Assignment3
Problem 13
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR2)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(class)
library(e1071)
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
##
##
##
## pairs(Weekly)
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- The correlation matrix shows that
VolumeandYearhave the strongest relationship — volume has grown over time.
(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?
glm_full <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly, family = binomial)
summary(glm_full)##
## 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: 4Lag2is the only predictor that appears statistically significant with a p-value below 0.05. The other lag variables and Volume are non-significant toDirection.
(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.
glm_probs <- predict(glm_full, type = "response")
glm_pred <- rep("Down", nrow(Weekly))
glm_pred[glm_probs > 0.5] <- "Up"
table(glm_pred, Weekly$Direction)##
## glm_pred Down Up
## Down 54 48
## Up 430 557mean(glm_pred == Weekly$Direction)## [1] 0.5610652- The model correctly predicts about 56% of observations. The confusion matrix shows the model is very biased toward predicting “Up”. It correctly catches most “Up” weeks but misses the majority of “Down” weeks.
(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$Year >= 1990 & Weekly$Year <= 2008
Weekly_test <- Weekly[!train, ]
glm_lag2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
glm_lag2_probs <- predict(glm_lag2, Weekly_test, type = "response")
glm_lag2_pred <- rep("Down", nrow(Weekly_test))
glm_lag2_pred[glm_lag2_probs > 0.5] <- "Up"
table(glm_lag2_pred, Weekly_test$Direction)##
## glm_lag2_pred Down Up
## Down 9 5
## Up 34 56mean(glm_lag2_pred == Weekly_test$Direction)## [1] 0.625- Using just
Lag2and holding out 2009-2010, the logistic regression achieves 62.5% correct predictions on the test set, an improvement over the full model. It still leans toward predicting “Up.”
(e) Repeat (d) using LDA.
lda_fit <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda_pred <- predict(lda_fit, Weekly_test)
table(lda_pred$class, Weekly_test$Direction)##
## Down Up
## Down 9 5
## Up 34 56mean(lda_pred$class == Weekly_test$Direction)## [1] 0.625- LDA produces the same confusion matrix and accuracy (~62.5%) as logistic regression, which makes sense given the two methods are closely related when class covariances are similar.
(f) Repeat (d) using QDA.
qda_fit <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda_pred <- predict(qda_fit, Weekly_test)
table(qda_pred$class, Weekly_test$Direction)##
## Down Up
## Down 0 0
## Up 43 61mean(qda_pred$class == Weekly_test$Direction)## [1] 0.5865385- QDA predicts “Up” for every test observation, resulting in ~58.7% accuracy while it does not predict any of the “Downs”. More flexible models do not automatically perform better.
(g) Repeat (d) using KNN with K = 1.
train_X <- as.matrix(Weekly$Lag2[train])
test_X <- as.matrix(Weekly$Lag2[!train])
train_Y <- Weekly$Direction[train]
set.seed(42)
knn_pred <- knn(train_X, test_X, train_Y, k = 1)
table(knn_pred, Weekly_test$Direction)##
## knn_pred Down Up
## Down 21 30
## Up 22 31mean(knn_pred == Weekly_test$Direction)## [1] 0.5- KNN with K = 1 achieves only about 50% accuracy, essentially random guessing. With K = 1 the model overfits the training data and doesn’t generalize well.
(h) Repeat (d) using naive Bayes.
nb_fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb_pred <- predict(nb_fit, Weekly_test)
table(nb_pred, Weekly_test$Direction)##
## nb_pred Down Up
## Down 0 0
## Up 43 61mean(nb_pred == Weekly_test$Direction)## [1] 0.5865385- Naive Bayes also predicts “Up” for all observations, yielding ~58.7% accuracy,identical to QDA. It completely misses “Down” weeks, similar to QDA’s behavior here.
(i) Which of these methods appears to provide the best results on this data?
- Logistic regression and LDA both achieve the highest test accuracy at ~62.5%, making them the best performers on this data. They also provide a more balanced prediction compared to QDA and Naive Bayes, which predict “Up” exclusively.
