Ch4
Victor Ramos
9/19/2021
#Chapter 4 HW
##Question 10
10a)Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5library(class)
library(MASS)
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
##
##
##
## 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 1.000000000plot(Weekly)
10b)Perform a logistic regression with Directions as the response & five lag variables plus Volume as predictors. Use summary to print results. Do any of the predictors appear to be statistically significant?
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: 4Lag2 is significant
10c)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>.5]="up"
table(glm.pred, Weekly$Direction)##
## glm.pred Down Up
## Down 54 48
## up 430 557When the market is going up, the logistic regression prediction is right 91.12% of the time. When the market goes down the regression prediction was right 52.94% of the time.
10d)Fit the logistic regression model using training data period 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct for the held out data.
attach(Weekly)
train = (Year < 2009)
Weekly.90 = Weekly[!train,]
glm.fit2 = glm(Direction~Lag2, data = Weekly, family = binomial, subset = train)
glm.prob1 = predict(glm.fit2, Weekly.90, type = "response")
glm.pred1 = rep("Down", length(glm.prob1))
glm.pred1[glm.prob1 > .5] = "Up"
Direction.90 = Direction[!train]
table(glm.pred1, Direction.90)## Direction.90
## glm.pred1 Down Up
## Down 9 5
## Up 34 56mean(glm.pred == Direction.90)## Warning in `==.default`(glm.pred, Direction.90): longer object length is not a
## multiple of shorter object length## Warning in is.na(e1) | is.na(e2): longer object length is not a multiple of
## shorter object length## [1] 0.03581267detach(Weekly)10e)Repeat d using LDA
attach(Weekly.90)
lda.fit = lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly.90)
table(lda.pred$class, Direction.90)## Direction.90
## Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == Direction.90)## [1] 0.625detach(Weekly.90)The overall accuracy for predicting up is FALSE, FALSE, TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, FALSE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, FALSE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, TRUE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE
10f)Repeat d using QDA
qda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.class = predict(qda.fit, Weekly.90)$class
table(qda.class, Direction.90)## Direction.90
## qda.class Down Up
## Down 0 0
## Up 43 61qda.class == Direction.90## [1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE
## [13] TRUE TRUE TRUE FALSE TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE
## [25] FALSE FALSE FALSE TRUE TRUE TRUE TRUE FALSE TRUE TRUE FALSE TRUE
## [37] TRUE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE FALSE TRUE TRUE
## [49] TRUE FALSE TRUE FALSE TRUE FALSE FALSE FALSE FALSE TRUE TRUE FALSE
## [61] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE FALSE TRUE FALSE
## [73] TRUE FALSE TRUE TRUE FALSE FALSE TRUE FALSE TRUE FALSE TRUE FALSE
## [85] FALSE FALSE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
## [97] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUEOverall accuracy is 70%
10g)Repeat d using KNN with K = 1
attach(Weekly)
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.90)## Direction.90
## knn.pred Down Up
## Down 21 30
## Up 22 31detach(Weekly)10h)
10i)
# Logistic regression with Lag2:Lag1
attach(Weekly.90)
glm.fit3 = glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
glm.probs1 = predict(glm.fit3, Weekly.90, type = "response")
glm.pred = rep("Down", length(glm.probs1))
glm.pred[glm.probs1 > 0.5] = "Up"
Direction.90 = Weekly$Direction[!train]
table(glm.pred, Direction.90)## Direction.90
## glm.pred Down Up
## Down 1 1
## Up 42 60# LDA with Lag2 interaction with Lag1
lda.fit3 = lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
lda.pred = predict(lda.fit3, Weekly.90)
mean(lda.pred$class == Direction.90)## [1] 0.576923111
library(ISLR)
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) :36511a)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.
mpg01 = rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] = 1
auto = data.frame(Auto, mpg01)11b)xplore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01?
cor(auto[,-9])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(auto)
11c)Split the data into a training set and test set
train = (auto$year%%2 == 0)
test = !train
auto.train = auto[train, ]
auto.test = auto[test, ]
mpg01.test = mpg01[test]11d)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.fita = lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = auto, subset = train)
lda.preda = predict(lda.fita, auto.test)
mean(lda.preda$class != mpg01.test)## [1] 0.126373611e)Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fita = qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = auto, subset = train)
qda.preda = predict(qda.fita, auto.test)
mean(qda.preda$class != mpg01.test)## [1] 0.131868111f)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.fita = glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = auto, subset = train)
glm.probsa = predict(glm.fita, auto.test, type = "response")
glm.preda = rep(0, length(glm.probsa))
glm.preda[glm.probsa > 0.5] = 1
mean(glm.preda != mpg01.test)## [1] 0.126373611g)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?
attach(auto)## The following object is masked _by_ .GlobalEnv:
##
## mpg01train.xa = cbind(cylinders, weight, displacement, horsepower)[train,]
test.XA = cbind(cylinders, weight, displacement, horsepower)[test,]
train.mpg01 = mpg01[train]
set.seed(1)
knn.preda = knn(train.xa, test.XA, train.mpg01, k = 1)
mean(knn.preda != mpg01.test)## [1] 0.1538462detach(auto)knn.preda2 = knn(train.xa, test.XA, train.mpg01, k = 50)
mean(knn.preda2 != mpg01.test)## [1] 0.1483516knn.preda3 = knn(train.xa, test.XA, train.mpg01, k = 1)
mean(knn.preda3 != mpg01.test)## [1] 0.1538462K with 50 seems to be the best at .148 error rate
13)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.
attach(Boston)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 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. :50.00crime01 = 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]#Logistic Regression
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] 0.1818182glm.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] 0.1857708# LDA
lda.fit = lda(crime01 ~ . - crime01 - crim, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1343874# LDA
lda.fit = lda(crime01 ~ . - crime01 - crim, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1343874lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax, data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1225296lda.fit = lda(crime01 ~ . - crime01 - crim - chas - tax - lstat - indus - age,
data = Boston, subset = train)
lda.pred = predict(lda.fit, Boston.test)
mean(lda.pred$class != crime01.test)## [1] 0.1185771train.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.pred = knn(train.X, test.X, train.crime01, k = 1)
mean(knn.pred != crime01.test)## [1] 0.458498knn.pred = knn(train.X, test.X, train.crime01, k = 1)
mean(knn.pred != crime01.test)## [1] 0.458498# KNN(k=10) with subset of variables
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] 0.2766798