Assignment3
Kashfia Sharmin
2025-03-01
library(ISLR2)
data("Weekly")head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Downstr(Weekly)## 'data.frame': 1089 obs. of 9 variables:
## $ Year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
## $ Lag1 : num 0.816 -0.27 -2.576 3.514 0.712 ...
## $ Lag2 : num 1.572 0.816 -0.27 -2.576 3.514 ...
## $ Lag3 : num -3.936 1.572 0.816 -0.27 -2.576 ...
## $ Lag4 : num -0.229 -3.936 1.572 0.816 -0.27 ...
## $ Lag5 : num -3.484 -0.229 -3.936 1.572 0.816 ...
## $ Volume : num 0.155 0.149 0.16 0.162 0.154 ...
## $ Today : num -0.27 -2.576 3.514 0.712 1.178 ...
## $ Direction: Factor w/ 2 levels "Down","Up": 1 1 2 2 2 1 2 2 2 1 ...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
##
##
##
## cor(Weekly[, -c(1, 9)])## Lag1 Lag2 Lag3 Lag4 Lag5
## Lag1 1.000000000 -0.07485305 0.05863568 -0.071273876 -0.008183096
## Lag2 -0.074853051 1.00000000 -0.07572091 0.058381535 -0.072499482
## Lag3 0.058635682 -0.07572091 1.00000000 -0.075395865 0.060657175
## Lag4 -0.071273876 0.05838153 -0.07539587 1.000000000 -0.075675027
## Lag5 -0.008183096 -0.07249948 0.06065717 -0.075675027 1.000000000
## Volume -0.064951313 -0.08551314 -0.06928771 -0.061074617 -0.058517414
## Today -0.075031842 0.05916672 -0.07124364 -0.007825873 0.011012698
## Volume Today
## Lag1 -0.06495131 -0.075031842
## Lag2 -0.08551314 0.059166717
## Lag3 -0.06928771 -0.071243639
## Lag4 -0.06107462 -0.007825873
## Lag5 -0.05851741 0.011012698
## Volume 1.00000000 -0.033077783
## Today -0.03307778 1.000000000hist(Weekly$Today, breaks = 20, col = "blue", main = "Histogram of Weekly Returns", xlab = "Return")
hist(Weekly$Volume, breaks = 20, col = "red", main = "Histogram of Weekly Returns", xlab = "Volume")
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Histogram for Volume shows a right-skewed distribution, meaning that the trading volume is low but occassionally has large values.
Volume has low correlation with Today so it may not be a strong predictor of market movement.
The of Up(605) vs Down(484) suggests some bias.
Weekly$Direction <- as.factor(Weekly$Direction)
model1 <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(model1)##
## 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: 4- 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?
- Lag2 is 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.
prob_pred <- predict(model1, type = "response")
predictions <- ifelse(prob_pred > 0.5, "Up", "Down")
predictions <- factor(predictions, levels = c("Down", "Up"))
conf_matrix <- table(Predicted = predictions, Actual = Weekly$Direction)
print(conf_matrix)## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557accuracy1 <- sum(diag(conf_matrix)) / sum(conf_matrix)
cat("Overall correct predictions:", accuracy1)## Overall correct predictions: 0.5610652train_data <- subset(Weekly, Year <= 2008)
test_data <- subset(Weekly, Year > 2008)
model2 <- glm(Direction ~ Lag2, data = train_data, family = binomial)
summary(model2)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = train_data)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4prob_pred2 <- predict(model2, newdata = test_data, type = "response")
