Assignment 3
Drew Hinck
2023-02-21
Problem 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.
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR2)
library(corrplot)attach(Weekly)
summary(Weekly)## 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
##
##
##
## str(Weekly)## 'data.frame': 1089 obs. of 9 variables:
## $ Year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
## $ Lag1 : num 0.816 -0.27 -2.576 3.514 0.712 ...
## $ Lag2 : num 1.572 0.816 -0.27 -2.576 3.514 ...
## $ Lag3 : num -3.936 1.572 0.816 -0.27 -2.576 ...
## $ Lag4 : num -0.229 -3.936 1.572 0.816 -0.27 ...
## $ Lag5 : num -3.484 -0.229 -3.936 1.572 0.816 ...
## $ Volume : num 0.155 0.149 0.16 0.162 0.154 ...
## $ Today : num -0.27 -2.576 3.514 0.712 1.178 ...
## $ Direction: Factor w/ 2 levels "Down","Up": 1 1 2 2 2 1 2 2 2 1 ...corrplot(cor(Weekly[,-9]), method = "number")
There is a very high correlation between Volume and
Year. Also the minimum value for each of the
Lag variables is identical, as well as the minimum for the
Today variable.
(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?
Weekly.fit=glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data = Weekly, family = binomial)
summary(Weekly.fit)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## 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: 4The only variable that is statistically significant is
Lag2.
(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.
glmprob.wk = predict(Weekly.fit, type = "response")
glmpred.wk = rep("Down", length(glmprob.wk))
glmpred.wk[glmprob.wk > 0.5] <- "Up"
table(glmpred.wk, Weekly$Direction)##
## glmpred.wk Down Up
## Down 54 48
## Up 430 557mean(glmpred.wk == Weekly$Direction)## [1] 0.5610652This confusion matrix shows an accuracy of only about 56%. The model is very good at predicting the positive class (Up) at about 92%, but is terrible at predicting the negative class (Down) at only 11% accuracy.
(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=(Year<2009)
Weekly.0910 <-Weekly[!train,]
Weekly.fit<-glm(Direction~Lag2, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit, Weekly.0910, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 9 5
## Up 34 56mean(logWeekly.pred == Direction.0910)## [1] 0.625(e) Repeat (d) using LDA.
library(MASS)
Weeklylda.fit<-lda(Direction~Lag2, data=Weekly,family=binomial, subset=train)
Weeklylda.pred<-predict(Weeklylda.fit, Weekly.0910)
table(Weeklylda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 9 5
## Up 34 56mean(Weeklylda.pred$class==Direction.0910)## [1] 0.625(f) Repeat (d) using QDA.
Weeklyqda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda.pred==Direction.0910)## [1] 0.5865385(g) Repeat (d) using KNN with K = 1.
library(class)
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=1)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 21 30
## Up 22 31mean(Weekknn.pred == Direction.0910)## [1] 0.5(h) Repeat (d) using naive Bayes.
library(e1071)
nbayes=naiveBayes(Direction~Lag2 ,data=Weekly ,subset=train)
nbayes##
## 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 2.317485nbayes.class=predict(nbayes ,Weekly.0910)
table(nbayes.class ,Direction.0910)## Direction.0910
## nbayes.class Down Up
## Down 0 0
## Up 43 61(i) Which of these methods appears to provide the best results on this data?
The regression model privided the highest level of accuracy compared to all other models.
(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.
#Log Reg with interaction
Weekly.fit2<-glm(Direction~Lag2:Lag3+Lag2+Lag3, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit2, Weekly.0910, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 8 4
## Up 35 57mean(logWeekly.pred == Direction.0910)## [1] 0.625# LDA with interaction
Weeklylda.fit2<-lda(Direction~Lag2:Lag3+Lag2+Lag3, data=Weekly,family=binomial, subset=train)
Weeklylda.pred2<-predict(Weeklylda.fit2, Weekly.0910)
table(Weeklylda.pred2$class, Direction.0910)## Direction.0910
## Down Up
## Down 8 4
## Up 35 57mean(Weeklylda.pred2$class==Direction.0910)## [1] 0.625# QDA with interaction
qda2=qda(Direction~Lag2:Lag3 ,data=Weekly ,subset=train)
classqda2=predict(qda2 ,Weekly.0910)$class
table(classqda2 ,Direction.0910)## Direction.0910
## classqda2 Down Up
## Down 6 8
## Up 37 53mean(classqda2==Direction.0910)## [1] 0.5673077# K=15
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=15)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 20 20
## Up 23 41mean(Weekknn.pred == Direction.0910)## [1] 0.5865385# K=50
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=50)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 20 23
## Up 23 38mean(Weekknn.pred == Direction.0910)## [1] 0.5576923detach(Weekly)Problem 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.
library(ISLR)
attach(Auto)
summary(Auto)## 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(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.
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto = data.frame(Auto, mpg01)(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.
corrplot(cor(Auto[,-9]), method="number")
pairs(Auto)
mpg01 has a high negative correlation with
cylinders, displacement,
horsepower, and weight. origin
and year also has a moderately high correlation with
mpg01.
