Assignment 3
John Shunk
2022-10-07
13.)
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.1.3View(Weekly)
Weekly=Weekly- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(corrplot)## corrplot 0.92 loadedsummary(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
##
##
##
## corrplot(cor(Weekly[,-9]), method="square")
The only variables with high correlation are Volume and Year.
- 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 statistically significant variable is Lag2.
- 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.
logWeekly.prob = predict(Weekly.fit, type='response')
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
table(logWeekly.pred, Weekly$Direction)##
## logWeekly.pred Down Up
## Down 54 48
## Up 430 557(557+54)/(54+48+430+557)## [1] 0.5610652557/(557+48)## [1] 0.920661254/(54+430)## [1] 0.1115702The model is correct roughly 56% of the time. Up weekly trends were correctly predicted 92% of the time, but down weekly trends were correctly predicted only 11% of the time.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
Weekly.train = (Weekly$Year<2009)
Weekly.0910 = Weekly[!Weekly.train,]
Weekly.fit = glm(Direction~Lag2, data=Weekly,family=binomial, subset=Weekly.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 = Weekly$Direction[!Weekly.train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 9 5
## Up 34 56mean(logWeekly.pred == Direction.0910)## [1] 0.625This model correctly predicts trends with 62.5% accuracy, a moderate improvement over the last model. This model is also better at predicting upward trends than downward trends.
- Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## BostonWeeklylda.fit = lda(Direction~Lag2, data=Weekly, family=binomial, subset=Weekly.train)
Weeklylda.pred<-predict(Weeklylda.fit, Weekly.0910)
table(Weeklylda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 9 5
## Up 34 56mean(Weeklylda.pred$class==Direction.0910)## [1] 0.62562.5% accurate.
- Repeat (d) using QDA.
Weeklyqda.fit = qda(Direction ~ Lag2, data = Weekly, subset = Weekly.train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda.pred==Direction.0910)## [1] 0.586538558.65% accurate.
- Repeat (d) using KNN with K = 1.
library(class)
Week.train=as.matrix(Weekly$Lag2[Weekly.train])
Week.test=as.matrix(Weekly$Lag2[!Weekly.train])
train.Direction = Weekly$Direction[Weekly.train]
set.seed(99)
Weekknn.pred = knn(Week.train,Week.test,train.Direction,k=1)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 21 30
## Up 22 31mean(Weekknn.pred == Direction.0910)## [1] 0.550% accuracy. Basically equal to random chance.
- Repeat (d) using naive Bayes.
library(e1071)
week.bayes = naiveBayes(Weekly$Direction ~ Lag2 ,data=Weekly ,subset=Weekly.train)
week.bayes.class = predict(week.bayes ,Weekly.0910)
table(week.bayes.class, Direction.0910)## Direction.0910
## week.bayes.class Down Up
## Down 0 0
## Up 43 61mean(week.bayes.class == Direction.0910)## [1] 0.586538558.65% accurate.
- Which of these methods appears to provide the best results on this data?
Logistic Regression and Linear Discriminant Analysis had the best accuracy, both having rates of 62.5%.
14.)
library(knitr)
library(tidyverse)## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --## v ggplot2 3.3.5 v purrr 0.3.4
## v tibble 3.1.6 v dplyr 1.0.7
## v tidyr 1.1.4 v stringr 1.4.0
## v readr 2.1.1 v forcats 0.5.1## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()
## x dplyr::select() masks MASS::select()View(Auto)
auto = Auto- 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.
set.seed(14)
df = auto
df = df[,1:8]
med = median(df$mpg)
df$mpg01 = ifelse(df$mpg > med, 1, 0)
df$mpg01 = as.factor(df$mpg01)- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
plot(df$mpg01, df$horsepower) #Good predictor
plot(df$mpg01, df$cylinders)
plot(df$mpg01, df$acceleration)
plot(df$mpg01, df$displacement) #Good
plot(df$mpg01, df$weight) #Good
plot(df$mpg01, df$year)
- Split the data into a training set and a test set.
rows = sample(1:nrow(df), 0.75*nrow(df))
auto.train = df[rows,]
auto.test = df[-rows,]- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.auto = lda(mpg01 ~ horsepower + displacement + weight, data = auto.train)
lda.pred = predict(lda.auto, auto.test)
lda.class = lda.pred$class
table(lda.class , auto.test$mpg01)##
## lda.class 0 1
## 0 38 4
## 1 9 4786.7% accurate. Test error is 13.3%.
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.auto = qda(mpg01 ~ horsepower + displacement + weight, data = auto.train)
qda.pred = predict(qda.auto, auto.test)
qda.class = qda.pred$class
table(qda.class, auto.test$mpg01)##
## qda.class 0 1
## 0 39 5
## 1 8 4686.7% accurate. Test error is 13.3%
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
log.auto = glm(mpg01 ~ horsepower + displacement + weight, family = 'binomial', data = auto.train)
log.pred = predict(log.auto, auto.test, type = 'response')
log.pred = as.factor(ifelse(log.pred > 0.5, 1, 0))
table(log.pred, auto.test$mpg01)##
## log.pred 0 1
## 0 39 4
## 1 8 4787.8% accurate. Test error is 12.2%
- 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?
