Lab 5
INFO-H 415/515
options(repos = "https://cran.r-project.org")Complete the following exercises from Introduction to Statistical Learning
Chapter 4: (13) and (14)
- 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.
install.packages("corrplot")## Installing package into 'C:/Users/admin/AppData/Local/R/win-library/4.3'
## (as 'lib' is unspecified)## package 'corrplot' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\admin\AppData\Local\Temp\RtmpOaYVia\downloaded_packageslibrary(ISLR)## Warning: package 'ISLR' was built under R version 4.3.2library(corrplot)## Warning: package 'corrplot' was built under R version 4.3.2## corrplot 0.92 loadedlibrary(MASS)
library(e1071)## Warning: package 'e1071' was built under R version 4.3.2library(class)- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
#loading data
data("Weekly")#printing summary of the data
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
##
##
##
## #plotting a plot
pairs(Weekly)
#plotting with respect to year and volume
plot(Weekly$Year, Weekly$Volume,
xlab ="Year",
ylab ="Volume")
There seems a relation between year and volume. As the year is increasing, volume seems to increase.
- 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?
#performing logistic regression
log_reg= glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, family= "binomial", data=Weekly)
summary(log_reg)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = "binomial", data = Weekly)
##
## 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: 4In the logistic regression model, only Lag2 is statistically significant since it has a p-value = 0.0296, while the intercept is also seems significant with p-value = 0.0019, indicating its importance in predicting market direction. Rest of the variables have p-values which are higher than the commonly used significance level of 0.05 Which suggests that they are not statistically significant.
- 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.
#Predicting probabilities
prob_c <-predict(log_reg, type ="response")
# Predicting based on probability threshold 0.5
pred_c <-ifelse(prob_c >0.5,"Up", "Down")
# Creating confusion matrix from the above prediction
conf_matrix_c <- table(Predicted= pred_c, Actual= Weekly$Direction)
# Printing confusion matrix
conf_matrix_c## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557#accuracy calculation
accuracy_c <- (sum(diag(conf_matrix_c))) / (sum(conf_matrix_c))
accuracy_c## [1] 0.5610652The confusion matrix shows that the logistic regression model correctly predicted “Down” 54 times and “Up” 557 times but made 48 false positive predictions and 430 false negative predictions. The overall accuracy of the model is approximately 62.5%, indicating its effectiveness in predicting market direction.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
#creating a dataset for training and testing
train_data<- Weekly[Weekly$Year<= 2008,]
test_data<-Weekly[Weekly$Year >2008,]
# fit a logistic regression model with the function with lag2
model_d <-glm(Direction ~Lag2, family=binomial, data= train_data)
# Predicting probabilities for test data
prob_d<- predict(model_d, newdata =test_data, type="response")
# Predicting classes based on probability threshold 0.5
pred_d <-ifelse(prob_d > 0.5,"Up", "Down")
# Creating confusion matrix
conf_matrix_d<- table(Predicted =pred_d, Actual= test_data$Direction)
conf_matrix_d## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56#accuracy calculation
accuracy <- (sum(diag(conf_matrix_d))) / (sum(conf_matrix_d))
accuracy## [1] 0.625The confusion matrix shows that out of 104 instances, 56 were correctly predicted as “Up” and 9 as “Down” but made 5 false positive predictions and 34 false negative predictions showcasing a moderate improvement over the model trained on the entire dataset. Notably, it displayed better accuracy in predicting upward trends (91.80%) compared to downward trends (20.93%). Accuracy is 62.5 percent.
- Repeat (d) using LDA.
#fitting LDA
model_e <-lda(Direction ~Lag2, data= train_data)
# Predicting
pred_e <-predict(model_e, newdata =test_data)$class
# Creating confusion matrix
conf_matrix_e <-table(Predicted =pred_e, Actual =test_data$Direction)
# Printing confusion matrix
conf_matrix_e## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56#accuracy calculation
accuracy_e <- (sum(diag(conf_matrix_e))) / (sum(conf_matrix_e))
accuracy_e## [1] 0.625It gave similar result as logic regression. They both have accuracy of 62 percent.
- Repeat (d) using QDA.
