Assignment 3
Sudeep Jacob
3/5/2021
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.3library(caret)## Warning: package 'caret' was built under R version 4.0.4## Loading required package: lattice## Loading required package: ggplot2library(MASS)
library(tidyverse)## -- Attaching packages -------------------------------------------------------------- tidyverse 1.3.0 --## v tibble 3.0.3 v dplyr 1.0.2
## v tidyr 1.1.2 v stringr 1.4.0
## v readr 1.3.1 v forcats 0.5.0
## v purrr 0.3.4## -- Conflicts ----------------------------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()
## x purrr::lift() masks caret::lift()
## x dplyr::select() masks MASS::select()library(class)- 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
data(Weekly)- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
pairs(Weekly[,-9])
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 1.000000000at first glance there is very little relationship between the lag variables, there appers to be a correlation between year and Volume.
- 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 ~. -Year -Today, data = Weekly, family = "binomial")
summary(glm_full)##
## Call:
## glm(formula = Direction ~ . - Year - Today, 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: 4- 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.
predict_glm <- predict(glm_full, type = "response")
predict_glm <- ifelse(predict_glm <0.5, "Down", "Up")
predict_glm<- factor(predict_glm)
confusionMatrix(predict_glm,Weekly$Direction, positive = "Up")## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 54 48
## Up 430 557
##
## Accuracy : 0.5611
## 95% CI : (0.531, 0.5908)
## No Information Rate : 0.5556
## P-Value [Acc > NIR] : 0.369
##
## Kappa : 0.035
##
## Mcnemar's Test P-Value : <2e-16
##
## Sensitivity : 0.9207
## Specificity : 0.1116
## Pos Pred Value : 0.5643
## Neg Pred Value : 0.5294
## Prevalence : 0.5556
## Detection Rate : 0.5115
## Detection Prevalence : 0.9063
## Balanced Accuracy : 0.5161
##
## 'Positive' Class : Up
## specificity is low aka the model does a bad job in predicting negative class
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train <- Weekly[Weekly$Year <= 2008,]
test <- Weekly[Weekly$Year >2008,]glm_train_test <- glm(Direction ~ Lag2, data = train, family ='binomial')
predict_test <- predict(glm_train_test, newdata = test, type = "response")
predict_test <- ifelse(predict_test <0.5, "Down", "Up")
predict_test <- factor(predict_test)
confusionMatrix(predict_test,test$Direction, positive = "Up")## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 9 5
## Up 34 56
##
## Accuracy : 0.625
## 95% CI : (0.5247, 0.718)
## No Information Rate : 0.5865
## P-Value [Acc > NIR] : 0.2439
##
## Kappa : 0.1414
##
## Mcnemar's Test P-Value : 7.34e-06
##
## Sensitivity : 0.9180
## Specificity : 0.2093
## Pos Pred Value : 0.6222
## Neg Pred Value : 0.6429
## Prevalence : 0.5865
## Detection Rate : 0.5385
## Detection Prevalence : 0.8654
## Balanced Accuracy : 0.5637
##
## 'Positive' Class : Up
## accuracy of 62.5% compared to 56.1%, but we can’t confirm the model is any better at prediction as the test sample was very small.
- Repeat (d) using LDA.
lda_lag2 <- lda(Direction ~ Lag2, data = train)
predict_lda <- predict(lda_lag2, newdata = test)
confusionMatrix(predict_lda$class,test$Direction, positive = "Up")## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 9 5
## Up 34 56
##
## Accuracy : 0.625
## 95% CI : (0.5247, 0.718)
## No Information Rate : 0.5865
## P-Value [Acc > NIR] : 0.2439
##
## Kappa : 0.1414
##
## Mcnemar's Test P-Value : 7.34e-06
##
## Sensitivity : 0.9180
## Specificity : 0.2093
## Pos Pred Value : 0.6222
## Neg Pred Value : 0.6429
## Prevalence : 0.5865
## Detection Rate : 0.5385
## Detection Prevalence : 0.8654
## Balanced Accuracy : 0.5637
##
## 'Positive' Class : Up
## accuracy didnt change, same assement as the earlier model
- Repeat (d) using QDA.
