Assignment #3
Briley Winge
#install.packages("GGally")
#install.packages("e1071")
#install.packages("modelr")
library(MASS)
library(ISLR)
library(GGally)## Loading required package: ggplot2library(class)
library(e1071)##
## Attaching package: 'e1071'## The following object is masked from 'package:ggplot2':
##
## elementlibrary(modelr)Applied
Question 13
This question should be answered using the Weekly data set, which is part of the ISLR2 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.
# loading in dataset
market_data <- Weekly- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
summary(market_data)## 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
##
##
##
## Based on the summary of the data set, Lags 1 through 5 and today have the same minimum and maximum return. We can also see that according to the data, there has been more up (positive) return then down (negative) returns.
ggpairs(
market_data,
ggplot2::aes(colour = Direction),
progress = FALSE
)## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value `binwidth`.
Based on the correlation scatterplot, we can see that Year and Volume are statistically significant and have a positive correlation of 0.842. Another thing to note is that the graph shows an upward trend in the points, meaning that as the years increase so does the volume of shares traded.
- 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?
market_glm <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = market_data,
family = "binomial")
summary(market_glm)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = "binomial", data = market_data)
##
## 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: 4From the summary of the logistic regression, there is only one variable that is statistically significant: Lag2.
- 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.
market_preds <- predict(market_glm , type = "response")
market_glm.pred <- rep("Down", 1089)
market_glm.pred[market_preds > .5] = "Up"
table(market_glm.pred, market_data$Direction)##
## market_glm.pred Down Up
## Down 54 48
## Up 430 557mean(market_glm.pred == market_data$Direction)## [1] 0.5610652From the confusion matrix we can see that the logistic regression model was able to correctly predict the direction of the market 56% of the time. In the confusion matrix, it shows that the model predicted the market to be up 987 weeks out of the 1,089 weeks in the data set. This could be any issue as it is showing the model has a high false positive rate. On the other hand, the model correctly predicted the market being up when it actually was up 92% of the time (557 / (557 + 48)).
- 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).
market_train <- (market_data$Year < 2009)
market_test <- market_data[!market_train,]
direction_test <- market_data$Direction[!market_train]train_glm <- glm(Direction ~ Lag2,
data = market_data,
family = binomial,
subset = market_train)train_glm_pred <- predict(train_glm, market_test, type = "response")
train_glm.pred <- rep("Down", 104)
train_glm.pred[train_glm_pred > .5] = "Up"
table(train_glm.pred, direction_test)## direction_test
## train_glm.pred Down Up
## Down 9 5
## Up 34 56mean(train_glm.pred == direction_test)## [1] 0.625- Repeat (d) using LDA.
market_lda <- lda(Direction ~ Lag2,
data = market_data,
subset = market_train)
market_lda## Call:
## lda(Direction ~ Lag2, data = market_data, subset = market_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda_pred <- predict(market_lda, market_test)
lda_class <- lda_pred$class
table(lda_class, direction_test)## direction_test
## lda_class Down Up
## Down 9 5
## Up 34 56mean(lda_class == direction_test)## [1] 0.625- Repeat (d) using QDA.
market_qda <- qda(Direction ~ Lag2,
data = market_data,
subset = market_train)
market_qda## Call:
## qda(Direction ~ Lag2, data = market_data, subset = market_train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda_preds <- predict(market_qda, market_test)
qda_class <- qda_preds$class
table(qda_class, direction_test)## direction_test
## qda_class Down Up
## Down 0 0
## Up 43 61mean(qda_class == direction_test)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
train_knn <- as.matrix(market_data$Lag2[market_train])
test_knn <- as.matrix(market_test$Lag2)
train_Direction <- market_data$Direction[market_train]set.seed(10)
knn_pred <- knn(train = train_knn,
test = test_knn,
cl = train_Direction,
k = 1)table(knn_pred, direction_test)## direction_test
## knn_pred Down Up
## Down 21 29
## Up 22 32mean(knn_pred == direction_test)## [1] 0.5096154- Repeat (d) using naive Bayes.
nb_fit <- naiveBayes(Direction ~ Lag2,
data = market_data,
subset = market_train)
nb_fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb_preds <- predict(nb_fit, market_test)
table(nb_preds, direction_test)## direction_test
## nb_preds Down Up
## Down 0 0
## Up 43 61mean(nb_preds == direction_test)## [1] 0.5865385- Which of these methods appears to provide the best results on this data?
