LAB 5
Sai Pranam
2025-03-10
Question 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.
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(MASS)
library(class)
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## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(corrplot)## Warning: package 'corrplot' was built under R version 4.4.3## corrplot 0.95 loadedlibrary(ISLR2)## Warning: package 'ISLR2' was built under R version 4.4.2##
## Attaching package: 'ISLR2'
##
## The following object is masked from 'package:MASS':
##
## Bostonlibrary(e1071)## Warning: package 'e1071' was built under R version 4.4.3summary(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]), type = "lower", diag = FALSE, method = "ellipse")
Volume is strongly positively correlated with Year. Other correlations are week, but Lag1 is negatively correlated with Lag2 but positively correlated with Lag3.
- 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?
fit <- glm(
Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = binomial
)
summary(fit)##
## 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: 4Lag2 is significant.
- 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.
contrasts(Weekly$Direction)## Up
## Down 0
## Up 1pred <- predict(fit, type = "response") > 0.5
(t <- table(ifelse(pred, "Up (pred)", "Down (pred)"), Weekly$Direction))##
## Down Up
## Down (pred) 54 48
## Up (pred) 430 557sum(diag(t)) / sum(t)## [1] 0.5610652The overall accuracy of the predictions is 0.56. While logistic regression is effective at predicting upward movements, it tends to mistakenly classify most downward movements as upward.
- 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$Year < 2009
fit <- glm(Direction ~ Lag2, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
(t <- table(ifelse(pred, "Up (pred)", "Down (pred)"), Weekly[!train, ]$Direction))##
## Down Up
## Down (pred) 9 5
## Up (pred) 34 56sum(diag(t)) / sum(t)## [1] 0.625The overall fraction of correct prediction is 0.625.
e)Repeat (d) using LDA.
fit <- lda(Direction ~ Lag2, data = Weekly[train, ])
pred <- predict(fit, Weekly[!train, ], type = "response")$class
(t <- table(pred, Weekly[!train, ]$Direction))##
## pred Down Up
## Down 9 5
## Up 34 56sum(diag(t)) / sum(t)## [1] 0.625f)Repeat (d) using QDA.
fit <- qda(Direction ~ Lag2, data = Weekly[train, ])
pred <- predict(fit, Weekly[!train, ], type = "response")$class
(t <- table(pred, Weekly[!train, ]$Direction))##
## pred Down Up
## Down 0 0
## Up 43 61sum(diag(t)) / sum(t)## [1] 0.5865385- Repeat (d) using KNN with K=1.
fit <- knn(
Weekly[train, "Lag2", drop = FALSE],
Weekly[!train, "Lag2", drop = FALSE],
Weekly$Direction[train]
)
(t <- table(fit, Weekly[!train, ]$Direction))##
## fit Down Up
## Down 21 29
## Up 22 32sum(diag(t)) / sum(t)## [1] 0.5096154- Repeat (d) using naive Bayes.
fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
pred <- predict(fit, Weekly[!train, ], type = "class")
(t <- table(pred, Weekly[!train, ]$Direction))##
## pred Down Up
## Down 0 0
## Up 43 61sum(diag(t)) / sum(t)## [1] 0.5865385- Which of these methods appears to provide the best results on this data?
