Assignment 7
Albert Uriegas
7/28/2020
library(ISLR)
library(caret)## Warning: package 'caret' was built under R version 4.0.2## Loading required package: lattice## Loading required package: ggplot2library(ggplot2)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(tibble)Exercise 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_Error <- 1-pmax(p, 1-p)
df <- data.frame(p, Gini, Entropy, Class_Error)
df1 <- data.frame(df[1], stack(df[2:ncol(df)]))
ggplot(df1, aes(x = p, y = values, color = ind)) +
geom_line(size = 2) +
labs(title = "Relationship of Quantities") +
ylab("Values") +
scale_color_brewer(palette = "Set1") +
scale_color_discrete(name = "", labels = c("Gini Coefficient", "Entropy",
"Classification Error")) +
theme_bw()## Scale for 'colour' is already present. Adding another scale for 'colour',
## which will replace the existing scale.
Exercise 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.
(a) Split the data set into a training set and a test set.
set.seed(1)
summary(Carseats)## Sales CompPrice Income Advertising
## Min. : 0.000 Min. : 77 Min. : 21.00 Min. : 0.000
## 1st Qu.: 5.390 1st Qu.:115 1st Qu.: 42.75 1st Qu.: 0.000
## Median : 7.490 Median :125 Median : 69.00 Median : 5.000
## Mean : 7.496 Mean :125 Mean : 68.66 Mean : 6.635
## 3rd Qu.: 9.320 3rd Qu.:135 3rd Qu.: 91.00 3rd Qu.:12.000
## Max. :16.270 Max. :175 Max. :120.00 Max. :29.000
## Population Price ShelveLoc Age Education
## Min. : 10.0 Min. : 24.0 Bad : 96 Min. :25.00 Min. :10.0
## 1st Qu.:139.0 1st Qu.:100.0 Good : 85 1st Qu.:39.75 1st Qu.:12.0
## Median :272.0 Median :117.0 Medium:219 Median :54.50 Median :14.0
## Mean :264.8 Mean :115.8 Mean :53.32 Mean :13.9
## 3rd Qu.:398.5 3rd Qu.:131.0 3rd Qu.:66.00 3rd Qu.:16.0
## Max. :509.0 Max. :191.0 Max. :80.00 Max. :18.0
## Urban US
## No :118 No :142
## Yes:282 Yes:258
##
##
##
## dim(Carseats)## [1] 400 11inTrain <- createDataPartition(Carseats$Sales, p = 0.5, list = FALSE)
train <- Carseats[inTrain,]
test <- Carseats[-inTrain,]
dim(train)## [1] 201 11(b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
The model ended up with 17 terminal nodes and 6 variables used. These included: ShelveLoc, Price, Income, Age, CompPrice and Advertising. The test MSE was 4.540765.
library(tree)## Warning: package 'tree' was built under R version 4.0.2tree_carseats <- tree(Sales ~ ., data = train)
summary(tree_carseats)##
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc" "Price" "Income" "Age" "CompPrice"
## [6] "Advertising"
## Number of terminal nodes: 17
## Residual mean deviance: 2.272 = 418 / 184
## Distribution of residuals:
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -4.76200 -0.95950 0.08568 0.00000 1.00600 3.55000plot(tree_carseats)
text(tree_carseats, pretty = 0)
pred_carseats <- predict(tree_carseats, test)
mean((test$Sales - pred_carseats)^2)## [1] 4.540765(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
The best level of complexity was at size 10. The test MSE was not improved; it increased to 4.700748.
cv_carseats <- cv.tree(tree_carseats, FUN = prune.tree)
# Create data frame for ggplot
dfcv_carseats <- data.frame(Size = cv_carseats$size, cv_dev = cv_carseats$dev,
K = cv_carseats$k)
ggplot(dfcv_carseats, aes(Size, cv_dev)) + geom_line(color = "red", size = 1.5) +
ylab("CV Deviation") + theme_bw()
pruned_carseats <- prune.tree(tree_carseats, best = 10)
plot(pruned_carseats)
text(pruned_carseats, pretty = 0)
pred_pruned <- predict(pruned_carseats, test)
mean((test$Sales - pred_pruned)^2)## [1] 4.700748(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.