predictions2 <- ifelse(prob_pred2 > 0.5, "Up", "Down")
predictions2 <- factor(predictions2, levels = c("Down", "Up"))
conf_matrix_test <- table(Predicted = predictions2, Actual = test_data$Direction)
print(conf_matrix_test)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56accuracy2 <- sum(diag(conf_matrix_test)) / sum(conf_matrix_test)
cat("Overall correct predictions:", accuracy2)## Overall correct predictions: 0.625library(MASS) ##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(class)
library(e1071)
train_X <- train_data$Lag2
train_Y <- train_data$Direction
test_X <- test_data$Lag2
test_Y <- test_data$Direction- LDA:
lda_model <- lda(Direction ~ Lag2, data = train_data)
lda_pred <- predict(lda_model, newdata = test_data)
lda_class <- lda_pred$class
lda_conf_matrix <- table(Predicted = lda_class, Actual = test_Y)
print(lda_conf_matrix)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56lda_accuracy <- sum(diag(lda_conf_matrix)) / sum(lda_conf_matrix)
cat("LDA Accuracy:", lda_accuracy, "\n")## LDA Accuracy: 0.625- QDA
qda_model <- qda(Direction ~ Lag2, data = train_data)
qda_pred <- predict(qda_model, newdata = test_data)
qda_class <- qda_pred$class
qda_conf_matrix <- table(Predicted = qda_class, Actual = test_Y)
print(qda_conf_matrix)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61qda_accuracy <- sum(diag(qda_conf_matrix)) / sum(qda_conf_matrix)
cat("QDA Accuracy:", qda_accuracy, "\n")## QDA Accuracy: 0.5865385- KNN with K = 1
knn_pred <- knn(train = matrix(train_X), test = matrix(test_X),
cl = train_Y, k = 1)
knn_conf_matrix <- table(Predicted = knn_pred, Actual = test_Y)
print(knn_conf_matrix)## Actual
## Predicted Down Up
## Down 21 29
## Up 22 32knn_accuracy <- sum(diag(knn_conf_matrix)) / sum(knn_conf_matrix)
cat("KNN with K=1 Accuracy:", knn_accuracy, "\n")## KNN with K=1 Accuracy: 0.5096154- Naive Bayes
nb_model <- naiveBayes(Direction ~ Lag2, data = train_data)
nb_pred <- predict(nb_model, newdata = test_data)
nb_conf_matrix <- table(Predicted = nb_pred, Actual = test_Y)
print(nb_conf_matrix)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61nb_accuracy <- sum(diag(nb_conf_matrix)) / sum(nb_conf_matrix)
cat("Naïve Bayes Accuracy:", nb_accuracy, "\n")## Naïve Bayes Accuracy: 0.5865385- The logistic regression model and the LDA model both have the highest accuracies of 0.625
transformed and interaction variables:
train_data$Lag2_sq <- train_data$Lag2^2
train_data$Lag1_Lag2 <- train_data$Lag1 * train_data$Lag2
train_data$Log_Volume <- log(train_data$Volume + 1)
test_data$Lag2_sq <- test_data$Lag2^2
test_data$Lag1_Lag2 <- test_data$Lag1 * test_data$Lag2
test_data$Log_Volume <- log(test_data$Volume + 1)log_model <- glm(Direction ~ Lag2 + Lag2_sq + Lag1_Lag2 + Log_Volume, data = train_data, family = binomial)
logit_pred_prob <- predict(log_model, newdata = test_data, type = "response")
logit_pred <- ifelse(logit_pred_prob > 0.5, "Up", "Down")
logit_pred <- factor(logit_pred, levels = c("Down", "Up"))
logit_conf_matrix <- table(Predicted = logit_pred, Actual = test_data$Direction)
