Assignment 3
Andre Vicente - ods882
6/22/2021
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.
library(ISLR)
library(tidyverse)## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --## v ggplot2 3.3.3 v purrr 0.3.4
## v tibble 3.1.2 v dplyr 1.0.6
## v tidyr 1.1.3 v stringr 1.4.0
## v readr 1.4.0 v forcats 0.5.1## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()as_tibble(Weekly)## # A tibble: 1,089 x 9
## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
## 1 1990 0.816 1.57 -3.94 -0.229 -3.48 0.155 -0.27 Down
## 2 1990 -0.27 0.816 1.57 -3.94 -0.229 0.149 -2.58 Down
## 3 1990 -2.58 -0.27 0.816 1.57 -3.94 0.160 3.51 Up
## 4 1990 3.51 -2.58 -0.27 0.816 1.57 0.162 0.712 Up
## 5 1990 0.712 3.51 -2.58 -0.27 0.816 0.154 1.18 Up
## 6 1990 1.18 0.712 3.51 -2.58 -0.27 0.154 -1.37 Down
## 7 1990 -1.37 1.18 0.712 3.51 -2.58 0.152 0.807 Up
## 8 1990 0.807 -1.37 1.18 0.712 3.51 0.132 0.041 Up
## 9 1990 0.041 0.807 -1.37 1.18 0.712 0.144 1.25 Up
## 10 1990 1.25 0.041 0.807 -1.37 1.18 0.134 -2.68 Down
## # ... with 1,079 more rows- 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
##
##
##
## There are more weeks ending in Up (605) than in Down (484). So overall, we would expect the &P 500 stock index to have increased between 1990 and 2010. That is confirmed that both the mean and median of percentage return for the week variable are positive.
The distribution of weekly percentage return for downs and up weeks are similar, both with means around 1.25 (up) and -1.25 (down). Most of the weekly percentage return range from -2% to 2%.
- 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_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: 4Only the lag2 predictor is significant in the model to explain the variation of the dependent variable Direction.
- 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.
predicted <- ifelse(predict(glm_fit, type = "response") < 0.5, "Down", "Up") %>%factor()
table(predicted, Weekly$Direction)##
## predicted Down Up
## Down 54 48
## Up 430 557- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train <- Weekly[Weekly$Year <= 2008, ]
test <- Weekly[Weekly$Year > 2008, ]
glm_fit <- glm(Direction ~ Lag2, data = train, family = "binomial")
glm_predicted <- ifelse(predict(glm_fit, newdata = test, type = "response") < 0.5, "Down", "Up") %>% factor()
table(glm_predicted, test$Direction)##
## glm_predicted Down Up
## Down 9 5
## Up 34 56#Accuracy
mean(glm_predicted == test$Direction)## [1] 0.625- Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectlda_fit <- lda(Direction ~ Lag2, data = train)
lad_predicted <- predict(lda_fit, newdata = test)$class
table(lad_predicted, test$Direction)##
## lad_predicted Down Up
## Down 9 5
## Up 34 56#Accuracy
mean(lad_predicted == test$Direction)## [1] 0.625- Repeat (d) using QDA.
qda_fit <- qda(Direction ~ Lag2, data = train)
qda_predicted <- predict(qda_fit, newdata = test)$class
table(qda_predicted, test$Direction)##
## qda_predicted Down Up
## Down 0 0
## Up 43 61#Accuracy
mean(qda_predicted == test$Direction)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
library(class)
train.X= data.frame(Lag2 = train$Lag2)
test.X=data.frame(Lag2 = test$Lag2)
train.Direction=train$Direction
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=1)
table(knn.pred,test$Direction)##
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred==test$Direction)## [1] 0.5- Which of these methods appears to provide the best results on this data?
In this particular dataset, the linear classifiers outperform the other ones. Both the logistic regression and the Linear discriminant analysis attain the accuracy of 62.25%.