Logistic regression and LDA are the best performing.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
fit <- glm(Direction ~ Lag1, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.5673077fit <- glm(Direction ~ Lag2, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.625fit <- glm(Direction ~ Lag3, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.5865385fit <- glm(Direction ~ Lag4, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.5865385fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.5865385fit <- glm(Direction ~ Lag1 * Lag2 * Lag3 * Lag4, data = Weekly[train, ], family = binomial)
pred <- predict(fit, Weekly[!train, ], type = "response") > 0.5
mean(ifelse(pred, "Up", "Down") == Weekly[!train, ]$Direction)## [1] 0.5961538fit <- lda(Direction ~ Lag1 + Lag2 + Lag3 + Lag4, data = Weekly[train, ])
pred <- predict(fit, Weekly[!train, ], type = "response")$class
mean(pred == Weekly[!train, ]$Direction)## [1] 0.5769231fit <- qda(Direction ~ Lag1 + Lag2 + Lag3 + Lag4, data = Weekly[train, ])
pred <- predict(fit, Weekly[!train, ], type = "response")$class
mean(pred == Weekly[!train, ]$Direction)## [1] 0.5192308fit <- naiveBayes(Direction ~ Lag1 + Lag2 + Lag3 + Lag4, data = Weekly[train, ])
pred <- predict(fit, Weekly[!train, ], type = "class")
mean(pred == Weekly[!train, ]$Direction)## [1] 0.5096154set.seed(1)
res <- sapply(1:30, function(k) {
fit <- knn(
Weekly[train, 2:4, drop = FALSE],
Weekly[!train, 2:4, drop = FALSE],
Weekly$Direction[train],
k = k
)
mean(fit == Weekly[!train, ]$Direction)
})
plot(1:30, res, type = "o", xlab = "k", ylab = "Fraction correct")
(k <- which.max(res))## [1] 26fit <- knn(
Weekly[train, 2:4, drop = FALSE],
Weekly[!train, 2:4, drop = FALSE],
Weekly$Direction[train],
k = k
)
table(fit, Weekly[!train, ]$Direction)##
## fit Down Up
## Down 23 18
## Up 20 43mean(fit == Weekly[!train, ]$Direction)## [1] 0.6346154KNN, when using the first three lag variables, slightly outperforms logistic regression with Lag2, provided that the value of k is tuned to 26.
Question 14
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
x <- cbind(Auto[, -1], data.frame("mpg01" = Auto$mpg > median(Auto$mpg)))- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
par(mfrow = c(2, 4))
for (i in 1:7) hist(x[, i], breaks = 20, main = colnames(x)[i])
par(mfrow = c(2, 4))
for (i in 1:7) boxplot(x[, i] ~ x$mpg01, main = colnames(x)[i])
pairs(x[, 1:7])
Most variables exhibit a relationship with the mpg01 category, and
several of them are highly collinear.
- Split the data into a training set and a test set.
set.seed(1)
train <- sample(seq_len(nrow(x)), nrow(x) * 2 / 3)- 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?
sort(sapply(1:7, function(i) {
setNames(abs(t.test(x[, i] ~ x$mpg01)$statistic), colnames(x)[i])
}))## acceleration year origin horsepower displacement weight
## 7.302430 9.403221 11.824099 17.681939 22.632004 22.932777
## cylinders
## 23.035328fit <- lda(mpg01 ~ cylinders + weight + displacement, data = x[train, ])
pred <- predict(fit, x[-train, ], type = "response")$class
mean(pred != x[-train, ]$mpg01)## [1] 0.1068702- 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?
fit <- qda(mpg01 ~ cylinders + weight + displacement, data = x[train, ])
pred <- predict(fit, x[-train, ], type = "response")$class
mean(pred != x[-train, ]$mpg01)## [1] 0.09923664- 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?
fit <- glm(mpg01 ~ cylinders + weight + displacement, data = x[train, ], family = binomial)
pred <- predict(fit, x[-train, ], type = "response") > 0.5
mean(pred != x[-train, ]$mpg01)## [1] 0.1145038- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit <- naiveBayes(mpg01 ~ cylinders + weight + displacement, data = x[train, ])
pred <- predict(fit, x[-train, ], type = "class")
mean(pred != x[-train, ]$mpg01)## [1] 0.09923664- 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?
res <- sapply(1:50, function(k) {
fit <- knn(x[train, c(1, 4, 2)], x[-train, c(1, 4, 2)], x$mpg01[train], k = k)
mean(fit != x[-train, ]$mpg01)
})
names(res) <- 1:50
plot(res, type = "o")
res[which.min(res)]## 3
## 0.1068702For the models tested, k=32 seems to yield the best performance. QDA has a lower overall error rate and performs slightly better than LDA.