Data Mining HW3
Roberto Cottle
March 3, 2020
Data Mining HW3
Problem 10
This question should be answered using the Weekly data set, which is part of the ISLR 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.
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library("ISLR")
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
##
##
##
## attach(Weekly)
plot(Volume)
The plot function helps us better determine if there is a pattern in the data. Plotting Volume, we can see a pattern in that it steadily increases until a much more dramatic increase towards the end.
(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?
HW3fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(HW3fit)##
## 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: 4At the 5% significance level, we can determine that only “Lag2” is statistically significant as it is the only predictor with a p value less than the .05 threshold.
(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.
probs <- predict(HW3fit, type = "response")
pred.glm <- rep("Down", length(probs))
pred.glm[probs > 0.5] <- "Up"
table(pred.glm, Direction)## Direction
## pred.glm Down Up
## Down 54 48
## Up 430 557The percentage of correct predictions on the training data is (54+557)/1089 which is equal to roughly 56%. We conclude that the training error rate is roughly 44%. We can also conclude that for weeks when the market goes up, the model is right roughly 92% of the time (557/(48+557)). And for weeks when the market goes down, the model is right only about 11% of the time (54/(54+430)).
(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 <- (Year < 2009)
Weekly.20092010 <- Weekly[!train, ]
Direction.20092010 <- Direction[!train]
fit.glm2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
summary(fit.glm2)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4probs2 <- predict(fit.glm2, Weekly.20092010, type = "response")
pred.glm2 <- rep("Down", length(probs2))
pred.glm2[probs2 > 0.5] <- "Up"
table(pred.glm2, Direction.20092010)## Direction.20092010
## pred.glm2 Down Up
## Down 9 5
## Up 34 56In this case, we may conclude that the percentage of correct predictions on the test data is (9+56)/104 wich is equal to 62.5%. In other words 37.5% is the test error rate. We could also say that for weeks when the market goes up, the model is right 91.8032787% of the time (56/(56+5)). For weeks when the market goes down, the model is right only 20.9302326% of the time (9/(9+34)).
(e) Repeat (d) using LDA.
library(MASS)
fit.lda <- lda(Direction ~ Lag2, data = Weekly, subset = train)
fit.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 0.4414162pred.lda <- predict(fit.lda, Weekly.20092010)
table(pred.lda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 9 5
## Up 34 56We can conclude that the percentage of correct predictions on the test data is 62.5%. This also means that 37.5% is the test error rate. We could also say that for weeks when the market goes up, the model is right roughly 91% of the time. For weeks when the market goes down, the model is right roughly 20% of the time. These results are very close to those obtained with the logistic regression model which is not surpising.
(f) Repeat (d) using QDA.
fit.qda <- qda(Direction ~ Lag2, data = Weekly, subset = train)
fit.qda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 0 0
## Up 43 61We can conclude that the percentage of correct predictions on the test data is roughly 59%. So roughly 41% is the test error rate. We could also say that for weeks when the market goes up, the model is right 100% of the time. For weeks when the market goes down, the model is right only 0% of the time.
(g) Repeat (d) using KNN with K = 1.
library(class)
train.X <- as.matrix(Lag2[train])
test.X <- as.matrix(Lag2[!train])
train.Direction <- Direction[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.Direction, k = 1)
table(pred.knn, Direction.20092010)## Direction.20092010
## pred.knn Down Up
## Down 21 30
## Up 22 31We can conclude that the percentage of correct predictions on the test data is 50%. So 50% is the test error rate. We could also say that for weeks when the market goes up, the model is right roughly 51% of the time. For weeks when the market goes down, the model is right approximately 49% of the time.
(h) Which of these methods appears to provide the best results on this data?
If we compare the test error rates, we see that logistic regression and LDA have the minimum error rates, followed by QDA and KNN.
