pnt130_Assignment3
Jongbum Choi
2025-06-30
Exercise 13.
This question should be answered using the Weekly data set, which is part of the ISLR2 package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1,089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
Exercise 13.(a)
Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Answer: There appears to be little correlation previous week and today. The only correlation is between Year and Volume.
library(ISLR2)
names(Weekly) # Predictor's Name## [1] "Year" "Lag1" "Lag2" "Lag3" "Lag4" "Lag5"
## [7] "Volume" "Today" "Direction"dim(Weekly) # Dimension of Data## [1] 1089 9summary(Weekly) # Summary of Data## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## pairs(Weekly)
cor(Weekly[, -9]) # Pairwise correlations## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1 -0.03228927 1.000000000 -0.07485305 0.05863568 -0.071273876
## Lag2 -0.03339001 -0.074853051 1.00000000 -0.07572091 0.058381535
## Lag3 -0.03000649 0.058635682 -0.07572091 1.00000000 -0.075395865
## Lag4 -0.03112792 -0.071273876 0.05838153 -0.07539587 1.000000000
## Lag5 -0.03051910 -0.008183096 -0.07249948 0.06065717 -0.075675027
## Volume 0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today -0.03245989 -0.075031842 0.05916672 -0.07124364 -0.007825873
## Lag5 Volume Today
## Year -0.030519101 0.84194162 -0.032459894
## Lag1 -0.008183096 -0.06495131 -0.075031842
## Lag2 -0.072499482 -0.08551314 0.059166717
## Lag3 0.060657175 -0.06928771 -0.071243639
## Lag4 -0.075675027 -0.06107462 -0.007825873
## Lag5 1.000000000 -0.05851741 0.011012698
## Volume -0.058517414 1.00000000 -0.033077783
## Today 0.011012698 -0.03307778 1.000000000attach(Weekly)
plot(Volume)
Exercise 13.(b)
Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
Answer: Lag2 has high corelation than other predictors
glm.fits <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(glm.fits)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4coef(glm.fits)## (Intercept) Lag1 Lag2 Lag3 Lag4 Lag5
## 0.26686414 -0.04126894 0.05844168 -0.01606114 -0.02779021 -0.01447206
## Volume
## -0.02274153summary(glm.fits)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4glm.probs <- predict(glm.fits, type = "response") # predict probability
glm.probs[1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Direction)## Up
## Down 0
## Up 1Exercise 13.(c)
Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
According to confusion matrix, the off-diagonals represent the incorrect prediction. (430+48)/1089 = 0.0489348
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
table(glm.pred, Direction) # build a Confusion Matrix## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557(54+557)/1089 # Correct prediction## [1] 0.5610652(430+48)/1089 # Incorrect Prediction## [1] 0.4389348Exercise 13.(d)
Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
Train <- Weekly$Year <= 2008
Test <- Weekly$Year > 2008
Weekly_Train <- Weekly[Train, ]
Weekly_Test <- Weekly[Test, ]
glm.fits_partial <- glm(Direction ~ Lag2, data = Weekly, subset = Train, family = binomial)
summary(glm.fits_partial)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly,
## subset = Train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4plot(glm.fits_partial)
glm.probs_partial <- predict(glm.fits_partial, Weekly_Test, type = "response")
glm.probs_partial[1:10]## 986 987 988 989 990 991 992 993
## 0.5261291 0.6447364 0.4862159 0.4852001 0.5197667 0.5401255 0.6233482 0.4809930
## 994 995
## 0.4512204 0.4848808contrasts(Weekly_Test$Direction)## Up
## Down 0
## Up 1glm.pred_partial <- ifelse(glm.probs_partial > 0.5, "Up", "Down")
table(glm.pred_partial, Weekly_Test$Direction) # build a Confusion Matrix##
## glm.pred_partial Down Up
## Down 9 5
## Up 34 56(9+56)/104## [1] 0.625Exercise 13.(e)
Repeat (d) using LDA
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlda.fit <- lda(Direction ~ Lag2, data = Weekly, subset = Train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = Train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162plot(lda.fit)
lda.pred <- predict(lda.fit, Weekly_Test)
names(lda.pred)## [1] "class" "posterior" "x"lda.class <- lda.pred$class
table(lda.class, Weekly_Test$Direction)##
## lda.class Down Up
## Down 9 5
## Up 34 56(9+56)/104## [1] 0.625Exercise 13.(f)
Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly, subset = Train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = Train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class <- predict(qda.fit, Weekly_Test)$class
table(qda.class, Weekly_Test$Direction)##
## qda.class Down Up
## Down 0 0
## Up 43 61(0+61)/104## [1] 0.5865385Exercise 13.(g)
Repeat (d) using KNN with K = 1.
