Assignment #3
2024-02-25
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?
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
##
##
##
## plot(Weekly, cex = 0.5)
- 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?
lr.weekly <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly, family = binomial)
summary(lr.weekly)##
## 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 appears to be statistically significant with a P-value of 0.0296.
- 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.
contrasts(Weekly$Direction)## Up
## Down 0
## Up 1lr.weekly.pred <- predict(lr.weekly, type = "response") > 0.5
(pred.table <- table(ifelse(lr.weekly.pred, "Up (pred)", "Down (pred)"), Weekly$Direction))##
## Down Up
## Down (pred) 54 48
## Up (pred) 430 557sum(diag(pred.table)) / sum(pred.table)## [1] 0.5610652Confusion matrix tells us things such as recall and precision. Overall accuracy was 0.56.
- 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).
sub.week <- Weekly$Year < 2009
new.model <- glm(Direction ~ Lag2, data = Weekly[sub.week, ], family = binomial)
new.predict <- predict(new.model, Weekly[!sub.week, ], type = "response") > 0.5
(new.pred.table <- table(ifelse(new.predict, "Up (pred)", "Down (pred)"), Weekly[!sub.week, ]$Direction))##
## Down Up
## Down (pred) 9 5
## Up (pred) 34 56sum(diag(new.pred.table)) / sum(new.pred.table)## [1] 0.625- Repeat (d) using LDA.
new.model2 <- lda(Direction ~ Lag2, data = Weekly[sub.week, ])
new.predict2 <- predict(new.model2, Weekly[!sub.week, ], type = "response")$class
(new.predict.table2 <- table(new.predict2, Weekly[!sub.week, ]$Direction))##
## new.predict2 Down Up
## Down 9 5
## Up 34 56sum(diag(new.predict.table2)) / sum(new.predict.table2)## [1] 0.625- Repeat (d) using QDA.
new.model3 <- qda(Direction ~ Lag2, data = Weekly[sub.week, ])
new.predict3 <- predict(new.model3, Weekly[!sub.week, ], type = "response")$class
(new.predict.table3 <- table(new.predict3, Weekly[!sub.week, ]$Direction))##
## new.predict3 Down Up
## Down 0 0
## Up 43 61sum(diag(new.predict.table3)) / sum(new.predict.table3)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
new.model4 <- knn(Weekly[sub.week, "Lag2", drop = FALSE],
Weekly[!sub.week, "Lag2", drop = FALSE],
Weekly$Direction[sub.week]
)
(new.predict.table4 <- table(new.model4, Weekly[!sub.week, ]$Direction))##
## new.model4 Down Up
## Down 21 30
## Up 22 31sum(diag(new.predict.table4)) / sum(new.predict.table4)## [1] 0.5- Repeat (d) using naive Bayes.
new.model5 <- naiveBayes(Direction ~ Lag2, data = Smarket, subset = sub.week)
new.predict5 <- predict(new.model5, Weekly[!sub.week, ], type = "class")
(new.predict.table5 <- table(new.predict5, Weekly[!sub.week, ]$Direction))##
## new.predict5 Down Up
## Down 27 29
## Up 16 32sum(diag(new.predict.table5)) / sum(new.predict.table5)## [1] 0.5673077- Which of these methods appears to provide the best results on this data?