bayes.auto = naiveBayes(mpg01 ~ horsepower + displacement + weight, data = auto.train)
bayes.class = predict(bayes.auto, auto.test)
table(bayes.class, auto.test$mpg01)##
## bayes.class 0 1
## 0 38 4
## 1 9 4786.7% accurate. Test error is 13.3%
- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
library(class)
knn_pred_y = knn(auto.train, auto.test, auto.train$mpg01, k = 1)
table(knn_pred_y, auto.test$mpg01)##
## knn_pred_y 0 1
## 0 39 7
## 1 8 44knn_pred_y = knn(auto.train, auto.test, auto.train$mpg01, k = 2)
table(knn_pred_y, auto.test$mpg01)##
## knn_pred_y 0 1
## 0 39 5
## 1 8 46knn_pred_y = knn(auto.train, auto.test, auto.train$mpg01, k = 3)
table(knn_pred_y, auto.test$mpg01)##
## knn_pred_y 0 1
## 0 39 5
## 1 8 46knn_pred_y = knn(auto.train, auto.test, auto.train$mpg01, k = 4)
table(knn_pred_y, auto.test$mpg01)##
## knn_pred_y 0 1
## 0 38 6
## 1 9 45K=1: 84.7% accurate. Test error is 15.3% K=2: 86.7% accurate. Test error is 13.3% K=3: 86.7% accurate. Test error is 13.3% K=4: 84.7% accurate. Test error is 15.3%
Model is most accurate when K=2 or K=3.
16.) Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
boston = Boston
boston$chas = as.factor(boston$chas)
boston$crime_factor = factor(ifelse(boston$crim > median(boston$crim),
'High', 'Low'),
levels = c('High', 'Low'))
cor_test <- boston %>%
select(-chas, -crime_factor) %>%
cor.mtest(conf.level = .95)
boston %>%
select(-chas, -crime_factor) %>%
cor %>%
corrplot(method = 'color',
order = 'hclust', addrect = 2,
tl.col = 'black', addCoef.col = 'black', number.cex = 0.65,
p.mat = cor_test$p, sig.level = .05)
These variables have a fairly high correlation with crime rate: lsat,
indus, nox, rad, tax.
set.seed(6)
rows = sample(1:nrow(boston), .75*nrow(boston))
bost.train = (boston[rows,])
bost.test = (boston[-rows,])
log.bost = glm(crime_factor ~ indus + nox + rad + tax + lstat, family = 'binomial', data = bost.train)
log.pred = predict(log.bost , bost.test, type = 'response')
log.pred = as.factor(ifelse(log.pred > 0.5, 1, 0))
table(log.pred , bost.test$crime_factor)##
## log.pred High Low
## 0 47 6
## 1 13 61Test error is 13.4%.
lda.bost = lda(crime_factor ~ indus + nox + rad + tax + lstat, data = bost.train)
lda.pred = predict(lda.bost, bost.test)
lda.class = lda.pred$class
table(lda.class, bost.test$crime_factor)##
## lda.class High Low
## High 41 2
## Low 19 65Test error is 18.9%.
qda.bost = qda(crime_factor ~ indus + nox + rad + tax + lstat, data = bost.train)
qda.pred = predict(qda.bost, bost.test)
qda.class <- qda.pred$class
table(qda.class, bost.test$crime_factor)##
## qda.class High Low
## High 44 2
## Low 16 65Test error is 15.0%.
trainKNN = bost.train[, c('indus','nox','rad','tax', 'lstat')]
testKNN = bost.test[, c('indus','nox','rad','tax', 'lstat')]
n = 20
testError = sapply(1:n,function(k){
knn.pred = knn(train = trainKNN, test = testKNN, cl = bost.train$crime_factor, k = k)
mean(knn.pred!=bost.test$crime_factor)
})
testError## [1] 0.07874016 0.09448819 0.06299213 0.04724409 0.03937008 0.03937008
## [7] 0.04724409 0.03937008 0.03937008 0.04724409 0.04724409 0.04724409
## [13] 0.04724409 0.06299213 0.04724409 0.04724409 0.06299213 0.06299213
## [19] 0.06299213 0.05511811When K=2, test error is 6.30%. This is by far the most accurate model.