library(class)
Train.X <- data.frame(Lag2=Weekly_Train$Lag2)
Test.X <- data.frame(Lag2=Weekly_Test$Lag2)
Train.y <- Weekly_Train$Direction
Test.y <- Weekly_Test$Direction
set.seed(1)
knn.pred <- knn(Train.X, Test.X, Weekly_Train$Direction, k = 1 )
table(Predicted = knn.pred, Actual = Test.y)## Actual
## Predicted Down Up
## Down 21 30
## Up 22 31(21+31)/104## [1] 0.5Exercise 13.(h)
Repeat (d) using naive Bayes.
library(e1071)
nb.fit <- naiveBayes(Direction ~ Lag2, data=Weekly, subset=Train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb.class <- predict(nb.fit, Weekly_Test)
table(nb.class, Weekly_Test$Direction)##
## nb.class Down Up
## Down 0 0
## Up 43 61(0+61)/104## [1] 0.5865385Exercise 13.(i)
Which of these methods appears to provide the best results on this data?
logistic regression and LDA have the best results
Exercise 13.(j)
Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
**KNN with K=4 is best case
library(class)
Train.X <- data.frame(Lag2=Weekly_Train$Lag2)
Test.X <- data.frame(Lag2=Weekly_Test$Lag2)
#Train.X <- cbind(Lag1, Lag2)[Train,]
#Test.X <- cbind(Lag1, Lag2)[Test,]
Train.y <- Weekly_Train$Direction
Test.y <- Weekly_Test$Direction
set.seed(1)
knn.pred <- knn(Train.X, Test.X, Weekly_Train$Direction, k = 4 )
table(Predicted = knn.pred, Actual = Test.y)## Actual
## Predicted Down Up
## Down 20 17
## Up 23 44(20+44)/104## [1] 0.6153846Add Predictor to lda => It does not work.
library(MASS)
lda.fit2 <- lda(Direction ~ Lag2+Lag1, data = Weekly, subset = Train)
lda.fit2## Call:
## lda(Direction ~ Lag2 + Lag1, data = Weekly, subset = Train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2 Lag1
## Down -0.03568254 0.289444444
## Up 0.26036581 -0.009213235
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.2982579
## Lag1 -0.3013148lda.pred2 <- predict(lda.fit2, Weekly_Test)
lda.class2 <- lda.pred2$class
table(lda.class2, Weekly_Test$Direction)##
## lda.class2 Down Up
## Down 7 8
## Up 36 53(9+56)/104## [1] 0.625Exercise 14
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
Exercise 14.(a)
Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg_median <- median(Auto$mpg)
mpg_median## [1] 22.75mpg_01 <- ifelse(Auto$mpg > mpg_median, 1, 0)
Auto_01 <- data.frame(Auto, mpg_01)
summary(Auto_01)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
## mpg_01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## Exercise 14.(b)
Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
Answer:displacement, horse power and weight are most useful predictor.
pairs(Auto_01)
library(ggplot2)
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:MASS':
##
## select## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionggplot(Auto_01, aes(x=factor(mpg_01), y=displacement)) + geom_boxplot()
ggplot(Auto_01, aes(x=factor(mpg_01), y=horsepower)) + geom_boxplot()
ggplot(Auto_01, aes(x=factor(mpg_01), y=weight)) + geom_boxplot()
ggplot(Auto_01, aes(x=factor(mpg_01), y=acceleration)) + geom_boxplot()
ggplot(Auto_01, aes(x=factor(mpg_01), y=year)) + geom_boxplot()
ggplot(Auto_01, aes(x=mpg, y=year)) + geom_point()
ggplot(Auto_01, aes(x=mpg, y=cylinders)) + geom_point()
ggplot(Auto_01, aes(x=mpg, y=origin)) + geom_point()
Exercise 14.(c)
Split the data into a training set and a test set.