qda_lag2 <- qda(Direction ~ Lag2, data = train)
predict_qda <- predict(qda_lag2, newdata = test)
confusionMatrix(predict_qda$class,test$Direction, positive = "Up")## Confusion Matrix and Statistics
##
## Reference
## Prediction Down Up
## Down 0 0
## Up 43 61
##
## Accuracy : 0.5865
## 95% CI : (0.4858, 0.6823)
## No Information Rate : 0.5865
## P-Value [Acc > NIR] : 0.5419
##
## Kappa : 0
##
## Mcnemar's Test P-Value : 1.504e-10
##
## Sensitivity : 1.0000
## Specificity : 0.0000
## Pos Pred Value : 0.5865
## Neg Pred Value : NaN
## Prevalence : 0.5865
## Detection Rate : 0.5865
## Detection Prevalence : 1.0000
## Balanced Accuracy : 0.5000
##
## 'Positive' Class : Up
## accuracy fell. qda predicts UP for every observation so indicated by the specificity of 0.
Repeat (d) using KNN with K = 1.
Which of these methods appears to provide the best results on this data?
LDA and Logit regression get the same test accuracy and similiar results.
- 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.
- 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.
- 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.
data(Auto)
Auto$mpg01 <- factor(as.numeric(Auto$mpg > median(Auto$mpg)))
Auto$origin <- factor(Auto$origin)- 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.
Auto_new <- subset(Auto , select = -c(mpg,name))
glimpse(Auto_new)## Rows: 392
## Columns: 8
## $ cylinders <dbl> 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 6, 6, 6, ...
## $ displacement <dbl> 307, 350, 318, 304, 302, 429, 454, 440, 455, 390, 383,...
## $ horsepower <dbl> 130, 165, 150, 150, 140, 198, 220, 215, 225, 190, 170,...
## $ weight <dbl> 3504, 3693, 3436, 3433, 3449, 4341, 4354, 4312, 4425, ...
## $ acceleration <dbl> 12.0, 11.5, 11.0, 12.0, 10.5, 10.0, 9.0, 8.5, 10.0, 8....
## $ year <dbl> 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70...
## $ origin <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, ...
## $ mpg01 <fct> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, ...g1 <- ggplot(Auto_new, aes(x = mpg01, y = cylinders, col = mpg01)) +
geom_jitter() +
theme(legend.position = "none") +
ggtitle("Cylinders vs mpg01 - Jitter Plot")
g2 <- ggplot(Auto_new, aes(x = mpg01, y = displacement, fill = mpg01)) +
geom_boxplot() +
theme(legend.position = "none") +
ggtitle("Displacement vs mpg01 - Boxplot")
g3 <- ggplot(Auto_new, aes(x = mpg01, y = horsepower, fill = mpg01)) +
geom_boxplot() +
theme(legend.position = "none") +
ggtitle("Horsepower vs mpg01 - Boxplot")
g4 <- ggplot(Auto_new, aes(x = origin, fill = mpg01)) +
geom_bar(position = "fill") +
scale_y_continuous(labels = scales::percent_format()) +
theme(axis.title.y = element_blank()) +
ggtitle("Origin vs mpg01 - Bar Plot")
g1 
g2
g3
g4
interesting find: American muscle car gas guzzlers are not a myth as seen by the last graph. At large american cars where had mpg below median.
pairs(Auto_new)
- Split the data into a training set and a test set.
set.seed(123)
index <- sample(1:nrow(Auto),.75*nrow(Auto),replace=FALSE)
train <- Auto_new[index,]
test <- Auto_new[-index,]- 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+weight+horsepower, train)lda_auto_predict <- predict(lda_auto, test)
mean(lda_auto_predict$class != test$mpg01)## [1] 0.1122449- 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 <-train(mpg01 ~cylinders+displacement+weight+horsepower, train, method ="qda")
qda_auto_predict <- predict(qda_auto, test)
mean(qda_auto_predict!= test$mpg01)## [1] 0.1020408- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
glm_auto <- train(mpg01 ~ cylinders+displacement+weight+horsepower, data= train, method = 'glm',family= 'binomial')
glm_auto_predict <- predict(glm_auto, test)
mean(glm_auto_predict != test$mpg01)## [1] 0.122449- 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_auto <- train(mpg01 ~ cylinders + displacement + weight + horsepower,
data = train,
method = "knn",
preProcess = c("center", "scale"),
tuneGrid = expand.grid(k = seq(1, 80, 1)))