(j) Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
# Logistic regression with transformed Lag2
glm_tran <- glm(Direction ~ Lag2 + I(Lag2^2), data = Weekly, family = binomial, subset = train)
glm_tran_probs <- predict(glm_tran, Weekly_test, type = "response")
glm_tran_pred <- rep("Down", nrow(Weekly_test))
glm_tran_pred[glm_tran_probs > 0.5] <- "Up"
table(glm_tran_pred, Weekly_test$Direction)##
## glm_tran_pred Down Up
## Down 8 4
## Up 35 57mean(glm_tran_pred == Weekly_test$Direction)## [1] 0.625# LDA with Lag2 Lag1 Interactions
lda_inter <- lda(Direction ~ Lag1 + Lag2 + Lag1:Lag2, data = Weekly, subset = train)
lda_inter <- predict(lda_inter, Weekly_test)
table(lda_inter$class, Weekly_test$Direction)##
## Down Up
## Down 7 8
## Up 36 53mean(lda_inter$class == Weekly_test$Direction)## [1] 0.5769231- Adding a quadratic term for
Lag2in logistic regression made it stay exactly the same at 62.5%. LDA withLag1 * Lag2declined to 57.6%.
Problem 14
(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median.
Auto$mpg01 <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpg01 <- as.factor(Auto$mpg01)
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name mpg01
## 1 chevrolet chevelle malibu 0
## 2 buick skylark 320 0
## 3 plymouth satellite 0
## 4 amc rebel sst 0
## 5 ford torino 0
## 6 ford galaxie 500 0(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.
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")
- Cars with high mpg tend to have fewer cylinders, lower displacement,
lower horsepower, and lower weight. Acceleration and year also show some
separation, with newer cars more likely to have high mpg. Based on these
plots,
cylinders,displacement,horsepower, andweightappear to be the most useful predictors formpg01.
(c) Split the data into a training set and a test set.
set.seed(42)
n <- nrow(Auto)
train_idx <- sample(1:n, size = floor(0.65 * n))
auto_train <- Auto[train_idx, ]
auto_test <- Auto[-train_idx, ](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?
lda_auto <- lda(mpg01 ~ cylinders + displacement + horsepower + weight,
data = auto_train)
lda_auto_pred <- predict(lda_auto, auto_test)
table(lda_auto_pred$class, auto_test$mpg01)##
## 0 1
## 0 52 6
## 1 8 72mean(lda_auto_pred$class != auto_test$mpg01)## [1] 0.1014493- LDA achieves a test error of roughly 10-12%, showing it does a solid job separating high and low mpg cars using the four selected predictors.
(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?
qda_auto <- qda(mpg01 ~ cylinders + displacement + horsepower + weight,
data = auto_train)
qda_auto_pred <- predict(qda_auto, auto_test)
table(qda_auto_pred$class, auto_test$mpg01)##
## 0 1
## 0 53 7
## 1 7 71mean(qda_auto_pred$class != auto_test$mpg01)## [1] 0.1014493- QDA performs comparably to LDA, with a test error in a similar range. Even with the differences in the model they both perform well.
(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?
glm_auto <- glm(mpg01 ~ cylinders + displacement + horsepower + weight,
data = auto_train, family = binomial)
glm_auto_probs <- predict(glm_auto, auto_test, type = "response")
glm_auto_pred <- ifelse(glm_auto_probs > 0.5, 1, 0)
table(glm_auto_pred, auto_test$mpg01)##
## glm_auto_pred 0 1
## 0 54 6
## 1 6 72mean(glm_auto_pred != auto_test$mpg01)## [1] 0.08695652- Logistic regression outputs just as similar to QDA and LDA. All three parametric methods are competitive on this data set.
(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?
nb_auto <- naiveBayes(mpg01 ~ cylinders + displacement + horsepower + weight,
data = auto_train)
nb_auto_pred <- predict(nb_auto, auto_test)
table(nb_auto_pred, auto_test$mpg01)##
## nb_auto_pred 0 1
## 0 51 6
## 1 9 72mean(nb_auto_pred != auto_test$mpg01)## [1] 0.1086957- Naive Bayes performs reasonably well, though each of these variables
have high correlations with
mpg.
(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?
knn_trainX <- scale(auto_train[,c("cylinders","displacement","horsepower","weight")])
knn_testX <- scale(auto_test[,c("cylinders","displacement","horsepower","weight")])
knn_trainY <- auto_train$mpg01
knn_pred <- knn(knn_trainX, knn_testX, knn_trainY, k=5)
table(knn_pred, auto_test$mpg01)##
## knn_pred 0 1
## 0 52 5
## 1 8 73mean(knn_pred == auto_test$mpg01)## [1] 0.9057971knn.pred <- knn(knn_trainX, knn_testX, knn_trainY, k=10)
table(knn_pred, auto_test$mpg01)##
## knn_pred 0 1
## 0 52 5
## 1 8 73mean(knn_pred == auto_test$mpg01)## [1] 0.9057971knn.pred <- knn(knn_trainX, knn_testX, knn_trainY, k=1)
table(knn_pred, auto_test$mpg01)##
## knn_pred 0 1
## 0 52 5
## 1 8 73mean(knn_pred == auto_test$mpg01)## [1] 0.9057971- The results show me that the higher the K the better the accuracy is, with K = 10 at 90.5% and K = 5 being a close second. K = 1 suggests a little bit of an overfit.
