Homework 3
Victor Feagins
3/3/2021
The following are my answers to questions 10,11,13 in Chapter 4 of the ISLR book
library(ISLR)
library(tidyverse)## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --## v ggplot2 3.3.3 v purrr 0.3.4
## v tibble 3.1.0 v dplyr 1.0.3
## v tidyr 1.1.3 v stringr 1.4.0
## v readr 1.4.0 v forcats 0.5.1## Warning: package 'tibble' was built under R version 4.0.4## Warning: package 'readr' was built under R version 4.0.4## Warning: package 'forcats' was built under R version 4.0.4## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()library(MASS)## Warning: package 'MASS' was built under R version 4.0.4##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectlibrary(class)Question 10
a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
corrplot<- function(numericdata){
cormat <- round(cor(numericdata),2)
melted_cormat <- reshape2::melt(cormat)
graph<- ggplot(data = melted_cormat, aes(Var2, Var1, fill = value))+
geom_tile(color = "white")+
scale_fill_gradient2(low = "blue", high = "red", mid = "white",
midpoint = 0, limit = c(-1,1), space = "Lab",
name="Pearson\nCorrelation")+ theme(axis.text.x = element_text(angle = 90, vjust = 0.5, hjust=1))
graph
}
corrplot(dplyr::select(Weekly,-Direction))
histogramplot <- function (numericdata){
graph <- numericdata %>%
gather() %>%
ggplot(aes(value)) +
facet_wrap(~ key, scales = "free") +
geom_density()
graph
}
histogramplot(dplyr::select(Weekly,-Direction))
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
##
##
##
## The lag and today variables seem to be a typical bell shaped curve. There only linear correlation is with year and volume. As technolgy advances people will do more trading.
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?
logit <- Weekly %>%
dplyr::select(-Today, -Year) %>%
glm(Direction ~., data = ., family = "binomial")
summary(logit)##
## Call:
## glm(formula = Direction ~ ., family = "binomial", data = .)
##
## 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: 4The only variable that seems significant at the .05 level is Lag 2
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.
Truth <- Weekly$Direction
Probs <- predict(logit, type = "response")
pred <- rep("Down", times = nrow(Weekly))
pred[Probs > .5] <- "Up"
mat <- table(pred,Truth)
mat## Truth
## pred Down Up
## Down 54 48
## Up 430 557(mat[1,1] + mat[2,2])/nrow(Weekly)## [1] 0.5610652The classifer is over classifying the up trend and not predicting the down trend correct at all.
d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
df <- Weekly %>%
filter(Year == 2009 | Year == 2010 )
logitd<- df %>%
glm(Direction ~Lag2, data = ., family = "binomial")
summary(logitd)##
## Call:
## glm(formula = Direction ~ Lag2, family = "binomial", data = .)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.5767 -1.3052 0.9242 1.0419 1.2978
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.32377 0.20136 1.608 0.108
## Lag2 0.08562 0.06707 1.277 0.202
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 141.04 on 103 degrees of freedom
## Residual deviance: 139.37 on 102 degrees of freedom
## AIC: 143.37
##
## Number of Fisher Scoring iterations: 4Truth <- df$Direction
Probs <- predict(logitd, type = "response")
pred <- rep("Down", times = nrow(df))
pred[Probs > .5] <- "Up"
mat <-table(pred,Truth)
mat## Truth
## pred Down Up
## Down 8 4
## Up 35 57(mat[1,1] + mat[2,2])/nrow(df)## [1] 0.625Repeat (d) using LDA.
ldae<- df %>%
lda(Direction ~Lag2, data = .)
ldae## Call:
## lda(Direction ~ Lag2, data = .)
##
## Prior probabilities of groups:
## Down Up
## 0.4134615 0.5865385
##
## Group means:
## Lag2
## Down -0.08904651
## Up 0.69591803
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.3262636Probs <- predict(ldae, type = "response")
mat <- table(Probs$class, Truth)
mat## Truth
## Down Up
## Down 8 4
## Up 35 57(mat[1,1] + mat[2,2])/nrow(df)## [1] 0.625(f) Repeat (d) using QDA.
qdaf<- df %>%
qda(Direction ~Lag2, data = .)
