Chapter 8: Tree-Based Methods
Maria E Morinigo
4/26/2021
Assignment # 7
Exercises
Conceptual
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 . The xaxis should display , 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, = 1− . You could make this plot by hand, but it will be much easier to make in R.
library(tidyverse)## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --## v ggplot2 3.3.3 v purrr 0.3.4
## v tibble 3.0.5 v dplyr 1.0.3
## v tidyr 1.1.2 v stringr 1.4.0
## v readr 1.4.0 v forcats 0.5.1## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()prop_class_1 <- seq(0, 1, 0.001)
prop_class_2 <- 1 - prop_class_1
classification_error <- 1 - pmax(prop_class_1, prop_class_2)
gini <- prop_class_1*(1-prop_class_1) + prop_class_2*(1-prop_class_2)
entropy <- -prop_class_1*log(prop_class_1) - prop_class_2*log(prop_class_2)
data.frame(prop_class_1, prop_class_2, classification_error, gini, entropy) %>%
pivot_longer(cols = c(classification_error, gini, entropy), names_to = "metric") %>%
ggplot(aes(x = prop_class_1, y = value, col = factor(metric))) +
geom_line(size = 1) +
scale_y_continuous(breaks = seq(0, 1, 0.1), minor_breaks = NULL) +
scale_color_hue(labels = c("Classification Error", "Entropy", "Gini")) +
labs(col = "Metric",
y = "Value",
x = "Proportion (of class '1')")## Warning: Removed 2 row(s) containing missing values (geom_path).
Applied
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.
library(tree)## Warning: package 'tree' was built under R version 4.0.4## Registered S3 method overwritten by 'tree':
## method from
## print.tree clilibrary(ISLR)
library(rpart)
library(caret)## Loading required package: lattice##
## Attaching package: 'caret'## The following object is masked from 'package:purrr':
##
## liftlibrary(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
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## selectlibrary(rattle)## Warning: package 'rattle' was built under R version 4.0.5## Loading required package: bitops## Rattle: A free graphical interface for data science with R.
## Version 5.4.0 Copyright (c) 2006-2020 Togaware Pty Ltd.
## Type 'rattle()' to shake, rattle, and roll your data.library(rpart.plot)## Warning: package 'rpart.plot' was built under R version 4.0.5library(randomForest)## Warning: package 'randomForest' was built under R version 4.0.5## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:rattle':
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## importance## The following object is masked from 'package:dplyr':
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## combine## The following object is masked from 'package:ggplot2':
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## marginattach(Carseats)(a) Split the data set into a training set and a test set.
set.seed(42)
inTrain = createDataPartition(Carseats$Sales, p = 0.5, list = FALSE)
train = Carseats[inTrain,]
test = Carseats[-inTrain,](b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
lm.tree = tree(Sales ~ ., train)plot(lm.tree)
text(lm.tree, pretty = 0, cex = 0.7)
The most important variables are ShelveLoc and Price. The tree has 17 terminal nods
summary(lm.tree)##
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Age" "CompPrice" "Education"
## [6] "Advertising"
## Number of terminal nodes: 19
## Residual mean deviance: 2.045 = 372.2 / 182
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3.23700 -0.93380 -0.03278 0.00000 1.04400 3.60500lm.tree_pred = predict(lm.tree, test)
mean((lm.tree_pred - test$Sales)^2)## [1] 5.1369(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
set.seed(3)
cv_tree_model <- cv.tree(lm.tree, K = 10)
CV_RSS = cv_tree_model$dev
n_leaves = cv_tree_model$size
df = data.frame(n_leaves, CV_RSS)
min_CV_RSS = as.numeric(min(CV_RSS) == CV_RSS)
ggplot(dat = df, aes(x = n_leaves, y = CV_RSS)) +
geom_line(col = "grey55") +
geom_point(size = 2, aes(col = factor(min_CV_RSS))) +
scale_x_continuous(breaks = seq(1, 17, 2)) +
scale_y_continuous(labels = scales::comma_format()) +
scale_color_manual(values = c("deepskyblue3", "green")) +
theme(legend.position = "none") +
labs(title = "Carseats Dataset - Regression Tree",
subtitle = "Selecting the complexity parameter with cross-validation",
x = "Terminal Nodes",
y = "CV RSS")
prune_tree = prune.tree(lm.tree, best = 8)
prune_tree_pred = predict(prune_tree, test)
mean((prune_tree_pred - test$Sales)^2)## [1] 4.736775Pruning the tree improved the test MSE
set.seed(1)
train_control <- trainControl(method="repeatedcv", number=10, repeats=3)
cv_lm.tree = train(Sales ~ .,data=train, trControl=train_control,method='rpart')## Warning in nominalTrainWorkflow(x = x, y = y, wts = weights, info = trainInfo, :
## There were missing values in resampled performance measures.cv_lm.tree$finalModel## n= 201
##
## node), split, n, deviance, yval
## * denotes terminal node
##
## 1) root 201 1530.09700 7.451592
## 2) ShelveLocGood< 0.5 164 990.38520 6.821098
## 4) Price>=89.5 141 715.82640 6.383546 *
## 5) Price< 89.5 23 82.07532 9.503478 *
## 3) ShelveLocGood>=0.5 37 185.55130 10.246220 *rpart.plot(cv_lm.tree$finalModel)
lm.tree_prune_pred = predict(cv_lm.tree, test)
mean((lm.tree_prune_pred - test$Sales)^2)## [1] 5.813624(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.