logit_accuracy <- sum(diag(logit_conf_matrix)) / sum(logit_conf_matrix)
print(logit_conf_matrix)## Actual
## Predicted Down Up
## Down 22 31
## Up 21 30cat("Logistic Regression Accuracy:", logit_accuracy, "\n")## Logistic Regression Accuracy: 0.5lda_model2 <- lda(Direction ~ Lag2 + Lag2_sq + Lag1_Lag2 + Log_Volume, data = train_data)
lda_pred2 <- predict(lda_model2, newdata = test_data)$class
lda_conf_matrix2 <- table(Predicted = lda_pred2, Actual = test_data$Direction)
lda_accuracy2 <- sum(diag(lda_conf_matrix2)) / sum(lda_conf_matrix2)
print(lda_conf_matrix2)## Actual
## Predicted Down Up
## Down 22 31
## Up 21 30cat("LDA Accuracy:", lda_accuracy2, "\n")## LDA Accuracy: 0.5qda_model2 <- qda(Direction ~ Lag2 + Lag2_sq + Lag1_Lag2 + Log_Volume, data = train_data)
qda_pred2 <- predict(qda_model2, newdata = test_data)$class
qda_conf_matrix2 <- table(Predicted = qda_pred2, Actual = test_data$Direction)
qda_accuracy2 <- sum(diag(qda_conf_matrix2)) / sum(qda_conf_matrix2)
print(qda_conf_matrix2)## Actual
## Predicted Down Up
## Down 25 38
## Up 18 23cat("QDA Accuracy:", qda_accuracy2, "\n")## QDA Accuracy: 0.4615385# scale predictors
train_knn <- scale(train_data[, c("Lag2", "Lag2_sq", "Lag1_Lag2", "Log_Volume")])
test_knn <- scale(test_data[, c("Lag2", "Lag2_sq", "Lag1_Lag2", "Log_Volume")],
center = attr(train_knn, "scaled:center"),
scale = attr(train_knn, "scaled:scale"))
train_Y <- train_data$Direction
test_Y <- test_data$Direction
k_values <- c(1, 3, 5, 7, 10)
knn_accuracies <- c()
for (k in k_values) {
knn_pred <- knn(train_knn, test_knn, train_Y, k = k)
knn_conf_matrix <- table(Predicted = knn_pred, Actual = test_Y)
knn_acc <- sum(diag(knn_conf_matrix)) / sum(knn_conf_matrix)
knn_accuracies <- c(knn_accuracies, knn_acc)
cat("K =", k, "Accuracy:", knn_acc, "\n")
}## K = 1 Accuracy: 0.5384615
## K = 3 Accuracy: 0.5
## K = 5 Accuracy: 0.5673077
## K = 7 Accuracy: 0.5192308
## K = 10 Accuracy: 0.5096154best_k <- k_values[which.max(knn_accuracies)]
cat("Best K:", best_k, "with accuracy:", max(knn_accuracies), "\n")## Best K: 5 with accuracy: 0.5673077nb_model <- naiveBayes(Direction ~ Lag2 + Lag2_sq + Lag1_Lag2 + Log_Volume, data = train_data)
nb_pred <- predict(nb_model, newdata = test_data)
nb_conf_matrix <- table(Predicted = nb_pred, Actual = test_data$Direction)
nb_accuracy <- sum(diag(nb_conf_matrix)) / sum(nb_conf_matrix)
print(nb_conf_matrix)## Actual
## Predicted Down Up
## Down 25 37
## Up 18 24cat("Naïve Bayes Accuracy:", nb_accuracy, "\n")## Naïve Bayes Accuracy: 0.4711538Experiment with different combinations of predictors, includ- ing possible transformations and interactions, for each of the methods. Report the variables, method, and associated confu- sion 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.
- Multiple Lag variables were used instead of just Lag2; Volume was included; interaction terms Lag1 * Lag2 was added
- Squared Lag2 to capture any nonlinear effects
- experimented with different values of K to find the best model.
14:
data("Auto")
str(Auto)## 'data.frame': 392 obs. of 9 variables:
