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?
cor(Weekly[,-9])
## 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
pairs(Weekly)
- Looking at the correlation matrix and the pairwise plots, it does
not look like their are any significant relations other than between
Year and Volume. Year and Volume have a correlation coefficient of .842
which means that there is a strong positive relationship between the
two.
- 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_model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(log_model)
##
## 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: 4
- We only see one predictor that is statistically significant. This is
Lag2 with a p-value of .0296.
- 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.
log_prob <- predict(log_model, type = "response")
log_pred <- rep("Down", length(log_prob))
log_pred[log_prob > 0.5] = "Up"
table(log_pred, Weekly$Direction)
##
## log_pred Down Up
## Down 54 48
## Up 430 557
- This means that we had 54 correct predictions when the market had
negative return, and 557 correct predictions for positive return. We
also see that we had 48 false negative and 430 false positive
predictions. This model mis-classifies 56.1% of the time ((48+430)/
(54+48+430+557)).
- Now fit the logistic regression model using a training data period
from 1990 to 2008, with Lag2 as the only predictor. Compute the
confusion matrix and the overall fraction of correct predictions for the
held out data (that is, the data from 2009 and 2010).
train_week <- (Weekly$Year < 2009)
Weekly.train <- Weekly[train_week,]
Weekly.test <- Weekly[!train_week,]
log_fit <- glm(Direction ~ Lag2, data = Weekly.train, family = binomial)
log_prob2 <- predict(log_fit, Weekly.test, type = "response")
log_pred2 <- rep("Down", length(log_prob2))
log_pred2[log_prob2> 0.5] = "Up"
direction_week <- Weekly$Direction[!train_week]
table(log_pred2, direction_week)
## direction_week
## log_pred2 Down Up
## Down 9 5
## Up 34 56
#Logistic Regression Accuracy
mean(log_pred2 == direction_week)
## [1] 0.625
- Repeat (d) using LDA.
lda_fit <- lda(Direction ~ Lag2, data = Weekly, subset = train_week)
lda_prob <- predict(lda_fit, Weekly.test)
lda_class <- lda_prob$class
table(lda_class, Weekly.test$Direction)
##
## lda_class Down Up
## Down 9 5
## Up 34 56
#LDA Accuracy
mean(lda_class == direction_week)
## [1] 0.625
- Repeat (d) using QDA.
qda_fit <- qda(Direction ~ Lag2, data = Weekly, subset = train_week)
qda_prob <- predict(qda_fit, Weekly.test)$class
table(qda_prob, Weekly.test$Direction)
##
## qda_prob Down Up
## Down 0 0
## Up 43 61
#QDA Accuracy
mean(qda_prob == direction_week)
## [1] 0.5865385
- Repeat (d) using KNN with K = 1.
train_data <- as.matrix(Weekly$Lag2[train_week])
test_data <- as.matrix(Weekly$Lag2[!train_week])
train_direction <- Weekly$Direction[train_week]
set.seed(1)
knn_pred <- knn(train_data, test_data, train_direction, k = 1)
table(knn_pred, Weekly.test$Direction)
##
## knn_pred Down Up
## Down 21 30
## Up 22 31
#KNN Accuracy
mean(knn_pred == direction_week)
## [1] 0.5
- Repeat (d) using naive Bayes.
nb_fit <- naiveBayes(Direction ~ Lag2, data = Weekly,
subset = train_week)
nb_class <- predict(nb_fit, Weekly.test)
table(nb_class, Weekly.test$Direction)
##
## nb_class Down Up
## Down 0 0
## Up 43 61
#Naive Bayes Accuracy
mean(nb_class == direction_week)
## [1] 0.5865385
- Which of these methods appears to provide the best results on this
data?
- Based on the created models and calculated accuracys, we see that
none of the models perform great. All accuracies are close to 0.5 which
is what we could expect by random chance. The second logistic regression
and LDA model perform the best with 62.5% accuracy. This is not great,
but one of these would be the best option of those calculated.
- Experiment with the 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 the values for K in the KNN classifier.
#Log Regression with interaction between Lag1 and Lag2
log_fits_3 <- glm(Direction~Lag1 + Lag2 + Lag1:Lag2, data = Weekly.train, family = binomial)
log_probs_3 <- predict(log_fits_3, Weekly.test, type = 'response')
log_pred_3 <- rep('Down', length(log_probs_3))
log_pred_3[log_probs_3>0.5] = "Up"
table(log_pred_3, Weekly.test$Direction)
##
## log_pred_3 Down Up
## Down 7 8
## Up 36 53
mean(log_pred_3 == direction_week)
## [1] 0.5769231
#Log Regression with Lag1 and Lag 2 squared
log_fits_4 <- glm(Direction~Lag1**2+Lag2**2, data = Weekly.train, family = binomial)
log_probs_4 <- predict(log_fits_4, Weekly.test, type = 'response')
log_pred_4 <- rep('Down', length(log_probs_4))
log_pred_4[log_probs_4>0.5] = "Up"
table(log_pred_4, Weekly.test$Direction)
##
## log_pred_4 Down Up
## Down 7 8
## Up 36 53
mean(log_pred_4 == direction_week)
## [1] 0.5769231
#LDA with 5 predictors
lda_fits_5 <- lda(Direction~Lag1+Lag2+Lag3+Lag4+Lag5, data = Weekly, subset = train_week)
lda_prob_5 <- predict(lda_fits_5, Weekly.test)
lda_class_5 <- lda_prob_5$class
table(lda_class_5, Weekly.test$Direction)
##
## lda_class_5 Down Up
## Down 9 13
## Up 34 48
mean(lda_class_5 == direction_week)
## [1] 0.5480769
#QDA with 3 predictors
qda_fits_6 <- qda(Direction~Lag1+Lag2+Lag3, data = Weekly, subset = train_week)
qda_prob_6 <- predict(qda_fits_6, Weekly.test)
qda_class_6 <- qda_prob_6$class
table(qda_class_6, Weekly.test$Direction)
##
## qda_class_6 Down Up
## Down 6 10
## Up 37 51
mean(qda_class_6 == direction_week)
## [1] 0.5480769
#K=3
set.seed(1)
knn_pred_7 <- knn(train_data, test_data, train_direction, k=3)
table(knn_pred_7, Weekly.test$Direction)
##
## knn_pred_7 Down Up
## Down 16 20
## Up 27 41
mean(knn_pred_7 == direction_week)
## [1] 0.5480769
#K=5
set.seed(1)
knn_pred_8 <- knn(train_data, test_data, train_direction, k=5)
table(knn_pred_8, Weekly.test$Direction)
##
## knn_pred_8 Down Up
## Down 16 21
## Up 27 40
mean(knn_pred_8 == direction_week)
## [1] 0.5384615
- Based on these models, the two logistic regression models performed
the best at 57.69% accuracy.