#fitting QDA
model_f <-qda(Direction ~Lag2, data =train_data)
# Predicting classes for test data
pred_f <-predict(model_f,newdata =test_data)$class
#creating confusion matrix
conf_matrix_f <-table(Predicted =pred_f, Actual= test_data$Direction)
# Printing confusion matrix
conf_matrix_f## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61#accuracy calculation
accuracy_f <-(sum(diag(conf_matrix_f))) / (sum(conf_matrix_f))
accuracy_f## [1] 0.5865385This shows less accuracy(58.65%) as compared to the upper model.The QDA model, in this case, seems to be biased towards predicting “Up” for all instances, resulting in lower accuracy.
- Repeat (d) using KNN with K = 1.
# Combining Lag2 with other predictors
train_features <-train_data[, c("Lag2", "Lag1", "Lag3", "Lag4","Lag5", "Volume")]
test_features <-test_data[, c("Lag2","Lag1", "Lag3", "Lag4","Lag5", "Volume")]
# Fitting KNN model using all predictors
model_g <- knn(train =train_features, test =test_features, cl =train_data$Direction, k = 1)
# Creating a confusion matrix
conf_matrix_g <-table(Predicted =model_g, Actual =test_data$Direction)
conf_matrix_g## Actual
## Predicted Down Up
## Down 21 32
## Up 22 29#accuracy calculation
accuracy_g <-(sum(diag(conf_matrix_g))) / (sum(conf_matrix_g))
accuracy_g## [1] 0.4807692In the K-Nearest Neighbors (KNN) model, the accuracy rate was 48.08%.
- Repeat (d) using naive Bayes.
# Fitting Naive Bayes model using Lag2
model_h <- naiveBayes(Direction ~Lag2, data =train_data)
# Predicting classes
pred_h <- predict(model_h,newdata =test_data)
# Creating a confusion matrix
conf_matrix_h <- table(Predicted =pred_h, Actual =test_data$Direction)
conf_matrix_h## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61#accuracy calculation
accuracy_h <-sum(diag(conf_matrix_h)) / sum(conf_matrix_h)
accuracy_h## [1] 0.5865385The accuracy in naive bayes is less than LDA and logistic regression but seems similar to QDA(58.65%).
- Which of these methods appears to provide the best results on this data?
The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis; both having rates of 62.5%.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
#predictor combination
predictors_combination <-c("Lag1", "Lag4")
# Subsetting data
training_data_sub <- subset(train_data,select = c("Direction", predictors_combination))
testing_data_sub <- subset(test_data,select= c("Direction", predictors_combination))
# Fitting a logistic regression model
logistic_model<- glm(Direction ~ ., data =training_data_sub, family = "binomial")
# Fitting LDA model
lda_model <-lda(Direction ~ ., data= training_data_sub)
# Fitting a QDA model
qda_model <-qda(Direction ~ ., data= training_data_sub)
# Fitting a KNN model
k <- 5 #value of K
knn_model <- knn(train =training_data_sub[, -1], test =testing_data_sub[, -1], cl= training_data_sub$Direction, k = k)
# Predicting using logistic regression model
logistic_pred <-ifelse(predict(logistic_model, newdata =testing_data_sub, type = "response") > 0.5, "Up", "Down")
# Predicting using LDA model
lda_pred <- predict(lda_model, newdata =testing_data_sub)$class
# Predicting using QDA model
qda_pred <-predict(qda_model, newdata =testing_data_sub)$class
# Calculating accuracy for logistic regression
logistic_accuracy <-mean(logistic_pred ==testing_data_sub$Direction)
# Calculating accuracy for LDA
lda_accuracy <- mean(lda_pred== testing_data_sub$Direction)
# Calculating accuracy for QDA
qda_accuracy <-mean(qda_pred== testing_data_sub$Direction)
# Calculating accuracy for KNN
knn_accuracy<- mean(knn_model== testing_data_sub$Direction)
# Print accuracies for each
cat("Logistic Regression Accuracy: ", logistic_accuracy)## Logistic Regression Accuracy: 0.5865385cat(" LDA Accuracy: ", lda_accuracy)## LDA Accuracy: 0.5865385cat(" QDA Accuracy: ", qda_accuracy )## QDA Accuracy: 0.5480769cat(" KNN Accuracy :" , knn_accuracy)## KNN Accuracy : 0.5The analysis utilized Lag1 and Lag4 as predictors to forecast market direction. Logistic regression and LDA achieved comparable accuracies of approximately 58.65%, while QDA yielded a slightly lower accuracy of around 54.81%. KNN classification, with K=5, resulted in an accuracy of approximately 50%, indicating its lower predictive performance compared to other methods.
- 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.