- 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 the interaction
glm_fit <- glm(Direction ~ Lag2:Lag1, data = train, family = "binomial")
glm_predicted <- ifelse(predict(glm_fit, newdata = test, type = "response") < 0.5, "Down", "Up") %>% factor()
table(glm_predicted, test$Direction)##
## glm_predicted Down Up
## Down 1 1
## Up 42 60#Accuracy
mean(glm_predicted == test$Direction)## [1] 0.5865385LDA regression with the interaction
lda_fit <- lda(Direction ~Lag2:Lag1, data = train)
lad_predicted <- predict(lda_fit, newdata = test)$class
table(lad_predicted, test$Direction)##
## lad_predicted Down Up
## Down 0 1
## Up 43 60#Accuracy
mean(lad_predicted == test$Direction)## [1] 0.5769231QDA with all the lagas as predictors
qda_fit <- qda(Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5, data = train)
qda_predicted <- predict(qda_fit, newdata = test)$class
table(qda_predicted, test$Direction)##
## qda_predicted Down Up
## Down 10 23
## Up 33 38#Accuracy
mean(qda_predicted == test$Direction)## [1] 0.4615385KNN with k = 3,5 and 10
train.X= data.frame(Lag2 = train$Lag2)
test.X=data.frame(Lag2 = test$Lag2)
train.Direction=train$Direction
set.seed(1)
#K1
knn.pred=knn(train.X,test.X,train.Direction,k=3)
table(knn.pred,test$Direction)##
## knn.pred Down Up
## Down 16 20
## Up 27 41mean(knn.pred==test$Direction)## [1] 0.5480769#K3
knn.pred=knn(train.X,test.X,train.Direction,k=5)
table(knn.pred,test$Direction)##
## knn.pred Down Up
## Down 15 20
## Up 28 41mean(knn.pred==test$Direction)## [1] 0.5384615#K5
knn.pred=knn(train.X,test.X,train.Direction,k=10)
table(knn.pred,test$Direction)##
## knn.pred Down Up
## Down 19 19
## Up 24 42mean(knn.pred==test$Direction)## [1] 0.5865385At end, the initial linear models (logistic and LDA) with a single predictor Lag2 still attaining the best prediction accuracy.
11. In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
library(ISLR)
as_tibble(Auto)## # A tibble: 392 x 9
## mpg cylinders displacement horsepower weight acceleration year origin
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 18 8 307 130 3504 12 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11 70 1
## 4 16 8 304 150 3433 12 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10 70 1
## 7 14 8 454 220 4354 9 70 1
## 8 14 8 440 215 4312 8.5 70 1
## 9 14 8 455 225 4425 10 70 1
## 10 15 8 390 190 3850 8.5 70 1
## # ... with 382 more rows, and 1 more variable: name <fct>- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg01 = rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] = 1
new_auto = data.frame(Auto, mpg01)
attach(new_auto)## The following object is masked _by_ .GlobalEnv:
##
## mpg01## The following object is masked from package:ggplot2:
##
## 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(new_auto[,-9])
The most evident correlation are displacement, horsepower, weight and acceleration. Since displacement, horsepower, weight are very correlated among themselves, horsepower and weight could better predictor for the model’s simplicity.
- Split the data into a training set and a test set.
auto_train_year = (year%%2 == 0)
auto_test_year = !auto_train_year
auto_train = new_auto[auto_train_year, ]
auto_test = new_auto[auto_test_year, ]
mpg01_test = mpg01[auto_test_year]
dim(auto_train)FALSE [1] 210 10dim(auto_test)FALSE [1] 182 10- 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?
auto_lda_fit = lda(mpg01 ~ acceleration + weight + horsepower, data = auto_train)
auto_lda_pred = predict(auto_lda_fit, auto_test)$class
mean(auto_lda_pred != mpg01_test)## [1] 0.1483516Error of 14.84%.
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto_qda_fit = qda(mpg01 ~acceleration + weight + horsepower, data = auto_train)
auto_qda_pred = predict(auto_qda_fit, auto_test)$class
mean(auto_qda_pred != mpg01_test)## [1] 0.1483516No change. Same error of 14.84%.
- 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?
auto_glm_fit = glm(mpg01 ~ acceleration + weight + horsepower, data = auto_train,family = binomial)
auto_glm_pred_values = predict(auto_glm_fit, auto_test, type = "response")
auto_glm_pred = rep(0, length(auto_glm_pred_values))
auto_glm_pred[auto_glm_pred_values > 0.5] = 1
mean(auto_glm_pred != mpg01_test)## [1] 0.1483516No change. Same error of 14.84%.
- 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(acceleration, weight, horsepower)[auto_train_year, ]
test.X = cbind(acceleration, weight, horsepower)[auto_test_year, ]
train.mpg01 = mpg01[auto_train_year]
set.seed(1)
# k=1
knn.pred = knn(train.X, test.X, train.mpg01, k = 1)
mean(knn.pred != mpg01_test)## [1] 0.1593407# k=5
knn.pred = knn(train.X, test.X, train.mpg01, k = 5)
mean(knn.pred != mpg01_test)## [1] 0.1483516# k=10
knn.pred = knn(train.X, test.X, train.mpg01, k = 10)
mean(knn.pred != mpg01_test)## [1] 0.1483516# k=20
knn.pred = knn(train.X, test.X, train.mpg01, k = 20)
mean(knn.pred != mpg01_test)## [1] 0.1538462# k=50
knn.pred = knn(train.X, test.X, train.mpg01, k = 50)
mean(knn.pred != mpg01_test)## [1] 0.1428571# k=100
knn.pred = knn(train.X, test.X, train.mpg01, k = 100)
mean(knn.pred != mpg01_test)## [1] 0.1428571# k=150
knn.pred = knn(train.X, test.X, train.mpg01, k = 150)
mean(knn.pred != mpg01_test)## [1] 0.1923077K=50 and k=100 have the smallest error of 14.29%. Also, that was the smallest error rate among the classifiers tested.
