Assignment 7
Arwa N
7/27/2020
Chapter 08 (page 332): 3, 8, 9
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 ˆpm1. The x axis should display ˆpm1, 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, ˆpm1 = 1− ˆpm2. You could make this plot by hand, but it will be much easier to make in R.
p = seq(0, 1, 0.01)
Gini = 2 * p * (1 - p)
class.err = 1 - pmax(p, 1 - p)
entropy = -(p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(Gini, class.err, entropy), 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 3.6.3data("Carseats")train = sample(dim(Carseats)[1], dim(Carseats)[1]/2)
traincar = Carseats[train, ]
testcar = Carseats[-train, ]- Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
MSe: 1.766985
library(tree)## Warning: package 'tree' was built under R version 3.6.3car.tree = tree(Sales ~ ., data = traincar)
summary(car.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = traincar)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "Income" "Population"
## [6] "Advertising" "CompPrice"
## Number of terminal nodes: 16
## Residual mean deviance: 2.36 = 434.3 / 184
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.84200 -0.88490 0.03929 0.00000 1.11400 3.94300plot(car.tree)
text(car.tree, pretty = 0)
car.pred <- predict(car.tree, traincar)
mean((car.pred - traincar$Sales)^2)## [1] 2.171461- Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
Yes, the MSE has improved to 5.381467
set.seed(1)
cv.car = cv.tree(car.tree)
par(mfrow = c(1, 2))
plot(cv.car$size, cv.car$dev, type = "b")
plot(cv.car$k, cv.car$dev, type = "b")
library(tree)prune.car <- prune.tree(car.tree, best = 11)
plot(prune.car)
text(prune.car, pretty = 0)
prune.pred = predict(prune.car, testcar)
mean((testcar$Sales - prune.pred)^2)## [1] 4.9178- Use the bagging approach in order to analyze this data. Whattest MSE do you obtain? Use the importance() function to determine which variables are most important.
The MSE is lower at 3.080276. Price and ShelveLoc are the most important variables.
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.set.seed(1)
bag.car = randomForest(Sales ~ ., data = traincar, mtry = 10, importance = TRUE)
bag.car##
## Call:
## randomForest(formula = Sales ~ ., data = traincar, mtry = 10, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.687721
## % Var explained: 68.81bag.predict = predict(bag.car, newdata = testcar)
mean((bag.predict - testcar$Sales)^2)## [1] 2.407913importance(bag.car)## %IncMSE IncNodePurity
## CompPrice 23.5369516 157.087753
## Income 6.7126892 77.661018
## Advertising 18.3777112 138.771448
## Population 1.1615591 53.900099
## Price 55.0070628 529.010544
## ShelveLoc 55.2464535 470.325180
## Age 19.7020453 181.323744
## Education 1.6094616 42.338034
## Urban 0.9951195 9.188756
## US 2.2685034 9.481156- Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
The MSE is even lower at 2.971218. Price and ShelveLoc are the most important variables.
set.seed(1)
rf.car = randomForest(Sales ~ ., data = traincar, mtry =5, importance = TRUE)
bag.car##
## Call:
## randomForest(formula = Sales ~ ., data = traincar, mtry = 10, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 10
##
## Mean of squared residuals: 2.687721
## % Var explained: 68.81rf.predict = predict(rf.car, newdata = testcar)
mean((rf.predict - testcar$Sales)^2)## [1] 2.54431importance(rf.car)## %IncMSE IncNodePurity
## CompPrice 15.9608098 145.37607
## Income 4.8632695 99.63118
## Advertising 16.7668132 146.49785
## Population 1.8207667 83.73252
## Price 40.6098614 454.00011
## ShelveLoc 47.3691355 420.61094
## Age 20.2536012 219.09658
## Education -0.0273898 57.01777
## Urban -1.4761741 12.22659
## US 2.9978563 19.89020- This problem involves the OJ data set which is part of the ISLR package.
library(ISLR)
data(OJ)- Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
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?
The error rate .1588, and there are 9 terminal nodes
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" "SpecialCH" "ListPriceDiff"
## [5] "PctDiscMM"
## Number of terminal nodes: 9
## Residual mean deviance: 0.7432 = 587.8 / 791
## Misclassification error rate: 0.1588 = 127 / 800- Type in the name of the tree object in order to get a detail et text output. Pick one of the terminal nodes, and interpret the information displayed.