(i) 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 Lag2:Lag1
fit.glm3 <- glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
probs3 <- predict(fit.glm3, Weekly.20092010, type = "response")
pred.glm3 <- rep("Down", length(probs3))
pred.glm3[probs3 > 0.5] = "Up"
table(pred.glm3, Direction.20092010)## Direction.20092010
## pred.glm3 Down Up
## Down 1 1
## Up 42 60mean(pred.glm3 == Direction.20092010)## [1] 0.5865385# LDA with Lag2 interaction with Lag1
fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)## [1] 0.5769231# QDA with sqrt(abs(Lag2))
fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 12 13
## Up 31 48mean(pred.qda2$class == Direction.20092010)## [1] 0.5769231# KNN k =10
pred.knn2 <- knn(train.X, test.X, train.Direction, k = 10)
table(pred.knn2, Direction.20092010)## Direction.20092010
## pred.knn2 Down Up
## Down 17 18
## Up 26 43mean(pred.knn2 == Direction.20092010)## [1] 0.5769231# KNN k = 100
pred.knn3 <- knn(train.X, test.X, train.Direction, k = 100)
table(pred.knn3, Direction.20092010)## Direction.20092010
## pred.knn3 Down Up
## Down 9 12
## Up 34 49mean(pred.knn3 == Direction.20092010)## [1] 0.5576923These comparisons show that the original logistic regression and LDA have the best performance in terms of test error rates.
Problem 11
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.
(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.
attach(Auto)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)(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(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 1.0000000pairs(Auto)
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")
We can conclude that there exists some association between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower”.
(c) Split the data into a training set and a test set.
train <- (year %% 2 == 0)
Auto.train <- Auto[train, ]
Auto.test <- Auto[!train, ]
mpg01.test <- mpg01[!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?
fit.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.lda## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.6741402638
## weight -0.0011465750
## displacement 0.0004481325
## horsepower 0.0059035377pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)## mpg01.test
## 0 1
## 0 86 9
## 1 14 73mean(pred.lda$class != mpg01.test)## [1] 0.1263736We can see that the test error rate is roughly 13%.
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)## mpg01.test
## 0 1
## 0 89 13
## 1 11 69mean(pred.qda$class != mpg01.test)## [1] 0.1318681We can see that the test error rate is roughly 13%.
(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?
fit.glm <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset = train)
summary(fit.glm)##
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower,
## family = binomial, data = Auto, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.48027 -0.03413 0.10583 0.29634 2.57584
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 17.658730 3.409012 5.180 2.22e-07 ***
## cylinders -1.028032 0.653607 -1.573 0.1158
## weight -0.002922 0.001137 -2.569 0.0102 *
## displacement 0.002462 0.015030 0.164 0.8699
## horsepower -0.050611 0.025209 -2.008 0.0447 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 289.58 on 209 degrees of freedom
## Residual deviance: 83.24 on 205 degrees of freedom
## AIC: 93.24
##
## Number of Fisher Scoring iterations: 7probs <- predict(fit.glm, Auto.test, 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 89 11
## 1 11 71mean(pred.glm != mpg01.test)## [1] 0.1208791We can see that the test error rate is roughly 12%.
(g) 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?
train.X <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[!train, ]
train.mpg01 <- mpg01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.mpg01, k = 1)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 83 11
## 1 17 71mean(pred.knn != mpg01.test)## [1] 0.1538462The test error rate is 15.3846154% for K=1.
pred.knn <- knn(train.X, test.X, train.mpg01, k = 10)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 77 7
## 1 23 75mean(pred.knn != mpg01.test)## [1] 0.1648352The test error rate is 16.4835165% for K=10.
pred.knn <- knn(train.X, test.X, train.mpg01, k = 100)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 81 7
## 1 19 75mean(pred.knn != mpg01.test)## [1] 0.1428571K=100 has the best test error rate at 14.2857143%, which is lower than K=1 and K-10, so K=100 seems to fit the model the best.
Problem 12
This problem involves writing functions.
Power <- function() {2^3}
Power()## [1] 8Power2 <- function(x, a) {x^a}
Power2(3, 8)## [1] 6561Power2(10, 3)## [1] 1000Power2(8, 17)## [1] 2.2518e+15Power2(131, 3)## [1] 2248091Power3 <- function(x , a) {result <- x^a
return(result)}x <- 1:10
plot(x, Power3(x, 2), log = "xy", xlab = "Log of x", ylab = "Log of x^2", main = "Log of x^2 vs Log of x")
PlotPower <- function(x, a) {plot(x, Power3(x, a))}
PlotPower(1:10, 3)