HW3
Michal Greenwood
March 1, 2019
Make sure to have ISLR, MASS and class installed
Q10This 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?
For Year and volume there appears to be en exponential relationshiop. Everything else does not indicate a clear pattern.
plot(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(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?
attach(Weekly)
glm.fit = glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly,
family = binomial)
summary(glm.fit)##
## 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: 4Given the P-value it looks that there is significance of 3%.
(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.fit, type = "response")
glm.pred = rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
table(glm.pred, Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557(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.0910 = Weekly[!train, ]
glm.fit = glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
glm.probs = predict(glm.fit, Weekly.0910, type = "response")
glm.pred = rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(glm.pred, Direction.0910)## Direction.0910
## glm.pred Down Up
## Down 9 5
## Up 34 56mean(glm.pred == Direction.0910)## [1] 0.625(e) Repeat (d) using LDA.
library(MASS)
lda.fit = lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly.0910)
table(lda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == Direction.0910)## [1] 0.625(f) Repeat (d) using QDA.
qda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.class = predict(qda.fit, Weekly.0910)$class
table(qda.class, Direction.0910)## Direction.0910
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == Direction.0910)## [1] 0.5865385An accuracy of 58.65%
(g) Repeat (d) using KNN with K = 1.
library(class)
train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.Direction = Direction[train]
set.seed(1)
knn.pred = knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Direction.0910)## Direction.0910
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Direction.0910)## [1] 0.5(h) Which of these methods appears to provide the best results on this data?
The LDA method and logistic regression do provide very similar test error rates.
(i) Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
glm.fit = glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
glm.probs = predict(glm.fit, Weekly.0910, type = "response")
glm.pred = rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(glm.pred, Direction.0910)## Direction.0910
## glm.pred Down Up
## Down 1 1
## Up 42 60mean(glm.pred == Direction.0910)## [1] 0.5865385lda.fit = lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly.0910)
mean(lda.pred$class == Direction.0910)## [1] 0.5769231qda.fit = qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
qda.class = predict(qda.fit, Weekly.0910)$class
table(qda.class, Direction.0910)## Direction.0910
## qda.class Down Up
## Down 12 13
## Up 31 48mean(qda.class == Direction.0910)## [1] 0.5769231knn.pred = knn(train.X, test.X, train.Direction, k = 10)
table(knn.pred, Direction.0910)## Direction.0910
## knn.pred Down Up
## Down 17 18
## Up 26 43mean(knn.pred == Direction.0910)## [1] 0.5769231knn.pred = knn(train.X, test.X, train.Direction, k = 100)
table(knn.pred, Direction.0910)## Direction.0910
## knn.pred Down Up
## Down 9 12
## Up 34 49mean(knn.pred == Direction.0910)## [1] 0.5576923Out of all the computations the original LDA and log. regression produced better results in the test error rates.
Q11 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.
library(ISLR)
attach(Auto)(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.
summary(Auto)## mpg cylinders displacement horsepower
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0
##
## weight acceleration year origin
## Min. :1613 Min. : 8.00 Min. :70.00 Min. :1.000
## 1st Qu.:2225 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000
## Median :2804 Median :15.50 Median :76.00 Median :1.000
## Mean :2978 Mean :15.54 Mean :75.98 Mean :1.577
## 3rd Qu.:3615 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000
## Max. :5140 Max. :24.80 Max. :82.00 Max. :3.000
##
## name
## amc matador : 5
## ford pinto : 5
## toyota corolla : 5
## amc gremlin : 4
## amc hornet : 4
## chevrolet chevette: 4
## (Other) :365mpg01 = 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 otherbfeatures 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 1.0000000plot(Auto)
(c) Split the data into a training set and a test set.
train = (year%%2 == 0)
test = !train
Auto.train = Auto[train, ]
Auto.test = Auto[test, ]
mpg01.test = mpg01[test](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?
library(MASS)
lda.fit = lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
subset = train)
lda.pred = predict(lda.fit, Auto.test)
mean(lda.pred$class != mpg01.test)## [1] 0.1263736There is a 12.64% test error rate
(e) Perform QDA on the training data in order to predict mpg01using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit = qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
subset = train)
qda.pred = predict(qda.fit, Auto.test)
mean(qda.pred$class != mpg01.test)## [1] 0.1318681There is a 13.19% test error rate
(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.fit = glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
family = binomial, subset = train)
glm.probs = predict(glm.fit, Auto.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != mpg01.test)## [1] 0.1208791There is a 12.09% test error rate
(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?
library(class)
train.X = cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X = cbind(cylinders, weight, displacement, horsepower)[test, ]
train.mpg01 = mpg01[train]
set.seed(1)
# KNN(k=1)
knn.pred = knn(train.X, test.X, train.mpg01, k = 1)
mean(knn.pred != mpg01.test)## [1] 0.1538462knn.pred = knn(train.X, test.X, train.mpg01, k = 10)
mean(knn.pred != mpg01.test)## [1] 0.1648352knn.pred = knn(train.X, test.X, train.mpg01, k = 100)
mean(knn.pred != mpg01.test)## [1] 0.1428571It appears with K=100 the error rate was the lowest with 14.29% using the 100 nearest neighbors. ## Q13 Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median.Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00attach(Boston)
crime01 = rep(0, length(crim))
crime01[crim > median(crim)] = 1
Boston = data.frame(Boston, crime01)
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]glm.fit = glm(crime01 ~ . - crime01 - crim, data = Boston, family = binomial,
subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredglm.probs = predict(glm.fit, Boston.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != crime01.test)## [1] 0.1818182There is a 18.18% test error rate
glm.fit = glm(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, family = binomial,
subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredglm.probs = predict(glm.fit, Boston.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
mean(glm.pred != crime01.test)## [1] 0.1857708There is a 18.58% test error rate
lda.fit = lda(crime01 ~ . - crime01 - crim, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1343874There is a 13.44% test error rate
lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1225296There is a 12.25% test error rate
lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax - lstat - indus - age,
data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1185771There is a 11.86% test error rate
library(class)
train.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black,
lstat, medv)[train, ]
test.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black,
lstat, medv)[test, ]
train.crime01 = crime01[train]
set.seed(1)
# KNN(k=1)
knn.pred = knn(train.X, test.X, train.crime01, k = 1)
mean(knn.pred != crime01.test)## [1] 0.458498There is a 45.85% test error rate
knn.pred = knn(train.X, test.X, train.crime01, k = 10)
mean(knn.pred != crime01.test)## [1] 0.1185771There is a 11.86% test error rate
knn.pred = knn(train.X, test.X, train.crime01, k = 100)
mean(knn.pred != crime01.test)## [1] 0.4901186There is a 49.01% test error rate
train.X = cbind(zn, nox, rm, dis, rad, ptratio, black, medv)[train, ]
test.X = cbind(zn, nox, rm, dis, rad, ptratio, black, medv)[test, ]
knn.pred = knn(train.X, test.X, train.crime01, k = 10)
mean(knn.pred != crime01.test)## [1] 0.2727273There is a 27.27% test error rate