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.
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. :50.00detach(new_auto)
attach(Boston)bos_lm <- lm(crim~., data = Boston)
summary(bos_lm)##
## Call:
## lm(formula = crim ~ ., data = Boston)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.924 -2.120 -0.353 1.019 75.051
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 17.033228 7.234903 2.354 0.018949 *
## zn 0.044855 0.018734 2.394 0.017025 *
## indus -0.063855 0.083407 -0.766 0.444294
## chas -0.749134 1.180147 -0.635 0.525867
## nox -10.313535 5.275536 -1.955 0.051152 .
## rm 0.430131 0.612830 0.702 0.483089
## age 0.001452 0.017925 0.081 0.935488
## dis -0.987176 0.281817 -3.503 0.000502 ***
## rad 0.588209 0.088049 6.680 6.46e-11 ***
## tax -0.003780 0.005156 -0.733 0.463793
## ptratio -0.271081 0.186450 -1.454 0.146611
## black -0.007538 0.003673 -2.052 0.040702 *
## lstat 0.126211 0.075725 1.667 0.096208 .
## medv -0.198887 0.060516 -3.287 0.001087 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 6.439 on 492 degrees of freedom
## Multiple R-squared: 0.454, Adjusted R-squared: 0.4396
## F-statistic: 31.47 on 13 and 492 DF, p-value: < 2.2e-16The predictors that are significant are zn, dis, rad, black, medv.
violent = rep(0, length(crim))
violent[crim > median(crim)] = 1
Boston = data.frame(Boston, violent)
trainb = 1:(dim(Boston)[1]/2)
testb = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston_train = Boston[trainb, ]
Boston_test = Boston[testb, ]
violent_test = violent[testb]
dim(Boston_train)## [1] 253 15dim(Boston_test)## [1] 253 15Logistic Regression
bos_glm_fit = glm(violent ~ zn+dis+ rad+ black+ medv, data = Boston_train, family = binomial)
bos_glm_pred_values = predict(bos_glm_fit, Boston_test, type = "response")
bos_glm_pred = rep(0, length(bos_glm_pred_values))
bos_glm_pred[bos_glm_pred_values > 0.5] = 1
mean(bos_glm_pred == violent_test)## [1] 0.8577075LDA - SIGNIFICANT
bos_lda_fit = lda(violent ~ zn+dis+ rad+ black+ medv, data = Boston_train)
bos_lda_pred = predict(bos_lda_fit, Boston_test)$class
mean(bos_lda_pred == violent_test)## [1] 0.8814229LDA - ALL
bos_lda_fit = lda(violent ~ zn+ indus+ chas+nox+rm+ age+ dis+rad+tax+ptratio+black+
lstat+ medv, data = Boston_train)
bos_lda_pred = predict(bos_lda_fit, Boston_test)$class
mean(bos_lda_pred == violent_test)## [1] 0.8656126QDA
bos_qda_fit = qda(violent ~ zn+dis+ rad+ black+ medv, data = Boston_train)
bos_qda_pred = predict(bos_qda_fit, Boston_test)$class
mean(bos_qda_pred == violent_test)## [1] 0.4940711KNN
bos_train_knn_X = cbind(zn, dis, rad, black, medv)[trainb, ]
bos_test_knn_X = cbind(zn, dis, rad, black, medv)[testb, ]
train_knn_violent = violent[trainb]
set.seed(1)
# k=1
knn.pred = knn(bos_train_knn_X, bos_test_knn_X, train_knn_violent, k = 1)
mean(knn.pred == violent_test)## [1] 0.743083# k=5
knn.pred = knn(bos_train_knn_X, bos_test_knn_X, train_knn_violent, k = 5)
mean(knn.pred == violent_test)## [1] 0.7905138# k=10
knn.pred = knn(bos_train_knn_X, bos_test_knn_X, train_knn_violent, k = 10)
mean(knn.pred == violent_test)## [1] 0.7391304# k=50
knn.pred = knn(bos_train_knn_X, bos_test_knn_X, train_knn_violent, k = 50)
mean(knn.pred == violent_test)## [1] 0.7509881# k=100
knn.pred = knn(bos_train_knn_X, bos_test_knn_X, train_knn_violent, k = 100)
mean(knn.pred == violent_test)## [1] 0.5652174The highest accuracy among the tested classifiers was of 88.14% attained by the Linear discriminant Analysis using a few significant predictors.