knn_auto## k-Nearest Neighbors
##
## 294 samples
## 4 predictor
## 2 classes: '0', '1'
##
## Pre-processing: centered (4), scaled (4)
## Resampling: Bootstrapped (25 reps)
## Summary of sample sizes: 294, 294, 294, 294, 294, 294, ...
## Resampling results across tuning parameters:
##
## k Accuracy Kappa
## 1 0.8819013 0.7632568
## 2 0.8846526 0.7688534
## 3 0.8846636 0.7687603
## 4 0.8900369 0.7795305
## 5 0.8921409 0.7837935
## 6 0.8970076 0.7934615
## 7 0.8989449 0.7973644
## 8 0.8994178 0.7983132
## 9 0.9042827 0.8080220
## 10 0.9031777 0.8056565
## 11 0.9061507 0.8115993
## 12 0.9068505 0.8130570
## 13 0.9053898 0.8100616
## 14 0.9049939 0.8092661
## 15 0.9047906 0.8088153
## 16 0.9047834 0.8087705
## 17 0.9036274 0.8065016
## 18 0.9033074 0.8058610
## 19 0.9022294 0.8037107
## 20 0.9018545 0.8029481
## 21 0.9011246 0.8014918
## 22 0.9007362 0.8007183
## 23 0.9014882 0.8022741
## 24 0.9014882 0.8022741
## 25 0.9014882 0.8022741
## 26 0.9014882 0.8022741
## 27 0.9014882 0.8022741
## 28 0.9014882 0.8022741
## 29 0.9014882 0.8022741
## 30 0.9014882 0.8022741
## 31 0.9011144 0.8015016
## 32 0.9011144 0.8015016
## 33 0.9011144 0.8015016
## 34 0.9011144 0.8015016
## 35 0.9011144 0.8015016
## 36 0.9011144 0.8015016
## 37 0.9007222 0.8007282
## 38 0.9007222 0.8007282
## 39 0.9007222 0.8007282
## 40 0.8999982 0.7992827
## 41 0.8999982 0.7992827
## 42 0.9003619 0.8000057
## 43 0.9000079 0.7992958
## 44 0.8999982 0.7992856
## 45 0.8996443 0.7985757
## 46 0.8996443 0.7985728
## 47 0.8999982 0.7992827
## 48 0.8999982 0.7992827
## 49 0.8996443 0.7985728
## 50 0.8996443 0.7985728
## 51 0.8996443 0.7985728
## 52 0.8988997 0.7970684
## 53 0.8988997 0.7970684
## 54 0.8992806 0.7978528
## 55 0.8992806 0.7978528
## 56 0.8992806 0.7978528
## 57 0.8992806 0.7978528
## 58 0.8988997 0.7970684
## 59 0.8985519 0.7963938
## 60 0.8985519 0.7963938
## 61 0.8981815 0.7956606
## 62 0.8981815 0.7956606
## 63 0.8981815 0.7956606
## 64 0.8981815 0.7956606
## 65 0.8981815 0.7956606
## 66 0.8981815 0.7956606
## 67 0.8981815 0.7956606
## 68 0.8981815 0.7956606
## 69 0.8981815 0.7956606
## 70 0.8981815 0.7956606
## 71 0.8981815 0.7956606
## 72 0.8974792 0.7942662
## 73 0.8974602 0.7942602
## 74 0.8978714 0.7950373
## 75 0.8978523 0.7950313
## 76 0.8974602 0.7942602
## 77 0.8974602 0.7942602
## 78 0.8974984 0.7943211
## 79 0.8971062 0.7935260
## 80 0.8974298 0.7941999
##
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was k = 12.knn_auto_predicted <- predict(knn_auto, test, type = "raw")
mean(knn_auto_predicted != test$mpg01)## [1] 0.1122449- Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
data(Boston)
Boston$crim <- factor(ifelse(Boston$crim > median(Boston$crim), 1, 0))train and split
set.seed(123)
index = sample(1:nrow(Boston),.75*nrow(Boston),replace=FALSE)
train_boston = Boston[index,]
test_boston = Boston[-index,]Glm:
glm_boston <- glm(crim ~.,train_boston, family = "binomial")
glm_boston_predict <-predict(glm_boston,test_boston, type = 'response')
glm_boston_predict <- ifelse(glm_boston_predict <0.5, 0, 1)
mean(glm_boston_predict != test_boston$crim)## [1] 0.1259843LDA:
lda_boston <- lda(crim ~ .,train_boston)
pred.lda_boston <- predict(lda_boston, test_boston)
mean(pred.lda_boston$class != test_boston$crim)## [1] 0.1811024KNN:
train.X <- subset(train_boston , select = -crim)
test.X <- subset(test_boston, select =-crim)
train_crim <- subset( train_boston , select =crim)