Logistic regression and LDA had the highest scores with 0.625 overall accuracy.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
sub.week <- Weekly$Year < 2009
new.model <- glm(Direction ~ Lag1 + Lag2, data = Weekly[sub.week, ], family = binomial)
new.predict <- predict(new.model, Weekly[!sub.week, ], type = "response") > 0.5
(new.pred.table <- table(ifelse(new.predict, "Up (pred)", "Down (pred)"), Weekly[!sub.week, ]$Direction))##
## Down Up
## Down (pred) 7 8
## Up (pred) 36 53sum(diag(pred.table)) / sum(pred.table)## [1] 0.5610652new.model2 <- lda(Direction ~ Lag1 + Lag2 + Lag3, data = Weekly[sub.week, ])
new.predict2 <- predict(new.model2, Weekly[!sub.week, ], type = "response")$class
(new.predict.table2 <- table(new.predict2, Weekly[!sub.week, ]$Direction))##
## new.predict2 Down Up
## Down 8 9
## Up 35 52sum(diag(new.predict.table2)) / sum(new.predict.table2)## [1] 0.5769231new.model3 <- qda(Direction ~ Lag2 + Lag3, data = Weekly[sub.week, ])
new.predict3 <- predict(new.model3, Weekly[!sub.week, ], type = "response")$class
(new.predict.table3 <- table(new.predict3, Weekly[!sub.week, ]$Direction))##
## new.predict3 Down Up
## Down 4 2
## Up 39 59sum(diag(new.predict.table3)) / sum(new.predict.table3)## [1] 0.6057692new.model4 <- knn(Weekly[sub.week, "Lag3", drop = FALSE],
Weekly[!sub.week, "Lag3", drop = FALSE],
Weekly$Direction[sub.week]
)
(new.predict.table4 <- table(new.model4, Weekly[!sub.week, ]$Direction))##
## new.model4 Down Up
## Down 21 27
## Up 22 34sum(diag(new.predict.table4)) / sum(new.predict.table4)## [1] 0.5288462new.model5 <- naiveBayes(Direction ~ Lag1 + Lag2, data = Smarket, subset = sub.week)
new.predict5 <- predict(new.model5, Weekly[!sub.week, ], type = "class")
(new.predict.table5 <- table(new.predict5, Weekly[!sub.week, ]$Direction))##
## new.predict5 Down Up
## Down 34 39
## Up 9 22sum(diag(new.predict.table5)) / sum(new.predict.table5)## [1] 0.5384615QUESTION 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.
m <- 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.
pairs(m[, 1:7])
par(mfrow = c(2, 4))
for (i in 1:7) hist(m[, i], breaks = 20, main = colnames(m)[i])
for (i in 1:7) boxplot(m[, i] ~ m$mpg01, main = colnames(m)[i])
Displacement, horsepower, weight, and acceleration seem to be useful in predicting mpg01. There seems to be linear relationships among the variables.
- Split the data into a training set and a test set.
set.seed(42)
train <- sample(seq_len(nrow(m)), nrow(m) * 4/5)- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
sort(sapply(1:7, function(i) {
setNames(abs(t.test(m[, i] ~ m$mpg01)$statistic), colnames(m)[i])
}))## acceleration year origin horsepower displacement weight
## 7.302430 9.403221 11.824099 17.681939 22.632004 22.932777
## cylinders
## 23.035328mpg.lda <- lda(mpg01 ~ displacement + horsepower + weight + acceleration, data = m[train, ])
mpg.pred <- predict(mpg.lda, m[-train, ], type = "response")$class
mean(mpg.pred != m[-train, ]$mpg01)## [1] 0.06329114- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
mpg.qda <- qda(mpg01 ~ displacement + horsepower + weight + acceleration, data = m[train, ])
mpg.pred.2 <- predict(mpg.qda, m[-train, ], type = "response")$class
mean(mpg.pred.2 != m[-train, ]$mpg01)## [1] 0.08860759- Perform logistic regression on the training data in order to pre- dict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
mpg.glm <- glm(mpg01 ~ displacement + horsepower + weight + acceleration, data = m[train, ], family = binomial)
mpg.pred.3 <- predict(mpg.glm, m[-train, ], type = "response") > 0.5
mean(mpg.pred.3 != m[-train, ]$mpg01)## [1] 0.06329114- 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?
mpg.bayes <- naiveBayes(mpg01 ~ displacement + horsepower + weight + acceleration, data = m[train, ])
mpg.pred.4 <- predict(mpg.bayes, m[-train, ], type = "class")
mean(mpg.pred.4 != m[-train, ]$mpg01)## [1] 0.06329114- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
res <- sapply(1:50, function(k) {
mpg.knn <- knn(m[train, c(2, 3, 4, 5)], m[-train, c(2, 3, 4, 5)], m$mpg01[train], k = k)
mean(mpg.knn != m[-train, ]$mpg01)
})
names(res) <- 1:50
plot(res, type = "o")