qdaf## Call:
## qda(Direction ~ Lag2, data = .)
##
## Prior probabilities of groups:
## Down Up
## 0.4134615 0.5865385
##
## Group means:
## Lag2
## Down -0.08904651
## Up 0.69591803Probs <- predict(qdaf, type = "response")
mat <- table(Probs$class, Truth)
mat## Truth
## Down Up
## Down 9 4
## Up 34 57(mat[1,1] + mat[2,2])/nrow(df)## [1] 0.6346154g) Repeat (d) using KNN with K = 1.
set.seed(123)
smp_size <- floor(0.50 * nrow(df))
train_ind <- sample(seq_len(nrow(df)), size = smp_size)
train.X <- df$Lag2[train_ind]
test.X <- df$Lag2[-train_ind]
train.Y <- df$Direction[train_ind]
test.Y <- df$Direction[-train_ind]
knn.pred<- knn(data.frame(train.X),data.frame(test.X),train.Y, k = 1)
mat <- table(knn.pred,train.Y)
mat## train.Y
## knn.pred Down Up
## Down 7 12
## Up 16 17(mat[1,1] + mat[2,2])/length(train.Y)## [1] 0.4615385h) Which of these methods appears to provide the best results on this data?
The method that provided the best result was QDA with the best correct prediction rate.
(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 forK in the KNN classifier.
smp_size <- floor(0.50 * nrow(df))
train_ind <- sample(seq_len(nrow(df)), size = smp_size)
train.X <- df$Lag2[train_ind]
test.X <- df$Lag2[-train_ind]
train.Y <- df$Direction[train_ind]
test.Y <- df$Direction[-train_ind]
knn.pred<- knn(data.frame(train.X),data.frame(test.X),train.Y, k = 5)
mat <- table(knn.pred,train.Y)
mat## train.Y
## knn.pred Down Up
## Down 10 14
## Up 13 15(mat[1,1] + mat[2,2])/length(train.Y)## [1] 0.4807692ldae<- df %>%
lda(Direction ~., data = .)
ldae## Call:
## lda(Direction ~ ., data = .)
##
## Prior probabilities of groups:
## Down Up
## 0.4134615 0.5865385
##
## Group means:
## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume
## Down 2009.465 0.2089767 -0.08904651 0.5855349 0.6180930 -0.08146512 5.111618
## Up 2009.525 0.5306230 0.69591803 0.2309344 0.1938361 0.60139344 5.034022
## Today
## Down -2.357488
## Up 2.230082
##
## Coefficients of linear discriminants:
## LD1
## Year 0.44977869
## Lag1 0.04901766
## Lag2 0.10000287
## Lag3 0.02479192
## Lag4 0.01118442
## Lag5 0.03928998
## Volume 0.07871424
## Today 0.51526811Probs <- predict(ldae, type = "response")
mat <- table(Probs$class, Truth)
mat## Truth
## Down Up
## Down 36 0
## Up 7 61(mat[1,1] + mat[2,2])/nrow(df)## [1] 0.932692311. In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
a) & b)
cardf<- Auto %>%
dplyr::select(-name) %>%
mutate(mpg01 = if_else(mpg > median(mpg),1,0))
corrplot(cardf)
histogramplot(cardf)
c) Split the data into a training set and a test set.
