STA 4133 Assignment#3
Jiaying Liang
March 4, 2021
10.This question should be answered using the Weekly data set, which is part of the ISLR 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.
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)
library(corrplot)## Warning: package 'corrplot' was built under R version 4.0.4## corrplot 0.84 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")
- 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?
attach(Weekly)
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: 4- 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, Direction)## Direction
## logWeekly.pred Down Up
## Down 54 48
## Up 430 557- 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- 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- Repeat (d) using QDA.
Weeklyqda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda.pred==Direction.0910)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
library(class)
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=1)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 21 30
## Up 22 31mean(Weekknn.pred == Direction.0910)## [1] 0.5Which of these methods appears to provide the best results on this data? The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis; both having rates of 62.5%.
Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
#Logistic Regression with Interaction Lag2:Lag4
Weekly.fit<-glm(Direction~Lag2:Lag4+Lag2, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit, Weekly.0910, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 3 4
## Up 40 57mean(logWeekly.pred == Direction.0910)## [1] 0.5769231#LDA with Interaction Lag2:Lag4
Weeklylda.fit<-lda(Direction~Lag2:Lag4+Lag2, data=Weekly,family=binomial, subset=train)
Weeklylda.pred<-predict(Weeklylda.fit, Weekly.0910)
table(Weeklylda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 3 3
## Up 40 58mean(Weeklylda.pred$class==Direction.0910)## [1] 0.5865385Weeklyqda.fit = qda(Direction ~ poly(Lag2,2), data = Weekly, subset = train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 7 3
## Up 36 58mean(Weeklyqda.pred==Direction.0910)## [1] 0.625#K=10
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=10)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 17 21
## Up 26 40mean(Weekknn.pred == Direction.0910)## [1] 0.5480769#K=100
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=100)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 10 11
## Up 33 50mean(Weekknn.pred == Direction.0910)## [1] 0.5769231Q11.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.
- 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.
require(ISLR); require(tidyverse); require(ggthemes); require(GGally)## Loading required package: tidyverse## Warning: package 'tidyverse' was built under R version 4.0.4## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --## v ggplot2 3.3.3 v purrr 0.3.4
## v tibble 3.0.6 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()
## x dplyr::select() masks MASS::select()## Loading required package: ggthemes## Warning: package 'ggthemes' was built under R version 4.0.4## Loading required package: GGally## Warning: package 'GGally' was built under R version 4.0.4## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2require(knitr); require(kableExtra); require(broom)## Loading required package: knitr## Loading required package: kableExtra## Warning: package 'kableExtra' was built under R version 4.0.4##
## Attaching package: 'kableExtra'## The following object is masked from 'package:dplyr':
##
## group_rows## Loading required package: broomtheme_set(theme_tufte(base_size = 14))
set.seed(1)
data('Auto')
Auto <- Auto %>%
filter(!cylinders %in% c(3,5)) %>%
mutate(mpg01 = factor(ifelse(mpg > median(mpg), 1, 0)),
cylinders = factor(cylinders,
levels = c(4,6,8),
ordered = TRUE),
origin = factor(origin,
levels = c(1,2,3),
labels = c('American', 'European', 'Asian')))
median(Auto$mpg)## [1] 23Auto %>%
dplyr::select(mpg, mpg01) %>%
sample_n(5)## mpg mpg01
## 333 29.8 1
## 171 23.0 0
## 133 25.0 1
## 306 28.4 1
## 276 17.0 0- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
Auto %>%
dplyr::select(-name, -mpg) %>%
ggpairs(aes(col = mpg01, fill = mpg01, alpha = 0.6),
upper = list(combo = 'box'),
diag = list(discrete = wrap('barDiag', position = 'fill')),
lower = list(combo = 'dot_no_facet')) +
theme(axis.text.x = element_text(angle = 90, hjust = 1))
Auto %>%
dplyr::select(-name, -mpg, - origin, -cylinders) %>%
gather(Variable, value, -mpg01) %>%
mutate(Variable = str_to_title(Variable)) %>%
ggplot(aes(mpg01, value, fill = mpg01)) +
geom_boxplot(alpha = 0.6) +
facet_wrap(~ Variable, scales = 'free', ncol = 1, switch = 'x') +
coord_flip() +
theme(legend.position = 'top') +
labs(x = '', y = '', title = 'Variable Boxplots by mpg01')## Warning: 'switch' is deprecated.