| Test Name | Accuracy |
|---|---|
| Logistic Regression | 0.625 |
| LDA | 0.625 |
| QDA | 0.587 |
| KNN with K = 1 | 0.510 |
| Naive Bayes | 0.587 |
Of these models, logistic regression and LDA had the bust accuracy in prediction the direction of the market.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
full_lda <- lda(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = market_data,
subset = market_train)
full_lda_pred <- predict(full_lda, market_test)
full_lda_class <- full_lda_pred$class
table(full_lda_class, direction_test)## direction_test
## full_lda_class Down Up
## Down 31 44
## Up 12 17mean(full_lda_class == direction_test)## [1] 0.4615385interaction_qda <- qda(Direction ~ Lag2:Volume,
data = market_data,
subset = market_train)
interaction_qda_preds <- predict(interaction_qda, market_test)
interaction_qda_class <- interaction_qda_preds$class
table(interaction_qda_class, direction_test)## direction_test
## interaction_qda_class Down Up
## Down 24 28
## Up 19 33mean(interaction_qda_class == direction_test)## [1] 0.5480769train_knn_5 <- as.matrix(market_data$Lag2[market_train])
test_knn_5 <- as.matrix(market_test$Lag2)
train_Direction_5 <- market_data$Direction[market_train]
set.seed(10)
knn_pred_5 <- knn(train = train_knn_5,
test = test_knn_5,
cl = train_Direction_5,
k = 5)
table(knn_pred_5, direction_test)## direction_test
## knn_pred_5 Down Up
## Down 16 20
## Up 27 41mean(knn_pred_5 == direction_test)## [1] 0.5480769nb_fit_more_pred <- naiveBayes(Direction ~ Lag2 + Volume + Year,
data = market_data,
subset = market_train)
nb_more_preds <- predict(nb_fit_more_pred, market_test)
table(nb_more_preds, direction_test)## direction_test
## nb_more_preds Down Up
## Down 43 59
## Up 0 2mean(nb_more_preds == direction_test)## [1] 0.4326923Updated Accuracies:
| Test Name | Accuracy |
|---|---|
| Logistic Regression | 0.625 |
| LDA | 0.461 |
| QDA | 0.548 |
| KNN with K = 5 | 0.548 |
| Naive Bayes | 0.433 |
After trying different variations of interactions, adding more predictors, and changing the K value on the LDA, QDA, KNN, and Naive Bayes model; the model performing the best is a tie between QDA and KNN.
Question 14
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.
# called in data set
auto <- read.csv("Auto.csv", header = TRUE)
# removed any rows where horsepower was ?
clean_auto <- auto[auto$horsepower != "?",]
# converting horsepower to a numeric value not that all ? have been removed
clean_auto$horsepower <- as.numeric(clean_auto$horsepower)- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
# found median of mpg
mpg_median <- median(auto$mpg)
# added new column to auto data set. Anything over the median is listed as 1
auto$mpg01 <- ifelse(auto$mpg > mpg_median, 1, 0)
auto$mpg01 <- as.factor(auto$mpg01)- 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.
plot(auto)
auto |>
ggplot(
aes(
x = mpg01,
y = displacement,
fill = mpg01
)
) +
geom_boxplot()
auto |>
ggplot(
aes(
x = mpg01,
y = weight,
fill = mpg01
)
) +
geom_boxplot()
auto |>
ggplot(
aes(
x = mpg01,
y = acceleration,
fill = mpg01
)
) +
geom_boxplot()
Based on the scatter plot, it seems that displacement, weight, and
acceleration would be good predictors for mpg01. According to the box
plots, of the three displacement and weight show the most distance
between 0 and 1 making them the most useful predictors.
- Split the data into a training set and a test set.
set.seed(42)
partitions <- resample_partition(auto, c(train = 0.75, test = 0.25))
auto_train <- as.data.frame(partitions$train)
auto_test <- as.data.frame(partitions$test)- 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 <- lda(mpg01 ~ displacement + weight + acceleration,
data = auto_train)
auto_lda_pred <- predict(auto_lda, auto_test)
auto_lda_class <- auto_lda_pred$class
# Test Error of the model
mean(auto_lda_class != auto_test$mpg01)## [1] 0.11- 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 <- qda(mpg01 ~ displacement + weight + acceleration,
data = auto_train)
auto_qda_pred <- predict(auto_qda, auto_test)
auto_qda_class <- auto_qda_pred$class
# Test Error of the model
mean(auto_qda_class != auto_test$mpg01)## [1] 0.1- 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 <- glm(mpg01 ~ displacement + weight + acceleration,
data = auto_train,
family = "binomial")
auto_glm_preds <- predict(auto_glm, newdata = auto_test, type = "response")
auto_glm.preds <- rep(0, nrow(auto_test))
auto_glm.preds[auto_glm_preds > .5] = 1
mean(auto_glm.preds != auto_test$mpg01)## [1] 0.14- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto_nb <- naiveBayes(mpg01 ~ displacement + weight + acceleration,
data = auto_train)
auto_nb_preds <- predict(auto_nb, auto_test)
mean(auto_nb_preds != auto_test$mpg01)## [1] 0.1- 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?