res[which.min(res)]## 3
## 0.07594937LDA, logistic regression, and Bayes had the lowest test error rates of 0.0633.
QUESTION 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
crime.med <- ifelse(Boston$crim > median(Boston$crim), 1, 0)
Boston$crime.med <- crime.medboston.glm <- glm(crime.med ~ ., data = Boston, family = "binomial")## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(boston.glm)##
## Call:
## glm(formula = crime.med ~ ., family = "binomial", data = Boston)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.437e+01 1.202e+05 0.000 1.000
## crim 1.083e+03 1.773e+04 0.061 0.951
## zn 2.194e+00 5.856e+01 0.037 0.970
## indus -2.510e+00 9.002e+02 -0.003 0.998
## chas 4.489e+00 1.014e+04 0.000 1.000
## nox -2.585e+02 1.458e+05 -0.002 0.999
## rm -3.953e+01 1.653e+03 -0.024 0.981
## age 3.437e-01 5.798e+01 0.006 0.995
## dis -1.742e+01 2.146e+03 -0.008 0.994
## rad -5.933e+00 2.642e+03 -0.002 0.998
## tax 1.639e-01 1.078e+02 0.002 0.999
## ptratio 5.525e+00 3.640e+03 0.002 0.999
## black 3.266e-02 1.208e+01 0.003 0.998
## lstat -1.687e+00 3.560e+02 -0.005 0.996
## medv 2.358e+00 5.382e+02 0.004 0.997
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 7.0146e+02 on 505 degrees of freedom
## Residual deviance: 2.8371e-05 on 491 degrees of freedom
## AIC: 30
##
## Number of Fisher Scoring iterations: 25boston.lda <- lda(crime.med ~ ., data = Boston)
boston.lda## Call:
## lda(crime.med ~ ., data = Boston)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## crim zn indus chas nox rm age dis
## 0 0.0955715 21.525692 7.002292 0.05138340 0.4709711 6.394395 51.31028 5.091596
## 1 7.1314756 1.201581 15.271265 0.08695652 0.6384190 6.174874 85.83953 2.498489
## rad tax ptratio black lstat medv
## 0 4.158103 305.7431 17.90711 388.7061 9.419486 24.94941
## 1 14.940711 510.7312 19.00395 324.6420 15.886640 20.11621
##
## Coefficients of linear discriminants:
## LD1
## crim 0.0046376592
## zn -0.0056431194
## indus 0.0126159626
## chas -0.0592836851
## nox 8.1826206579
## rm 0.0874007870
## age 0.0112829040
## dis 0.0453643651
## rad 0.0699133176
## tax -0.0008444666
## ptratio 0.0513806507
## black -0.0009892799
## lstat 0.0143945059
## medv 0.0386990631boston.bayes <- naiveBayes(as.factor(crime.med) ~ ., data = Boston)
boston.bayes##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5 0.5
##
## Conditional probabilities:
## crim
## Y [,1] [,2]
## 0 0.0955715 0.06281773
## 1 7.1314756 11.10912294
##
## zn
## Y [,1] [,2]
## 0 21.525692 29.319808
## 1 1.201581 4.798611
##
## indus
## Y [,1] [,2]
## 0 7.002292 5.514454
## 1 15.271265 5.439010
##
## chas
## Y [,1] [,2]
## 0 0.05138340 0.2212161
## 1 0.08695652 0.2823299
##
## nox
## Y [,1] [,2]
## 0 0.4709711 0.05559789
## 1 0.6384190 0.09870365
##
## rm
## Y [,1] [,2]
## 0 6.394395 0.5556856
## 1 6.174874 0.8101381
##
## age
## Y [,1] [,2]
## 0 51.31028 25.88190
## 1 85.83953 17.87423
##
## dis
## Y [,1] [,2]
## 0 5.091596 2.081304
## 1 2.498489 1.085521
##
## rad
## Y [,1] [,2]
## 0 4.158103 1.659121
## 1 14.940711 9.529843
##
## tax
## Y [,1] [,2]
## 0 305.7431 87.4837
## 1 510.7312 167.8553
##
## ptratio
## Y [,1] [,2]
## 0 17.90711 1.811216
## 1 19.00395 2.346947
##
## black
## Y [,1] [,2]
## 0 388.7061 22.83774
## 1 324.6420 118.83084
##
## lstat
## Y [,1] [,2]
## 0 9.419486 4.923497
## 1 15.886640 7.546922
##
## medv
## Y [,1] [,2]
## 0 24.94941 7.232047
## 1 20.11621 10.270362sample <- sample(1:nrow(Boston), 0.8*nrow(Boston))
train <- Boston[sample, ]
test <- Boston[-sample, ]
boston.knn <- knn(train[, -1], test[, -1], train$crime.med, k = 7)
table(boston.knn, test$crime.med)##
## boston.knn 0 1
## 0 48 2
## 1 2 50