set.seed(1)
n <- length(Auto_01$mpg)
Select_Idx <- sample(1:n, size = 0.7 * n)
Auto_Train <- Auto_01[Select_Idx,]
Auto_Test <- Auto_01[-Select_Idx,]Exercise 14.(d)
Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Answer: Test error rate is 12.71%
lda.mpg.fit <- lda(Auto_Train$mpg_01 ~ displacement + horsepower + weight + acceleration + year, data = Auto_Train)
lda.mpg.fit## Call:
## lda(Auto_Train$mpg_01 ~ displacement + horsepower + weight +
## acceleration + year, data = Auto_Train)
##
## Prior probabilities of groups:
## 0 1
## 0.4927007 0.5072993
##
## Group means:
## displacement horsepower weight acceleration year
## 0 271.9333 129.13333 3611.052 14.71556 74.47407
## 1 116.8129 79.27338 2342.165 16.57338 77.60432
##
## Coefficients of linear discriminants:
## LD1
## displacement -0.0102367198
## horsepower 0.0124774848
## weight -0.0010441618
## acceleration -0.0005208118
## year 0.1116760488lda.mpg.pred <- predict(lda.mpg.fit, Auto_Test)
lda.mpg.class <- lda.mpg.pred$class
table(lda.mpg.class, Auto_Test$mpg_01)##
## lda.mpg.class 0 1
## 0 47 1
## 1 14 56mean(lda.mpg.class != Auto_Test$mpg_01)## [1] 0.1271186Exercise 14.(e)
Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Answer: Test Error Rate is 12.71%
qda.mpg.fit <- qda(mpg_01 ~ displacement + horsepower + weight + acceleration + year, data = Auto_Train)
qda.mpg.fit## Call:
## qda(mpg_01 ~ displacement + horsepower + weight + acceleration +
## year, data = Auto_Train)
##
## Prior probabilities of groups:
## 0 1
## 0.4927007 0.5072993
##
## Group means:
## displacement horsepower weight acceleration year
## 0 271.9333 129.13333 3611.052 14.71556 74.47407
## 1 116.8129 79.27338 2342.165 16.57338 77.60432qda.mpg.pred <- predict(qda.mpg.fit, Auto_Test)
qda.mpg.class <- qda.mpg.pred$class
table(qda.mpg.class, Auto_Test$mpg_01)##
## qda.mpg.class 0 1
## 0 51 5
## 1 10 52mean(qda.mpg.class != Auto_Test$mpg_01)## [1] 0.1271186Exercise 14.(f)
Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Answer: Test Error Rate is 10.17%
glm.mpg.fit <- glm(mpg_01 ~ displacement + horsepower + weight + acceleration + year, data=Auto_Train, family = binomial)
glm.mpg.prob <- predict(glm.mpg.fit, Auto_Test, type = "response")
glm.mpg.class <- ifelse(glm.mpg.prob > 0.5, 1, 0)
table(glm.mpg.class, Auto_Test$mpg_01)##
## glm.mpg.class 0 1
## 0 53 4
## 1 8 53mean(glm.mpg.class != Auto_Test$mpg_01)## [1] 0.1016949Exercise 14.(g)
Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Answer: Test Error Rate: 11.02%
nb.mpg.fit <- naiveBayes(mpg_01 ~ displacement + horsepower + weight + acceleration + year, data = Auto_Train)
nb.mpg.pred <- predict(nb.mpg.fit, Auto_Test)
table(nb.mpg.pred, Auto_Test$mpg_01)##
## nb.mpg.pred 0 1
## 0 49 1
## 1 12 56mean(nb.mpg.pred != Auto_Test$mpg_01)## [1] 0.1101695Exercise 14.(h)
Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
Answer: Test Error Rate is 10.17% with K=3
Train.mpg.X <- Auto_Train[, c("displacement", "horsepower", "weight", "acceleration", "year")]
Test.mpg.X <- Auto_Test[, c("displacement", "horsepower", "weight", "acceleration", "year")]
Train.mpg.X <- as.data.frame(Train.mpg.X)
Test.mpg.X <- as.data.frame(Test.mpg.X)
Train.mpg.Y <- Auto_Train$mpg_01
Test.mpg.Y <- Auto_Test$mpg_01
# k = 1 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 1)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 51 6
## 1 10 51mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1355932# k = 2 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 2)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 52 9
## 1 9 48mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1525424# k = 3 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 3)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 52 3
## 1 9 54mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1016949# k = 4 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 4)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 51 4
## 1 10 53mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1186441# k = 5 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 5)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 51 5
## 1 10 52mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1271186# k = 6 case
knn.mpg.pred <- knn(Train.mpg.X, Test.mpg.X, cl = Train.mpg.Y, k = 6)