(c) Split the data into a training set and a test set.
train <- (year %% 2 == 0)
train.auto <- Auto[train,]
test.auto <- Auto[-train,](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?
autolda.fit <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
autolda.pred <- predict(autolda.fit, test.auto)
table(autolda.pred$class, test.auto$mpg01)##
## 0 1
## 0 169 7
## 1 26 189mean(autolda.pred$class != test.auto$mpg01)## [1] 0.08439898The test error in the LDA model is 0.0844, or about 8.44%.
(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?
autoqda.fit <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
autoqda.pred <- predict(autoqda.fit, test.auto)
table(autoqda.pred$class, test.auto$mpg01)##
## 0 1
## 0 176 20
## 1 19 176mean(autoqda.pred$class != test.auto$mpg01)## [1] 0.09974425The test error rate for the QDA model is 9.97%.
(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?
auto.fit<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto,family=binomial)
auto.probs = predict(auto.fit, test.auto, type = "response")
auto.pred = rep(0, length(auto.probs))
auto.pred[auto.probs > 0.5] = 1
table(auto.pred, test.auto$mpg01)##
## auto.pred 0 1
## 0 174 12
## 1 21 184mean(auto.pred != test.auto$mpg01)## [1] 0.08439898The test error rate for the logistic regression model is 8.44%.
(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?
nbayes=naiveBayes(mpg01~displacement+horsepower+weight+year+cylinders+origin ,data=Auto, subset = train)
nbayes.class = predict(nbayes, test.auto)
table(nbayes.class, test.auto$mpg01)##
## nbayes.class 0 1
## 0 171 15
## 1 24 181mean(nbayes.class != test.auto$mpg01)## [1] 0.09974425The test error rate for the Naive Bayes test is 9.97%
(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?
# K=1
train.K= cbind(displacement,horsepower,weight,cylinders,year, origin)[train,]
test.K=cbind(displacement,horsepower,weight,cylinders, year, origin)[-train,]
set.seed(1)
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=1)
mean(autok.pred != test.auto$mpg01)## [1] 0.07161125# K=20
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=20)
mean(autok.pred != test.auto$mpg01)## [1] 0.1150895# K=30
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=30)
mean(autok.pred != test.auto$mpg01)## [1] 0.1227621The value of K=1 performs the best on this data with an error variance of 7.16%.
detach(Auto)Problem 16
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.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00attach(Boston)Binary crime variable
crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] <- 1
Boston= data.frame(Boston,crime01)Training/testing split
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]corrplot(cor(Boston), method="number")
The variables indus, nox, age,
dis, rad, and tax are most highly
correlated with the dependent variable crime01.
# Logistic regression
set.seed(1)
Boston.fit <-glm(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Boston.probs = predict(Boston.fit, Boston.test, type = "response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1
table(Boston.pred, crime01.test)## crime01.test
## Boston.pred 0 1
## 0 75 8
## 1 15 155mean(Boston.pred != crime01.test)## [1] 0.09090909summary(Boston.fit)##
## Call:
## glm(formula = crime01 ~ indus + nox + age + dis + rad + tax,
## family = binomial, data = Boston.train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.97810 -0.21406 -0.03454 0.47107 3.04502
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -42.214032 7.617440 -5.542 2.99e-08 ***
## indus -0.213126 0.073236 -2.910 0.00361 **
## nox 80.868029 16.066473 5.033 4.82e-07 ***
## age 0.003397 0.012032 0.282 0.77772
## dis 0.307145 0.190502 1.612 0.10690
## rad 0.847236 0.183767 4.610 4.02e-06 ***
## tax -0.013760 0.004956 -2.777 0.00549 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.37 on 252 degrees of freedom
## Residual deviance: 144.44 on 246 degrees of freedom
## AIC: 158.44
##
## Number of Fisher Scoring iterations: 8# LDA
Boston.ldafit <-lda(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Bostonlda.pred = predict(Boston.ldafit, Boston.test)
table(Bostonlda.pred$class, crime01.test)## crime01.test
## 0 1
## 0 81 18
## 1 9 145mean(Bostonlda.pred$class != crime01.test)## [1] 0.1067194# Naive Bayes
nbayes=naiveBayes(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train, subset = train)
nbayes.class = predict(nbayes, Boston.test)
table(nbayes.class, Boston.test$crime01)##
## nbayes.class 0 1
## 0 77 13
## 1 13 150mean(nbayes.class != Boston.test$crime01)## [1] 0.1027668# KNN K=1
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=1)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 31 155
## 1 59 8mean(Bosknn.pred !=crime01.test)## [1] 0.8458498# KNN K=50
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=50)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 37 14
## 1 53 149mean(Bosknn.pred !=crime01.test)## [1] 0.2648221Logistic regression provided the lowest error rate of all the models
at 9.09%. The error rate for the LDA and Naive Bayes were similar at
10.67% and 10.28% respectively, while the KNN classifier was much higher
at K=1 with and error rate of 84.58% and K=50 having an error rate of
26.48%. In order to compare the models accurately, the same variables
were used for each model. The variables that were chosen were
indus, nox, age,
dis, rad, and tax because of
their high correlation with the dependent variable
crime01.