smp_size <- floor(0.50 * nrow(cardf))
train_ind <- sample(seq_len(nrow(cardf)), size = smp_size)
train.X <- cardf[train_ind,]
test.X <- cardf[-train_ind,]
train.Y <- cardf$mpg01[train_ind]
test.Y <- cardf$mpg01[-train_ind]ldacars <- lda(mpg01 ~cylinders + displacement + horsepower + weight, data = train.X)
ldacars## Call:
## lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train.X)
##
## Prior probabilities of groups:
## 0 1
## 0.4540816 0.5459184
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.651685 264.3708 124.97753 3548.685
## 1 4.186916 117.3364 80.11215 2358.654
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.5744023711
## displacement -0.0027758506
## horsepower 0.0108161523
## weight -0.0009902655testpredict <- predict(ldacars, test.X)$class
mat <- table(testpredict, test.Y)
1- (mat[1,1] + mat[2,2])/length(test.Y)## [1] 0.09693878Test error rate of .112 #### 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?
qdacars <- qda(mpg01 ~cylinders + displacement + horsepower + weight, data = train.X)
qdacars## Call:
## qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train.X)
##
## Prior probabilities of groups:
## 0 1
## 0.4540816 0.5459184
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.651685 264.3708 124.97753 3548.685
## 1 4.186916 117.3364 80.11215 2358.654testpredict <- predict(qdacars, test.X)$class
mat <- table(testpredict, test.Y)
1 - (mat[1,1] + mat[2,2])/length(test.Y)## [1] 0.1020408Test error rate of .107 #### 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?
logcars <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, data = train.X, family = "binomial")
summary(logcars)##
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower +
## weight, family = "binomial", data = train.X)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.6574 -0.2590 0.1760 0.4568 3.1329
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 11.5474104 2.3202958 4.977 6.47e-07 ***
## cylinders -0.2820324 0.4652243 -0.606 0.5444
## displacement -0.0126407 0.0107870 -1.172 0.2413
## horsepower -0.0247602 0.0189312 -1.308 0.1909
## weight -0.0019241 0.0009547 -2.015 0.0439 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 270.06 on 195 degrees of freedom
## Residual deviance: 107.89 on 191 degrees of freedom
## AIC: 117.89
##
## Number of Fisher Scoring iterations: 7testpredict <- predict(logcars,test.X, type = "response")
predicted.classes <- ifelse(testpredict > 0.5, 1, 0)
1 - mean(predicted.classes == test.Y)## [1] 0.1071429Test error rate of .112
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?
set.seed(123)
knn.pred<- knn(data.frame(train.X),data.frame(test.X),train.Y, k = 5)
mat <- table(knn.pred,test.Y)
mat## test.Y
## knn.pred 0 1
## 0 91 6
## 1 16 831- (mat[1,1] + mat[2,2])/length(test.Y)## [1] 0.1122449K of 5 seems to have the lowest error rate so I will say k of 5 is the best. With a test error rate of .107
Question 12
a) Write a function, Power(), that prints out the result of raising 2 to the 3rd power. In other words, your function should compute 23 and print out the results.
Power <- function(num){
print(num ^ 3)
}
Power(2)## [1] 8b) Create a new function, Power2(), that allows you to pass any two numbers, x and a, and prints out the value of x^a. You can do this by beginning your function with the line
Power2 <- function(x,a){
print(x ^ a)
}
Power2(3,8)## [1] 6561c) Using the Power2() function that you just wrote, compute 10^3, 81^7, and 131^3.
Power2(10,3)## [1] 1000Power2(81,7)## [1] 2.287679e+13Power2(131,3)## [1] 2248091d)Now create a new function, Power3(), that actually returns the result x^a as an R object, rather than simply printing it to the screen. That is, if you store the value x^a in an object called result within your function, then you can simply return() this return() result, using the following line:
Power3 <- function(x,a){
result <- x ^ a
result
}e)
x <-1:100
plot(x,Power3(x,2), xlab = "x", ylab = "x^2",main = "Using the Power3 Function", log="xy")
f) Create a function, PlotPower(), that allows you to create a plot of x against x^a for a fixed a and for a range of values of x.
PlotPower <- function(x,a){
plot(x,Power3(x,a))
}
PlotPower(1:10,3)