Assignment #3: Resampling Methods
library(ISLR2)
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## Bostonlibrary(class)
library(e1071)
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.
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
##
##
##
##
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
pairs(Weekly)
cor(subset(Weekly, select = -Direction))## 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 1.000000000- Based on the results Year and Volume are the only variables with a relationship.
(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?
log.reg<- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,data = Weekly,family = binomial)
summary(log.reg)##
## 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- Based on the results Lag 2 is statistically significant based on the p-value of <.05
(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.
Weekly.prob= predict(log.reg, type='response')
Weekly.pred =rep("Down", length(Weekly.prob))
Weekly.pred[Weekly.prob > 0.5] = "Up"table(Weekly.pred, Weekly$Direction)##
## Weekly.pred Down Up
## Down 54 48
## Up 430 557mean(Weekly.pred == Weekly$Direction)## [1] 0.5610652- The results show that 56.1% of the responses were predicted correctly, but the training error rate of 43.89% suggests that the predictions may be overly optimistic. Specifically, out of 484 down days, only 54 were accurately predicted, which represents an accuracy rate of approximately 11%. In contrast, 557 out of 605 up days were predicted correctly, resulting in an accuracy rate of approximately 92%.
(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$Year < 2009)
train_set <- Weekly[train, ]
test_set <- Weekly[!train, ]
train.logit <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)predict_glm_direction <- function(model, newdata = NULL) {
predictions <- predict(model, newdata, type="response")
return(as.factor(ifelse(predictions < 0.5, "Down", "Up")))
}
direction.pred <- predict_glm_direction(train.logit, test_set)
table(direction.pred, test_set$Direction)##
## direction.pred Down Up
## Down 9 5
## Up 34 56mean(direction.pred == test_set$Direction)## [1] 0.625- Based on the model created on the data from 1990 through 2008 using only Lag2 as the predictor, 62.5% of the weeks correctly predicted the market direction.
(e) Repeat (d) using LDA.
weekly.lda <- lda(Direction ~ Lag2, data = Weekly, subset = train)
weekly.lda## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda.pred = predict(weekly.lda, Weekly[!train, ])
table(lda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == Weekly[!train, ]$Direction)## [1] 0.625- 62.5% correctly predicted.
(f) Repeat (d) using QDA.
weekly.qda = qda(Direction ~ Lag2, data = Weekly, subset = train)
weekly.qda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.pred = predict(weekly.qda, Weekly[!train, ])
table(qda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Weekly[!train, ]$Direction)## [1] 0.5865385- 58.65 were correcly predicted.
(g) Repeat (d) using KNN with K = 1.
train.knn= data.frame(Weekly[train, ]$Lag2)
test.knn= data.frame(Weekly[!train, ]$Lag2)
direction.train = Weekly[train, ]$Directionset.seed(1)
knn.pred = knn(train.knn, test.knn, direction.train, k = 1)
table(knn.pred, Weekly[!train, ]$Direction)##
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Weekly[!train, ]$Direction)## [1] 0.5- 50% correctly predicted.
(h) Repeat (d) using naive Bayes.
weekly.nbayes = naiveBayes(Direction~Lag2 ,data=Weekly ,subset=train)
weekly.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.pred=predict(weekly.nbayes ,Weekly[!train ,])
table(nbayes.pred ,Weekly$Direction[!train])##
## nbayes.pred Down Up
## Down 0 0
## Up 43 61mean(nbayes.pred == Weekly[!train, ]$Direction)## [1] 0.5865385- 58.65% were correctly predicted.
(i) Which of these methods appears to provide the best results on this data?
- The Logistic Regression model predicted 62.5% of the market correctly. This was the highest out of all the models. Therefore, the Logistic Regression method provides the best results.
(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.
train.knn= data.frame(Weekly[train, ]$Lag2)
test.knn= data.frame(Weekly[!train, ]$Lag2)
direction.train = Weekly[train, ]$Directionset.seed(1)
knn.pred2 = knn(train.knn, test.knn, direction.train, k = 10)
table(knn.pred2, Weekly[!train, ]$Direction)##
## knn.pred2 Down Up
## Down 17 21
## Up 26 40mean(knn.pred == Weekly[!train, ]$Direction)## [1] 0.5Question 14: In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
Auto$mpg01 <- Auto$mpg > median(Auto$mpg)
head(Auto$mpg01, n = 20)## [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [13] FALSE FALSE TRUE FALSE FALSE FALSE TRUE TRUEhead(Auto[,c(1,10)],5)## mpg mpg01
## 1 18 FALSE
## 2 15 FALSE
## 3 18 FALSE
## 4 16 FALSE
## 5 17 FALSE
(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.
cor(subset(Auto, select = -name))## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(Auto)
- based on these results: cylinders, weight, displacement, horsepower, and mpg seem to be most likely to be useful in predicting mpg01.
