 149 15
## MM 30 76Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis. this code would not knit
Which tree size corresponds to the lowest cross-validated classification error rate? 7
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 = 7)- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.pruned)##
## 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 / 800misclassification is still 0.155
- 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] 0.1666667They have equal missclassification rates