table(Predicted = knn.mpg.pred, Actual = Test.mpg.Y)## Actual
## Predicted 0 1
## 0 50 4
## 1 11 53mean(knn.mpg.pred != Test.mpg.Y)## [1] 0.1271186Exercise 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
library(MASS) # for Boston dataset and LDA/QDA
library(class) # for KNN
library(e1071) # for Naive Bayes
library(dplyr) # for data manipulation
data(Boston)
median_crim <- median(Boston$crim) # Calc Median Value
Boston$crim_01 <- ifelse(Boston$crim > median_crim, 1, 0)
pairs(Boston)
all_predictors <- c('zn', 'indus', 'chas', 'nox', 'rm', 'age', 'dis', 'rad', 'tax', 'ptratio', 'black', 'lstat')
cor_related <- c('nox', 'rad', 'tax', 'dis', 'lstat', 'black')
small_subset <- c('nox', 'rad', 'tax', 'dis')
Boston_wo <- Boston %>% select(-crim)set.seed(1)
n <- nrow(Boston)
train_idx <- sample(1:n, size = 0.7 * n)
Boston_train <- Boston[train_idx, ]
Boston_test <- Boston[-train_idx, ]Logistic Regression
# ===== case all_predictors ==== #
predictor_set <- all_predictors
# Build formula dynamically
formula <- as.formula(
paste("crim_01 ~", paste(predictor_set, collapse = " + "))
)
# Train logistic regression model
glm.fit <- glm(formula, data = Boston_train, family = binomial)
# Predict probabilities on test set
glm.prob <- predict(glm.fit, Boston_test, type = "response")
# Convert probabilities to class labels
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
# Confusion matrix and accuracy
table(Predicted = glm.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 61 6
## 1 12 73mean(glm.pred == Boston_test$crim_01) # Accuracy## [1] 0.8815789# ===== case cor_related ==== #
predictor_set <- cor_related
# Build formula dynamically
formula <- as.formula(
paste("crim_01 ~", paste(predictor_set, collapse = " + "))
)
# Train logistic regression model
glm.fit <- glm(formula, data = Boston_train, family = binomial)
# Predict probabilities on test set
glm.prob <- predict(glm.fit, Boston_test, type = "response")
# Convert probabilities to class labels
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
# Confusion matrix and accuracy
table(Predicted = glm.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 58 13
## 1 15 66mean(glm.pred == Boston_test$crim_01) # Accuracy## [1] 0.8157895# ===== case Small Subset ==== #
predictor_set <- small_subset
# Build formula dynamically
formula <- as.formula(
paste("crim_01 ~", paste(predictor_set, collapse = " + "))
)
# Train logistic regression model
glm.fit <- glm(formula, data = Boston_train, family = binomial)
# Predict probabilities on test set
glm.prob <- predict(glm.fit, Boston_test, type = "response")
# Convert probabilities to class labels
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
# Confusion matrix and accuracy
table(Predicted = glm.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 58 14
## 1 15 65mean(glm.pred == Boston_test$crim_01) # Accuracy## [1] 0.8092105LDA
# ===== case all_predictors ==== #
predictor_set <- all_predictors
formula <- as.formula( paste("crim_01 ~", paste(predictor_set, collapse = " + ")))
lda.fit <- lda(formula, data = Boston_train)
lda.pred <- predict(lda.fit, Boston_test)$class
# Confusion matrix & accuracy
table(Predicted = lda.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 71 19
## 1 2 60mean(lda.pred == Boston_test$crim_01)## [1] 0.8618421# ===== case cor_related ==== #
predictor_set <- cor_related