(c) Split the data into a training set and a test set.
train <- sample(nrow(Auto) * 0.7)
train_set <- Auto[train, ]
test_set <- Auto[-train, ]
Auto.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,subset = 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?
show_model_performance <- function(predicted_status, observed_status) {
confusion_matrix <- table(predicted_status,
observed_status,
dnn = c("Predicted Status", "Observed Status"))
print(confusion_matrix)
error_rate <- mean(predicted_status != observed_status)
cat("\n") # \n means newline so it just prints a blank line
cat(" Error Rate:", 100 * error_rate, "%\n")
cat("Correctly Predicted:", 100 * (1-error_rate), "%\n")
cat("False Positive Rate:", 100 * confusion_matrix[2,1] / sum(confusion_matrix[,1]), "%\n")
cat("False Negative Rate:", 100 * confusion_matrix[1,2] / sum(confusion_matrix[,2]), "%\n")
}
predictions <- predict(Auto.lda, test_set)
show_model_performance(predictions$class, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 18 16
## TRUE 2 82
##
## Error Rate: 15.25424 %
## Correctly Predicted: 84.74576 %
## False Positive Rate: 10 %
## False Negative Rate: 16.32653 %
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Auto.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
predictions <- predict(Auto.qda, test_set)
show_model_performance(predictions$class, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 18 19
## TRUE 2 79
##
## Error Rate: 17.79661 %
## Correctly Predicted: 82.20339 %
## False Positive Rate: 10 %
## False Negative Rate: 19.38776 %
(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Auto.logit <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset =train)
predictions <- predict(Auto.logit, test_set, type = "response")
show_model_performance(predictions > 0.5, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 20 21
## TRUE 0 77
##
## Error Rate: 17.79661 %
## Correctly Predicted: 82.20339 %
## False Positive Rate: 0 %
## False Negative Rate: 21.42857 %
(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?
Auto.nb <- naiveBayes(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset =train)
predictions.nb <- predict(Auto.nb, test_set)
table(predictions.nb, test_set$mpg01)##
## predictions.nb FALSE TRUE
## FALSE 18 14
## TRUE 2 84mean(predictions.nb == test_set$mpg01)## [1] 0.8644068-86.44% correctly predicted.
(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?
data = scale(Auto[,-c(9,10)])
set.seed(55)
train.auto <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
test.auto <- -train
training_data = data[train.auto,c("cylinders","horsepower","weight","acceleration")]
testing_data = data[test.auto, c("cylinders", "horsepower","weight","acceleration")]
train.mpg01 = Auto$mpg01[train.auto]
test.mpg01= Auto$mpg01[test.auto]set.seed(50)
knn.auto.pred = knn(training_data, testing_data, train.mpg01, k = 1)
table(knn.auto.pred, test.mpg01)## test.mpg01
## knn.auto.pred FALSE TRUE
## FALSE 18 8
## TRUE 2 90mean(knn.auto.pred != test.mpg01)## [1] 0.08474576- Test error rate is 8.47% for K=1
set.seed(50)
knn.auto.pred2 = knn(training_data, testing_data, train.mpg01, k = 10)
table(knn.auto.pred2, test.mpg01)## test.mpg01
## knn.auto.pred2 FALSE TRUE
## FALSE 18 15
## TRUE 2 83mean(knn.auto.pred2 != test.mpg01)## [1] 0.1440678Test error rate is 14.41% for K=10
based on the results: k=1 is the best value of K since it is the least amount of error.
Question 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.
data("Boston")
attach(Boston)
crime1 = rep(0, length(crim))
crime1[crim>median(crim)] = 1
Boston = data.frame(Boston, crime1)train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2+1):dim(Boston)[1]
Boston_train = Boston[train,]
Boston_test = Boston[test,]
crime1.test = crime1[test]
plot(Boston)
Logistic Regression
set.seed(1)
Boston.logit = glm(crime1~nox + tax + dis + medv + lstat, data = Boston, family = binomial)
Boston.probs = predict(Boston.logit, Boston_test, type ="response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1
table(Boston.pred, crime1.test)## crime1.test
## Boston.pred 0 1
## 0 75 15
## 1 15 148mean(Boston.pred == crime1.test)## [1] 0.8814229- 88% correctly predicted.
LDA
Boston.lda = lda(crime1~nox+tax+dis+medv+lstat, data = Boston_train)
Boston.pred.lda = predict(Boston.lda, Boston_test)
table(Boston.pred.lda$class, crime1.test)## crime1.test
## 0 1
## 0 80 16
## 1 10 147mean(Boston.pred.lda$class == crime1.test)## [1] 0.8972332-89.7% was correctly predicted
K-Nearest Neighbor
train.B2= cbind(nox,tax,dis,lstat)[train,]
test.B2 = cbind(nox,tax,dis,lstat)[test,]
train.crime = crime1.test
set.seed(1)
Boston.pred.knn = knn(train.B2, test.B2, train.crime, k=10)
table(Boston.pred.knn, crime1.test)## crime1.test
## Boston.pred.knn 0 1
## 0 43 21
## 1 47 142mean(Boston.pred.knn == crime1.test)## [1] 0.7312253- 73% was correctly predicted.
Naive Bayes
Bostob.nb = naiveBayes(crime1~nox+tax+dis+lstat, data = Boston, subset = train)
Boston.pred.nb = predict(Bostob.nb, Boston_test)
table(Boston.pred.nb, crime1.test)## crime1.test
## Boston.pred.nb 0 1
## 0 75 18
## 1 15 145mean(Boston.pred.nb == crime1.test)## [1] 0.8695652- 86.96% was correctly predicted.