set.seed(1)
lm.tree.bag = randomForest(y = train$Sales, x = train[,-1], mtry = ncol(train)-1, importance = TRUE)lm.tree.bag_pred = predict(lm.tree.bag, test)
mean((lm.tree.bag_pred - test$Sales)^2)## [1] 2.586186importance(lm.tree.bag)## %IncMSE IncNodePurity
## CompPrice 25.28278837 142.851284
## Income 8.16235954 90.864390
## Advertising 16.65584832 135.921010
## Population 2.38962533 55.623128
## Price 53.17554101 472.501798
## ShelveLoc 53.54289290 403.629127
## Age 14.83193349 120.339334
## Education 0.56849853 50.394925
## Urban 2.57147047 9.961402
## US -0.06388945 6.480824Based on this result, ShelveLoc and `Price are the most important variables.
(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.
lm.rf = train(Sales ~., data = train,method='rf',trControl = trainControl("cv", number = 10),importance = TRUE)
lm.rf$bestTune## mtry
## 3 11lm.rf$finalModel##
## Call:
## randomForest(x = x, y = y, mtry = param$mtry, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 11
##
## Mean of squared residuals: 2.732217
## % Var explained: 64.11varImp(lm.rf)## rf variable importance
##
## Overall
## ShelveLocGood 100.000
## Price 99.044
## CompPrice 46.028
## ShelveLocMedium 38.275
## Advertising 32.968
## Age 28.614
## Income 16.520
## UrbanYes 5.608
## USYes 4.859
## Education 1.911
## Population 0.000plot(varImp(lm.rf))
The MSE obtained is 2.73 a little higher than the bagging method.
9. This problem involves the OJ data set which is part of the ISLR package.
str(OJ)## 'data.frame': 1070 obs. of 18 variables:
## $ Purchase : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
## $ WeekofPurchase: num 237 239 245 227 228 230 232 234 235 238 ...
## $ StoreID : num 1 1 1 1 7 7 7 7 7 7 ...
## $ PriceCH : num 1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
## $ PriceMM : num 1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
## $ DiscCH : num 0 0 0.17 0 0 0 0 0 0 0 ...
## $ DiscMM : num 0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
## $ SpecialCH : num 0 0 0 0 0 0 1 1 0 0 ...
## $ SpecialMM : num 0 1 0 0 0 1 1 0 0 0 ...
## $ LoyalCH : num 0.5 0.6 0.68 0.4 0.957 ...
## $ SalePriceMM : num 1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
## $ SalePriceCH : num 1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
## $ PriceDiff : num 0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
## $ Store7 : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
## $ PctDiscMM : num 0 0.151 0 0 0 ...
## $ PctDiscCH : num 0 0 0.0914 0 0 ...
## $ ListPriceDiff : num 0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
## $ STORE : num 1 1 1 1 0 0 0 0 0 0 ...attach(OJ)(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
index_train = sample(1:nrow(OJ), 800)
OJ_train = OJ[index_train,]
OJ_test = OJ[-index_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?
OJ_tree = tree(Purchase ~ ., 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 / 800Four variable were used to construct the tree: LoyalCH,PriceDiff,SpecialCH,ListPriceDiff, and PctDiscMM. There are 9 terminal nodes. The training error rate is 0.1588.
(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 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 ) *Terminal node 8: LoyalCH < 0.0356415 59 10.14 MM ( 0.01695 0.98305 ). It took two split to produce node 8, a split at LoyalCH = 0.5036 and LoyalCH = 0.2808. The overall prediction is CH and the split is 1.695%CH` and 98.31% MM. There are 59 observations in the node.
(d) Create a plot of the tree, and interpret the results.
plot(OJ_tree)
text(OJ_tree, pretty = 0, cex = 0.7)
(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_tree_pred = predict(OJ_tree, OJ_test, type = "class")table(OJ_tree_pred, OJ_test$Purchase)##
## OJ_tree_pred CH MM
## CH 160 38
## MM 8 641 - mean(OJ_tree_pred == OJ_test$Purchase)## [1] 0.17037041 - mean(OJ_tree_pred == "CH")## [1] 0.2666667The test error rate is 0.1704 comparing to the predicted CH for all the observations in OJ_test.
(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
set.seed(1)
OJ_tree_cv = cv.tree(OJ_tree, K = 10, FUN = prune.misclass)
OJ_tree_cv## $size
## [1] 9 8 7 4 2 1
##
## $dev
## [1] 145 145 146 146 167 315
##
## $k
## [1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(OJ_tree_cv$size, OJ_tree_cv$dev)
min(OJ_tree_cv$dev)## [1] 145(h) Which tree size corresponds to the lowest cross-validated classification error rate?
Tree sizes 9 and 8 have the lowest same 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.
pruned_OJ_tree <- prune.tree(OJ_tree, best = 8)
pruned_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 ) *
## 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 ) *(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
mean(predict(OJ_tree, type = "class") != OJ_train$Purchase)## [1] 0.15875mean(predict(pruned_OJ_tree, type = "class") != OJ_train$Purchase)## [1] 0.1625The training error for the pruned tree is higher.
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
mean(predict(OJ_tree, type = "class", newdata = OJ_test) != OJ_test$Purchase)## [1] 0.1703704mean(predict(pruned_OJ_tree, type = "class", newdata = OJ_test) != OJ_test$Purchase)## [1] 0.162963Now,the pruned tree performed better on the test data set.