The test MSE was 2.558038. The 5 most important variables were CompPrice, Income, Advertising, Population and Price.
library(randomForest)## randomForest 4.6-14## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:dplyr':
##
## combine## The following object is masked from 'package:ggplot2':
##
## marginset.seed(1)
bag_carseats <- randomForest(Sales ~ ., data = train, mtry = 10, importance = TRUE)
bag_pred <- predict(bag_carseats, test)
mean((test$Sales - bag_pred)^2)## [1] 2.558038importance(bag_carseats)## %IncMSE IncNodePurity
## CompPrice 21.1831118 163.914465
## Income 11.5870916 96.312530
## Advertising 16.6612943 121.613296
## Population -0.5761569 48.296635
## Price 49.8770064 423.588145
## ShelveLoc 48.6842318 366.493993
## Age 17.4970778 138.454912
## Education -1.0620839 53.131551
## Urban -1.9187461 7.154914
## US 2.7621595 9.673692(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
The test MSE was 2.646104. Price, ShelveLocGood, CompPrice, ShelveLocMedium and Age were the 5 most important variables. As the number of variables considered at each split increased toward 11, the error rate decreased in the training model. The reason for 11 variables is that ShelveLoc was split into 2 binary variables: ShelveLocGood and ShelveLocMedium.
library(caret)
set.seed(1)
rf_carseats <- train(Sales ~ ., data = train, method = "rf", trControl =
trainControl("cv", number = 10), importance = TRUE)
rf_carseats## Random Forest
##
## 201 samples
## 10 predictor
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 181, 181, 181, 180, 181, 181, ...
## Resampling results across tuning parameters:
##
## mtry RMSE Rsquared MAE
## 2 1.945632 0.6131705 1.589400
## 6 1.734240 0.6477697 1.394678
## 11 1.727223 0.6328361 1.394093
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was mtry = 11.rf_carseats$bestTune## mtry
## 3 11rf_carseats$finalModel##
## Call:
## randomForest(x = x, y = y, mtry = param$mtry, importance = TRUE)
## Type of random forest: regression
## Number of trees: 500
## No. of variables tried at each split: 11
##
## Mean of squared residuals: 2.770705
## % Var explained: 62.27rf_pred <- predict(rf_carseats, test)
mean((test$Sales - rf_pred)^2)## [1] 2.646104plot(varImp(rf_carseats))
Exercise 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.
library(tree)
set.seed(1)
summary(OJ)## Purchase WeekofPurchase StoreID PriceCH PriceMM
## CH:653 Min. :227.0 Min. :1.00 Min. :1.690 Min. :1.690
## MM:417 1st Qu.:240.0 1st Qu.:2.00 1st Qu.:1.790 1st Qu.:1.990
## Median :257.0 Median :3.00 Median :1.860 Median :2.090
## Mean :254.4 Mean :3.96 Mean :1.867 Mean :2.085
## 3rd Qu.:268.0 3rd Qu.:7.00 3rd Qu.:1.990 3rd Qu.:2.180
## Max. :278.0 Max. :7.00 Max. :2.090 Max. :2.290
## DiscCH DiscMM SpecialCH SpecialMM
## Min. :0.00000 Min. :0.0000 Min. :0.0000 Min. :0.0000
## 1st Qu.:0.00000 1st Qu.:0.0000 1st Qu.:0.0000 1st Qu.:0.0000
## Median :0.00000 Median :0.0000 Median :0.0000 Median :0.0000
## Mean :0.05186 Mean :0.1234 Mean :0.1477 Mean :0.1617
## 3rd Qu.:0.00000 3rd Qu.:0.2300 3rd Qu.:0.0000 3rd Qu.:0.0000
## Max. :0.50000 Max. :0.8000 Max. :1.0000 Max. :1.0000
## LoyalCH SalePriceMM SalePriceCH PriceDiff Store7
## Min. :0.000011 Min. :1.190 Min. :1.390 Min. :-0.6700 No :714
## 1st Qu.:0.325257 1st Qu.:1.690 1st Qu.:1.750 1st Qu.: 0.0000 Yes:356
## Median :0.600000 Median :2.090 Median :1.860 Median : 0.2300
## Mean :0.565782 Mean :1.962 Mean :1.816 Mean : 0.1465
## 3rd Qu.:0.850873 3rd Qu.:2.130 3rd Qu.:1.890 3rd Qu.: 0.3200
## Max. :0.999947 Max. :2.290 Max. :2.090 Max. : 0.6400
## PctDiscMM PctDiscCH ListPriceDiff STORE
## Min. :0.0000 Min. :0.00000 Min. :0.000 Min. :0.000
## 1st Qu.:0.0000 1st Qu.:0.00000 1st Qu.:0.140 1st Qu.:0.000
## Median :0.0000 Median :0.00000 Median :0.240 Median :2.000
## Mean :0.0593 Mean :0.02731 Mean :0.218 Mean :1.631
## 3rd Qu.:0.1127 3rd Qu.:0.00000 3rd Qu.:0.300 3rd Qu.:3.000
## Max. :0.4020 Max. :0.25269 Max. :0.440 Max. :4.000dim(OJ)## [1] 1070 18train <- sample(dim(OJ)[1], size = 800)
oj_train <- OJ[train,]
oj_test <- OJ[-train,]
dim(oj_test)## [1] 270 18(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?