## Use 'strip.position' instead.
## See help("Deprecated")
- Split the data into a training set and a test set.
set.seed(1)
num_train <- nrow(Auto) * 0.75
inTrain <- sample(nrow(Auto), size = num_train)
training <- Auto[inTrain,]
testing <- Auto[-inTrain,]- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in
require(MASS)
fmla <- as.formula('mpg01 ~ displacement + horsepower + weight + year + cylinders')
lda_model <- lda(fmla, data = training)
pred <- predict(lda_model, testing)
table(pred$class, testing$mpg01)##
## 0 1
## 0 48 5
## 1 5 39mean(pred$class == testing$mpg01)## [1] 0.8969072- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda_model <- qda(fmla, data = training)
pred <- predict(qda_model, testing)
table(pred$class, testing$mpg01)##
## 0 1
## 0 48 5
## 1 5 39mean(pred$class == testing$mpg01)## [1] 0.8969072- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
log_reg <- glm(fmla, data = training, family = binomial)
pred <- predict(log_reg, testing, type = 'response')
pred_values <- round(pred)
table(pred_values, testing$mpg01)##
## pred_values 0 1
## 0 49 3
## 1 4 41mean(pred_values == testing$mpg01)## [1] 0.9278351- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
require(class)
set.seed(1)
acc <- list()
x_train <- training[,c('cylinders', 'displacement', 'horsepower', 'weight', 'year')]
y_train <- training$mpg0
x_test <- testing[,c('cylinders', 'displacement', 'horsepower', 'weight', 'year')]
for (i in 1:20) {
knn_pred <- knn(train = x_train, test = x_test, cl = y_train, k = i)
acc[as.character(i)] = mean(knn_pred == testing$mpg01)
}
acc <- unlist(acc)Q13.Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
attach(Boston)
crim01 <- rep(0, length(crim)) # create binary variable for "crim"
crim01[crim > median(crim)] <- 1
Boston <- data.frame(Boston, crim01)train <- 1:(length(crim) / 2) # split first half of data to train set
test <- (length(crim) / 2 + 1):length(crim)
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]# logistic regression
glm.fit <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim, family = binomial,
## data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.83229 -0.06593 0.00000 0.06181 2.61513
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -91.319906 19.490273 -4.685 2.79e-06 ***
## zn -0.815573 0.193373 -4.218 2.47e-05 ***
## indus 0.354172 0.173862 2.037 0.04164 *
## chas 0.167396 0.991922 0.169 0.86599
## nox 93.706326 21.202008 4.420 9.88e-06 ***
## rm -4.719108 1.788765 -2.638 0.00833 **
## age 0.048634 0.024199 2.010 0.04446 *
## dis 4.301493 0.979996 4.389 1.14e-05 ***
## rad 3.039983 0.719592 4.225 2.39e-05 ***
## tax -0.006546 0.007855 -0.833 0.40461
## ptratio 1.430877 0.359572 3.979 6.91e-05 ***
## black -0.017552 0.006734 -2.606 0.00915 **
## lstat 0.190439 0.086722 2.196 0.02809 *
## medv 0.598533 0.185514 3.226 0.00125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 69.568 on 239 degrees of freedom
## AIC: 97.568
##
## Number of Fisher Scoring iterations: 10glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] <- 1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 68 24
## 1 22 139mean(glm.pred != crim01.test)## [1] 0.1818182# logistic regression (-chas -nox -tax)
glm.fit <- glm(crim01 ~ . - crim01 - crim -chas -nox -tax, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim - chas - nox - tax,
## family = binomial, data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.04443 -0.24461 -0.00114 0.38919 2.72999
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -17.291707 6.019497 -2.873 0.004071 **
## zn -0.478891 0.104276 -4.593 4.38e-06 ***
## indus 0.362719 0.082969 4.372 1.23e-05 ***
## rm -2.364642 0.967625 -2.444 0.014535 *
## age 0.063371 0.015457 4.100 4.14e-05 ***
## dis 1.494535 0.397249 3.762 0.000168 ***
## rad 1.756498 0.357330 4.916 8.85e-07 ***
## ptratio 0.575045 0.161917 3.551 0.000383 ***
## black -0.018916 0.006754 -2.801 0.005102 **
## lstat 0.057632 0.053051 1.086 0.277326
## medv 0.237282 0.081326 2.918 0.003527 **
## ---
## 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: 139.59 on 242 degrees of freedom
## AIC: 161.59
##
## Number of Fisher Scoring iterations: 9glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] <- 1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 78 28
## 1 12 135mean(glm.pred != crim01.test)## [1] 0.1581028We may conclude that, for this logistic regression, we have a test error rate of 15.8102767%.
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test, ]
train.crim01 <- crim01[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.crim01, k = 1)
table(knn.pred, crim01.test)## crim01.test
## knn.pred 0 1
## 0 85 111
## 1 5 52We may conclude that, for this KNN (k=1), we have a test error rate of 45.8498024%.
knn.pred <- knn(train.X, test.X, train.crim01, k = 10)
table(knn.pred, crim01.test)## crim01.test
## knn.pred 0 1
## 0 83 23
## 1 7 140We may conclude that, for this KNN (k=10), we have a test error rate of 11.8577075%.
knn.pred <- knn(train.X, test.X, train.crim01, k = 100)
table(knn.pred, crim01.test)## crim01.test
## knn.pred 0 1
## 0 86 120
## 1 4 43We may conclude that, for this KNN (k=100), we have a test error rate of 49.0118577%. Proceed to picking the model with the lowest test error rate, but there are more exploration left to do.