formula <- as.formula( paste("crim_01 ~", paste(predictor_set, collapse = " + ")))
lda.fit <- lda(formula, data = Boston_train)
lda.pred <- predict(lda.fit, Boston_test)$class
# Confusion matrix & accuracy
table(Predicted = lda.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 69 20
## 1 4 59mean(lda.pred == Boston_test$crim_01)## [1] 0.8421053# ===== case Small Subset ==== #
predictor_set <- small_subset
formula <- as.formula( paste("crim_01 ~", paste(predictor_set, collapse = " + ")))
lda.fit <- lda(formula, data = Boston_train)
lda.pred <- predict(lda.fit, Boston_test)$class
# Confusion matrix & accuracy
table(Predicted = lda.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 69 19
## 1 4 60mean(lda.pred == Boston_test$crim_01)## [1] 0.8486842Naive Bayes
# ===== case all_predictors ==== #
predictor_set <- all_predictors
nb.fit <- naiveBayes(crim_01 ~ . -crim_01, data = Boston_train)
nb.pred <- predict(nb.fit, newdata = Boston_test)
table(Predicted = nb.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 70 5
## 1 3 74mean(nb.pred == Boston_test$crim_01)## [1] 0.9473684# ===== case cor_related ==== #
predictor_set <- cor_related
nb.fit <- naiveBayes(crim_01 ~ . -crim_01, data = Boston_train)
nb.pred <- predict(nb.fit, newdata = Boston_test)
table(Predicted = nb.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 70 5
## 1 3 74mean(nb.pred == Boston_test$crim_01)## [1] 0.9473684# ===== case Small Subset ==== #
predictor_set <- small_subset
nb.fit <- naiveBayes(crim_01 ~ . -crim_01, data = Boston_train)
nb.pred <- predict(nb.fit, newdata = Boston_test)
table(Predicted = nb.pred, Actual = Boston_test$crim_01)## Actual
## Predicted 0 1
## 0 70 5
## 1 3 74mean(nb.pred == Boston_test$crim_01)## [1] 0.9473684KNN
# ===== case all_predictors ==== #
predictor_set <- all_predictors
train.X <- scale(Boston_train[, predictor_set])
test.X <- scale(
Boston_test[, predictor_set],
center = attr(train.X, "scaled:center"),
scale = attr(train.X, "scaled:scale")
)
train.y <- Boston_train$crim_01
test.y <- Boston_test$crim_01
for (k in 1:5) {
knn.pred <- knn(train = train.X,
test = test.X,
cl = train.y,
k = k)
cat("all_predictors ","K =", k, "Accuracy =", mean(knn.pred == test.y), "\n")
}## all_predictors K = 1 Accuracy = 0.9276316
## all_predictors K = 2 Accuracy = 0.9078947
## all_predictors K = 3 Accuracy = 0.9605263
## all_predictors K = 4 Accuracy = 0.9539474
## all_predictors K = 5 Accuracy = 0.9407895# ===== case cor_related ==== #
predictor_set <- cor_related
train.X <- scale(Boston_train[, predictor_set])
test.X <- scale(
Boston_test[, predictor_set],
center = attr(train.X, "scaled:center"),
scale = attr(train.X, "scaled:scale")
)
train.y <- Boston_train$crim_01
test.y <- Boston_test$crim_01
for (k in 1:5) {
knn.pred <- knn(train = train.X,
test = test.X,
cl = train.y,
k = k)
cat("cor_related ","K =", k, "Accuracy =", mean(knn.pred == test.y), "\n")
}## cor_related K = 1 Accuracy = 0.9473684
## cor_related K = 2 Accuracy = 0.9342105
## cor_related K = 3 Accuracy = 0.9407895
## cor_related K = 4 Accuracy = 0.9144737
## cor_related K = 5 Accuracy = 0.9473684# ===== case Small Subset ==== #
predictor_set <- small_subset
train.X <- scale(Boston_train[, predictor_set])
test.X <- scale(
Boston_test[, predictor_set],
center = attr(train.X, "scaled:center"),
scale = attr(train.X, "scaled:scale")
)
train.y <- Boston_train$crim_01
test.y <- Boston_test$crim_01
for (k in 1:5) {
knn.pred <- knn(train = train.X,
test = test.X,
cl = train.y,
k = k)
cat("Small Subset ","K =", k, "Accuracy =", mean(knn.pred == test.y), "\n")
}## Small Subset K = 1 Accuracy = 0.9605263
## Small Subset K = 2 Accuracy = 0.9539474
## Small Subset K = 3 Accuracy = 0.9605263
## Small Subset K = 4 Accuracy = 0.9539474
## Small Subset K = 5 Accuracy = 0.9605263