The tree used the following 5 variables: LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, and PctDiscMM. The training error rate was 0.1588 and the tree has 9 terminal nodes.
tree_oj <- tree(Purchase ~ ., data = oj_train)
summary(tree_oj)##
## 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(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.
Using the example of node 8, the prediction is split according to LoyalCH (loyalty to Citrus Hill) being below 3.56%, the number of observations meeting this criterion is 59, the deviance is 10.14, and the probability of selecting Minute Maid (MM) being 98.3% (the probability of selecting Citrus Hill (CH) is 1.7%).
tree_oj## 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 ) *(d) Create a plot of the tree, and interpret the results.
LoyalCH is the most important variable as it shows up as the top, second and one of the third splits. PriceDiff and ListPriceDiff show in the 3rd split, while SpecialCH and PctDiscMM are in the 4th level.
plot(tree_oj)
text(tree_oj, pretty = 0)
(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?
The test error rate is 0.1704.
tree_pred = predict(tree_oj, oj_test, type = "class")
confusionMatrix(oj_test$Purchase, tree_pred)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 160 8
## MM 38 64
##
## Accuracy : 0.8296
## 95% CI : (0.7794, 0.8725)
## No Information Rate : 0.7333
## P-Value [Acc > NIR] : 0.0001259
##
## Kappa : 0.6154
##
## Mcnemar's Test P-Value : 1.904e-05
##
## Sensitivity : 0.8081
## Specificity : 0.8889
## Pos Pred Value : 0.9524
## Neg Pred Value : 0.6275
## Prevalence : 0.7333
## Detection Rate : 0.5926
## Detection Prevalence : 0.6222
## Balanced Accuracy : 0.8485
##
## 'Positive' Class : CH
## 1-0.8296## [1] 0.1704(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv_oj <- cv.tree(tree_oj, FUN = prune.misclass)
cv_oj## $size
## [1] 9 8 7 4 2 1
##
## $dev
## [1] 150 150 149 158 172 315
##
## $k
## [1] -Inf 0.000000 3.000000 4.333333 10.500000 151.000000
##
## $method
## [1] "misclass"
##
## attr(,"class")
## [1] "prune" "tree.sequence"(g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
# create data frame for ggplot
cv_oj1 <- data.frame(Size = cv_oj$size, cv_error = cv_oj$dev/800)
ggplot(cv_oj1, aes(Size, cv_error))+ geom_line(color = "purple", size = 1.5) +
ylab("CV Classification Error Rate") +
scale_x_continuous(breaks = c(0:10)) + theme_bw()
(h) Which tree size corresponds to the lowest cross-validated classification error rate?
The tree with size 7 has the lowest cross-validated classification error rate.
(i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
prune_oj <- prune.tree(tree_oj, best = 7)
summary(prune_oj)##
## Classification tree:
## snip.tree(tree = tree_oj, nodes = c(10L, 4L))
## Variables actually used in tree construction:
## [1] "LoyalCH" "PriceDiff" "ListPriceDiff" "PctDiscMM"
## Number of terminal nodes: 7
## Residual mean deviance: 0.7748 = 614.4 / 793
## Misclassification error rate: 0.1625 = 130 / 800(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?
The training error for the pruned tree (0.1625) is higher than the error rate of the unpruned tree (0.1588).
(k) Compare the test error rates between the pruned and unpruned trees. Which is higher?
The test error rate of the unpruned tree (0.1704) is higher than the error rate of the pruned tree (0.163).
ojprune_pred <- predict(prune_oj, oj_test, type = "class")
confusionMatrix(oj_test$Purchase, ojprune_pred)## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 160 8
## MM 36 66
##
## Accuracy : 0.837
## 95% CI : (0.7875, 0.879)
## No Information Rate : 0.7259
## P-Value [Acc > NIR] : 1.185e-05
##
## Kappa : 0.6336
##
## Mcnemar's Test P-Value : 4.693e-05
##
## Sensitivity : 0.8163
## Specificity : 0.8919
## Pos Pred Value : 0.9524
## Neg Pred Value : 0.6471
## Prevalence : 0.7259
## Detection Rate : 0.5926
## Detection Prevalence : 0.6222
## Balanced Accuracy : 0.8541
##
## 'Positive' Class : CH
## 1-0.837## [1] 0.163