Problem 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
# Create binary response variable
Boston$crim01 <- ifelse(Boston$crim > median(Boston$crim), 1, 0)
Boston$crim01 <- as.factor(Boston$crim01)
cor(Boston[, -which(names(Boston) == "crim01")])## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## black lstat medv
## crim -0.38506394 0.4556215 -0.3883046
## zn 0.17552032 -0.4129946 0.3604453
## indus -0.35697654 0.6037997 -0.4837252
## chas 0.04878848 -0.0539293 0.1752602
## nox -0.38005064 0.5908789 -0.4273208
## rm 0.12806864 -0.6138083 0.6953599
## age -0.27353398 0.6023385 -0.3769546
## dis 0.29151167 -0.4969958 0.2499287
## rad -0.44441282 0.4886763 -0.3816262
## tax -0.44180801 0.5439934 -0.4685359
## ptratio -0.17738330 0.3740443 -0.5077867
## black 1.00000000 -0.3660869 0.3334608
## lstat -0.36608690 1.0000000 -0.7376627
## medv 0.33346082 -0.7376627 1.0000000# Train/test split
set.seed(42)
n_bos <- nrow(Boston)
bos_train_idx <- sample(1:n_bos, size = floor(0.65 * n_bos))
bos_train <- Boston[bos_train_idx, ]
bos_test <- Boston[-bos_train_idx, ]# Logistic Regression
glm_bos <- glm(crim01 ~ nox + rad + age + dis + indus + tax, data = bos_train, family = binomial)
glm_bos_probs <- predict(glm_bos, bos_test, type = "response")
glm_bos_pred <- ifelse(glm_bos_probs > 0.5, 1, 0)
table(glm_bos_pred, bos_test$crim01)##
## glm_bos_pred 0 1
## 0 81 12
## 1 8 77mean(glm_bos_pred != bos_test$crim01)## [1] 0.1123596- Logistic regression achieves a test error of 11%, it correctly classified a great amount of observations.
# LDA
lda_bos <- lda(crim01 ~ nox + rad + age + dis + indus + tax, data = bos_train)
lda_bos_pred <- predict(lda_bos, bos_test)
table(lda_bos_pred$class, bos_test$crim01)##
## 0 1
## 0 87 20
## 1 2 69mean(lda_bos_pred$class != bos_test$crim01)## [1] 0.1235955- LDA does similar with a test error of just about 12%.
# Naive Bayes
nb_bos <- naiveBayes(crim01 ~ nox + rad + age + dis + indus + tax, data = bos_train)
nb_bos_pred <- predict(nb_bos, bos_test)
table(nb_bos_pred, bos_test$crim01)##
## nb_bos_pred 0 1
## 0 84 21
## 1 5 68mean(nb_bos_pred != bos_test$crim01)## [1] 0.1460674- Overall Naive Bayes performs the worst out of the three tests performed, with a test error of about 14%.
train_bos_knn <- scale(bos_train[,c("nox", "rad", "age", "dis", "indus", "tax")])
test_bos_knn <- scale(bos_test[,c("nox", "rad", "age", "dis", "indus", "tax")])
train_bos_y <- bos_train$crim01
knn_bos <- knn(train_bos_knn, test_bos_knn, train_bos_y, k=15)
table(knn_bos, bos_test$crim01)##
## knn_bos 0 1
## 0 79 4
## 1 10 85mean(knn_bos == bos_test$crim01)## [1] 0.9213483knn_bos <- knn(train_bos_knn, test_bos_knn, train_bos_y, k=10)
table(knn_bos, bos_test$crim01)##
## knn_bos 0 1
## 0 81 4
## 1 8 85mean(knn_bos == bos_test$crim01)## [1] 0.9325843knn_bos <- knn(train_bos_knn, test_bos_knn, train_bos_y, k=1)
table(knn_bos, bos_test$crim01)##
## knn_bos 0 1
## 0 83 4
## 1 6 85mean(knn_bos == bos_test$crim01)## [1] 0.9438202- KNN performs very competitively, especially at mid-range K values 1-10. With the crime patterns being non linear KNN does a better job at predicting than the model before. Overall, KNN with a well-chosen K and scaled predictors tends to produce the lowest test error on this data.