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:MASS':
##
## select## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionsummary(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) :365head(Auto)library(dplyr)
# Calculating the median of mpg
mpg_median<-median(Auto$mpg)
# Creating mpg01 variable
Auto <- Auto %>%
mutate(mpg01 =ifelse(mpg >mpg_median, 1, 0))
# Print few rows
head(Auto)- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
pairs(Auto)
par(mfrow=c(3,3))
boxplot(mpg~mpg01,data=Auto)
boxplot(cylinders~mpg01,data=Auto)
boxplot(displacement~mpg01,data=Auto)
boxplot(horsepower~mpg01,data=Auto)
boxplot(weight~mpg01,data=Auto)
boxplot(acceleration~mpg01,data=Auto)
boxplot(year~mpg01,data=Auto)
boxplot(origin~mpg01,data=Auto)
cylinders,displacement,horsepower,weight,acceleration,year seem most
likely to be useful in predicting mpg01.
- Split the data into a training set and a test set.
set.seed(123)
#split the data
indices <-sample(1:nrow(Auto), nrow(Auto)*0.7)
train_data <-Auto[indices, ]
test_data <- Auto[-indices, ]
#dimensions
dim(train_data)## [1] 274 10dim(test_data)## [1] 118 10- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
# Choosing variables associated with mpg01
selected_vars <- c("cylinders","displacement","horsepower","weight","acceleration","year")
# Building a LDA model
lda_model <- lda(mpg01 ~ .,data =train_data[, c("mpg01",selected_vars)])
# Predicting mpg01 on test data
predictions <- predict(lda_model,newdata =test_data[,selected_vars])
#test error
test_error <-mean(predictions$class!= test_data$mpg01)
cat("Test Error:", test_error, "\n")## Test Error: 0.1016949- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
# Choosing variables associated with mpg01
selected_vars <- c("cylinders","displacement","horsepower", "weight", "acceleration", "year")
# Building QDA model
qda_model <-qda(mpg01 ~ ., data =train_data[, c("mpg01", selected_vars)])
# Predicting mpg01 on test data
predictions_qda <- predict(qda_model,newdata =test_data[, selected_vars])
#test error
test_error_qda <-mean(predictions_qda$class !=test_data$mpg01)
cat("Test Error (QDA):", test_error_qda, "\n")## Test Error (QDA): 0.1016949- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
# Building logistic regression model
logistic_model <- glm(mpg01 ~ ., data = train_data[, c("mpg01",selected_vars)],family = binomial)
# Predicting mpg01 on test data
predictions_logistic <- predict(logistic_model,newdata = test_data[, selected_vars], type ="response")
# Converting predicted probabilities to class labels (0 or 1)
predictions_logistic_class <-ifelse(predictions_logistic > 0.5, 1, 0)
# test error
test_error_logistic <-mean(predictions_logistic_class!= test_data$mpg01)
cat("Test Error (Logistic Regression):",test_error_logistic, "\n")## Test Error (Logistic Regression): 0.1186441- 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?
# Using variables that seemed most associated with mpg01
selected_vars <- c("cylinders","displacement","horsepower","weight","acceleration")
# using naive bayes
nb_model <-naiveBayes(mpg01 ~ ., data = train_data[, c("mpg01",selected_vars)])
#predicting mpg01 on test data
predictions_nb <-predict(nb_model, newdata= test_data[, selected_vars])
#test error
test_error_nb <- mean(predictions_nb != test_data$mpg01)
cat("Test Error (Naive Bayes):", test_error_nb)## Test Error (Naive Bayes): 0.1016949- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
library(class)
#using the variables that seemed most associated with mpg01
selected_vars <- c("cylinders","displacement", "horsepower","weight","acceleration")
# Defining a function to perform KNN with varying values of K
perform_knn <- function(k_value) {
# training
knn_model <- knn(train =train_data[, selected_vars],
test= test_data[, selected_vars],
cl =train_data$mpg01,
k= k_value)
# test error
test_error_knn <- mean(knn_model != test_data$mpg01)
return(test_error_knn)
}
# Specify values
k_values <-c(1, 3, 5, 7, 9)
#use KNN for each value of K and evaluate test errors
test_errors <-sapply(k_values, perform_knn)
# Print
cat("Test Errors (KNN):\n")## Test Errors (KNN):print(test_errors)## [1] 0.1694915 0.1271186 0.1101695 0.1016949 0.1101695# best value of K
best_k <-k_values[which.min(test_errors)]
cat("\nBest K value:", best_k, "\n")##
## Best K value: 7