Node 20 denoted by the * tells us that this is a terminal node. The split criterion isSpecialCH < 0.5
oj.tree## node), split, n, deviance, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 1073.00 CH ( 0.60625 0.39375 )
## 2) LoyalCH < 0.5036 365 441.60 MM ( 0.29315 0.70685 )
## 4) LoyalCH < 0.280875 177 140.50 MM ( 0.13559 0.86441 )
## 8) LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ) *
## 9) LoyalCH > 0.0356415 118 116.40 MM ( 0.19492 0.80508 ) *
## 5) LoyalCH > 0.280875 188 258.00 MM ( 0.44149 0.55851 )
## 10) PriceDiff < 0.05 79 84.79 MM ( 0.22785 0.77215 )
## 20) SpecialCH < 0.5 64 51.98 MM ( 0.14062 0.85938 ) *
## 21) SpecialCH > 0.5 15 20.19 CH ( 0.60000 0.40000 ) *
## 11) PriceDiff > 0.05 109 147.00 CH ( 0.59633 0.40367 ) *
## 3) LoyalCH > 0.5036 435 337.90 CH ( 0.86897 0.13103 )
## 6) LoyalCH < 0.764572 174 201.00 CH ( 0.73563 0.26437 )
## 12) ListPriceDiff < 0.235 72 99.81 MM ( 0.50000 0.50000 )
## 24) PctDiscMM < 0.196197 55 73.14 CH ( 0.61818 0.38182 ) *
## 25) PctDiscMM > 0.196197 17 12.32 MM ( 0.11765 0.88235 ) *
## 13) ListPriceDiff > 0.235 102 65.43 CH ( 0.90196 0.09804 ) *
## 7) LoyalCH > 0.764572 261 91.20 CH ( 0.95785 0.04215 ) *- Create a plot of the tree, and interpret the results.
Loyal CH is the most important variable of the tree. If LoyalCH<0.28, the tree predicts MM. If LoyalCH>0.76, the tree predicts CH.
plot(oj.tree)
text(oj.tree, pretty = 0)
- 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?
pred.oj = predict(oj.tree, oj.test, type = "class")
table(oj.test$Purchase, pred.oj)## pred.oj
## CH MM
## CH 160 8
## MM 38 64pred.unprune = predict(oj.tree, oj.test, type = "class")
misclass.unprune = sum(oj.test$Purchase !=pred.unprune)
misclass.unprune/length(pred.unprune)## [1] 0.1703704- Apply the cv.tree() function to the training set in order to determine the optimal tree size.
oj.cv = cv.tree(oj.tree, FUN = prune.tree)
oj.cv## $size
## [1] 9 8 7 6 5 4 3 2 1
##
## $dev
## [1] 685.6493 698.8799 702.8083 702.8083 714.1093 725.4734 780.2099
## [8] 790.0301 1074.2062
##
## $k
## [1] -Inf 12.62207 13.94616 14.35384 26.21539 35.74964 43.07317
## [8] 45.67120 293.15784
##
## $method
## [1] "deviance"
##
## attr(,"class")
## [1] "prune" "tree.sequence"- Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(oj.cv$size, oj.cv$dev, type = "b", xlab = "Tree size", ylab = "Deviance")
- Which tree size corresponds to the lowest cross-validated classification error rate?
Size of 6th gives lowest cross-validation error.
- Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
oj.prune = prune.tree(oj.tree, best = 6)- Compare the training error rates between the pruned and unpruned trees. Which is higher?
summary(oj.prune)##
## Classification tree:
## snip.tree(tree = oj.tree, nodes = c(10L, 4L, 12L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff"
## Number of terminal nodes: 6
## Residual mean deviance: 0.7919 = 628.8 / 794
## Misclassification error rate: 0.1788 = 143 / 800- Compare the test error rates between the pruned and unpruned trees. Which is higher?
pruned is higher.
pred.prune = predict(oj.prune, oj.test, type = "class")
misclass.prune = sum(oj.test$Purchase != pred.prune)
misclass.prune/length(pred.prune)## [1] 0.1851852