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.
data("Auto")
mpg01 <- rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] <- 1
Auto <- data.frame(Auto, mpg01)
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
## mpg01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
##
- Explore the data graphically in order to investigate the association
between mpg01 and the other features. Which of the other features seems
most likely to be useful in predicting mpg01? Scatterplots and boxplots
may be useful tools to answer this question. Describe your
findings.
cor(Auto[, -9])
## 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 1.0000000
pairs(Auto)
par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")
- The graphs indicate that the strongest predictors of mpg01 are
cylinders, displacement, and weight which all have negative
relationships around -0.75.
- Split the data into a training set and a test set.
set.seed(123)
train <- sample(1:dim(Auto)[1], dim(Auto)[1]*.7, rep=FALSE)
test <- -train
training_data<- Auto[train, ]
testing_data= Auto[test, ]
mpg01.test <- mpg01[test]
- 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_model <- lda(mpg01 ~ cylinders + displacement + weight, data = training_data)
pred_lda <- predict(lda_model, testing_data)
table(pred_lda$class, mpg01.test)
## mpg01.test
## 0 1
## 0 50 4
## 1 10 54
#LDA Test Error
mean(pred_lda$class != mpg01.test)
## [1] 0.1186441
- The test error of this model is 11.86%.
- Perform QDA on the training data in order to predict mpg01 using the
variables that seemed most associated with mpg01 in (b). What is the
test error of the model obtained?
qda_model <- qda(mpg01 ~ cylinders + displacement + weight, data = training_data)
pred_qda <- predict(qda_model, testing_data)
table(pred_qda$class, mpg01.test)
## mpg01.test
## 0 1
## 0 54 4
## 1 6 54
#QDA Test Error
mean(pred_qda$class != mpg01.test)
## [1] 0.08474576
- The test error of this model is 8.47%.
- Perform logistic regression on the training data in order to predict
mpg01 using the variables that seemed most associated with mpg01 in (b).
What is the test error of the model obtained?
glm_model <- glm(mpg01 ~ cylinders + displacement + weight, data = training_data, family = binomial)
probs <- predict(glm_model, testing_data, type = "response")
pred_glm <- rep(0, length(probs))
pred_glm[probs > 0.5] <- 1
table(pred_glm, mpg01.test)
## mpg01.test
## pred_glm 0 1
## 0 52 5
## 1 8 53
#Logistic Regression Test Error
mean(pred_glm != mpg01.test)
## [1] 0.1101695
- The test error of this model is 11.02%.
- Perform naive Bayes on the training data in order to predict mpg01
using the variables that seemed most associated with mpg01 in (b). What
is the test error of the model obtained?
nb_model <- naiveBayes(mpg01 ~ cylinders + displacement + weight, data = training_data)
nb_pred <- predict(nb_model, testing_data)
table(nb_pred, mpg01.test)
## mpg01.test
## nb_pred 0 1
## 0 52 4
## 1 8 54
#Naive Bayes Test Error
mean(nb_pred != mpg01.test)
## [1] 0.1016949
- The test error of this model is 10.17%.
- perform KNN on the training data, with several values of K, in order
to predict mpg01. Use only the variables that seemed most associated
with mpg01 in (b). What test errors do you obtain? Which value of K
seems to perform the best on this data set?
data = scale(Auto[,-c(9,10)])
set.seed(1234)
train <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
test <- -train
training_data = data[train,c("cylinders","displacement","weight")]
testing_data = data[test, c("cylinders", "displacement","weight")]
train.mpg01 = Auto$mpg01[train]
test.mpg01= Auto$mpg01[test]
set.seed(1234)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 1)
table(knn_pred_y, test.mpg01)
## test.mpg01
## knn_pred_y 0 1
## 0 55 8
## 1 9 46
#KNN Test Error with K = 1
mean(knn_pred_y != test.mpg01)
## [1] 0.1440678
- The test error for this model is 14.41%.
knn_pred_y = NULL
error_rate = NULL
for(i in 1:dim(testing_data)[1]){
set.seed(1234)
knn_pred_y = knn(training_data,testing_data,train.mpg01,k=i)
error_rate[i] = mean(test.mpg01 != knn_pred_y)
}
min_error_rate = min(error_rate)
print(min_error_rate)
## [1] 0.1101695
K = which(error_rate == min_error_rate)
print(K)
## [1] 10 112 113
- The lowest test error is 11.02% when K = 10.