preds <- c("displacement", "weight", "acceleration")
auto_train_x <- auto_train[, preds]
auto_test_x <- auto_test[, preds]
auto_knn_1 <- knn(train = auto_train_x,
test = auto_test_x,
cl = auto_train$mpg01,
k = 1)
auto_knn_3 <- knn(train = auto_train_x,
test = auto_test_x,
cl = auto_train$mpg01,
k = 3)
auto_knn_5 <- knn(train = auto_train_x,
test = auto_test_x,
cl = auto_train$mpg01,
k = 5)
auto_knn_10 <- knn(train = auto_train_x,
test = auto_test_x,
cl = auto_train$mpg01,
k = 10)
mean(auto_knn_1 != auto_test$mpg01)## [1] 0.14mean(auto_knn_3 != auto_test$mpg01)## [1] 0.12mean(auto_knn_5 != auto_test$mpg01)## [1] 0.13mean(auto_knn_10 != auto_test$mpg01)## [1] 0.16| KNN Value | Test Erro |
|---|---|
| K = 1 | 0.14 |
| K = 3 | 0.12 |
| K = 5 | 0.13 |
| K = 10 | 0.15 |
Of the KNN tests ran, when K = 3 it had the lowest test error.
Question 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
boston_data <- Bostonmedian_crime <- median(boston_data$crim)
boston_data$high_crime <- ifelse(boston_data$crim > median_crime, 1, 0)full_boston_glm <- glm(high_crime ~ ., data = boston_data)
summary(full_boston_glm)##
## Call:
## glm(formula = high_crime ~ ., data = boston_data)
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.4811899 0.3596766 -4.118 4.48e-05 ***
## crim 0.0011350 0.0022288 0.509 0.610817
## zn -0.0013810 0.0009315 -1.483 0.138837
## indus 0.0030875 0.0041258 0.748 0.454618
## chas -0.0145084 0.0583661 -0.249 0.803793
## nox 2.0025143 0.2618145 7.649 1.08e-13 ***
## rm 0.0213894 0.0303113 0.706 0.480736
## age 0.0027612 0.0008862 3.116 0.001941 **
## dis 0.0111019 0.0141047 0.787 0.431598
## rad 0.0171097 0.0045460 3.764 0.000188 ***
## tax -0.0002067 0.0002550 -0.810 0.418098
## ptratio 0.0125743 0.0092372 1.361 0.174055
## black -0.0002421 0.0001824 -1.328 0.184948
## lstat 0.0035227 0.0037541 0.938 0.348518
## medv 0.0094707 0.0030244 3.131 0.001843 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.1013345)
##
## Null deviance: 126.500 on 505 degrees of freedom
## Residual deviance: 49.755 on 491 degrees of freedom
## AIC: 294.34
##
## Number of Fisher Scoring iterations: 2# creating test and train
set.seed(42)
boston_partitions <- resample_partition(boston_data, c(train = 0.75, test = 0.25))
boston_train <- as.data.frame(boston_partitions$train)
boston_test <- as.data.frame(boston_partitions$test)# Logistic Regression
boston_glm <- glm(high_crime ~ nox + age + rad + medv,
data = boston_train,
family = "binomial")
boston_glm_preds <- predict(boston_glm, newdata = boston_test, type = "response")
boston_glm.preds <- rep(0, nrow(boston_test))
boston_glm.preds[boston_glm_preds > .5] = 1
mean(boston_glm.preds == boston_test$high_crime)## [1] 0.8818898# LDA
boston_lda <- lda(high_crime ~ nox + age + rad + medv,
data = boston_train)
boston_lda_pred <- predict(boston_lda, boston_test)
boston_lda_class <- boston_lda_pred$class
# Test Error of the model
mean(boston_lda_class == boston_test$high_crime)## [1] 0.8897638# Naive Bayes
boston_nb <- naiveBayes(high_crime ~ nox + age + rad + medv,
data = boston_train)
boston_nb_preds <- predict(boston_nb, boston_test)
mean(boston_nb_preds == boston_test$high_crime)## [1] 0.8897638boston_preds <- c("nox", "age", "rad", "medv")
boston_train_x <- boston_train[, boston_preds]
boston_test_x <- boston_test[, boston_preds]
boston_knn_1 <- knn(train = boston_train_x,
test = boston_test_x,
cl = boston_train$high_crime,
k = 1)
boston_knn_3 <- knn(train = boston_train_x,
test = boston_test_x,
cl = boston_train$high_crime,
k = 3)
boston_knn_5 <- knn(train = boston_train_x,
test = boston_test_x,
cl = boston_train$high_crime,
k = 5)
boston_knn_10 <- knn(train = boston_train_x,
test = boston_test_x,
cl = boston_train$high_crime,
k = 10)
mean(boston_knn_1 == boston_test$high_crime)## [1] 0.7795276mean(boston_knn_3 == boston_test$high_crime)## [1] 0.8503937mean(boston_knn_5 == boston_test$high_crime)## [1] 0.8740157mean(boston_knn_10 == boston_test$high_crime)## [1] 0.8897638