## $ mpg : num 18 15 18 16 17 15 14 14 14 15 ...
## $ cylinders : int 8 8 8 8 8 8 8 8 8 8 ...
## $ displacement: num 307 350 318 304 302 429 454 440 455 390 ...
## $ horsepower : int 130 165 150 150 140 198 220 215 225 190 ...
## $ weight : int 3504 3693 3436 3433 3449 4341 4354 4312 4425 3850 ...
## $ acceleration: num 12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
## $ year : int 70 70 70 70 70 70 70 70 70 70 ...
## $ origin : int 1 1 1 1 1 1 1 1 1 1 ...
## $ name : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
## - attr(*, "na.action")= 'omit' Named int [1:5] 33 127 331 337 355
## ..- attr(*, "names")= chr [1:5] "33" "127" "331" "337" ...mpg_median <- median(Auto$mpg)
Auto$mpg01 <- ifelse(Auto$mpg > mpg_median, 1, 0)
Auto$mpg01 <- as.factor(Auto$mpg01)
table(Auto$mpg01)##
## 0 1
## 196 196summary(Auto$mpg01)## 0 1
## 196 196Auto_new <- data.frame(mpg01 = Auto$mpg01, Auto)
head(Auto_new)## mpg01 mpg cylinders displacement horsepower weight acceleration year origin
## 1 0 18 8 307 130 3504 12.0 70 1
## 2 0 15 8 350 165 3693 11.5 70 1
## 3 0 18 8 318 150 3436 11.0 70 1
## 4 0 16 8 304 150 3433 12.0 70 1
## 5 0 17 8 302 140 3449 10.5 70 1
## 6 0 15 8 429 198 4341 10.0 70 1
## name mpg01.1
## 1 chevrolet chevelle malibu 0
## 2 buick skylark 320 0
## 3 plymouth satellite 0
## 4 amc rebel sst 0
## 5 ford torino 0
## 6 ford galaxie 500 0library(ggplot2)ggplot(Auto, aes(x = mpg01, y = cylinders, fill = mpg01)) +
geom_boxplot() +
ggtitle("Cylinders vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Cylinders")
ggplot(Auto, aes(x = mpg01, y = displacement, fill = mpg01)) +
geom_boxplot() +
ggtitle("Displacement vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Displacement")
ggplot(Auto, aes(x = mpg01, y = horsepower, fill = mpg01)) +
geom_boxplot() +
ggtitle("Horsepower vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Horsepower")
ggplot(Auto, aes(x = mpg01, y = weight, fill = mpg01)) +
geom_boxplot() +
ggtitle("Weight vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Weight")
ggplot(Auto, aes(x = mpg01, y = acceleration, fill = mpg01)) +
geom_boxplot() +
ggtitle("Acceleration vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Acceleration")
ggplot(Auto, aes(x = mpg01, y = year, fill = mpg01)) +
geom_boxplot() +
ggtitle("Year vs mpg01") +
xlab("mpg01 (0 = Low, 1 = High)") +
ylab("Year")
ggplot(Auto, aes(x = displacement, y = mpg, color = mpg01)) +
geom_point(alpha = 0.7) +
ggtitle("Displacement vs MPG") +
xlab("Displacement") +
ylab("MPG")
ggplot(Auto, aes(x = horsepower, y = mpg, color = mpg01)) +
geom_point(alpha = 0.7) +
ggtitle("Horsepower vs MPG") +
xlab("Horsepower") +
ylab("MPG")
ggplot(Auto, aes(x = weight, y = mpg, color = mpg01)) +
geom_point(alpha = 0.7) +
ggtitle("Weight vs MPG") +
xlab("Weight") +
ylab("MPG")
cor_matrix <- cor(Auto[, sapply(Auto, is.numeric)])
print(cor_matrix["mpg",])## mpg cylinders displacement horsepower weight acceleration
## 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442 0.4233285
## year origin
## 0.5805410 0.5652088- Explore the data graphically in order to investigate the associ- ation between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scat- terplots and boxplots may be useful tools to answer this ques- tion. Describe your findings.
Higher displacments leads to lower mpg
higiher horsepower tend ot be in lower mpg category
higher weights have lower mpg
set.seed(123)
train_indices <- sample(1:nrow(Auto), size = 0.7 * nrow(Auto))
train_data2 <- Auto[train_indices, ]
test_data2 <- Auto[-train_indices, ]
lda_model3 <- lda(mpg01 ~ weight + displacement + horsepower + cylinders + year, data = train_data2)
print(lda_model3)## Call:
## lda(mpg01 ~ weight + displacement + horsepower + cylinders +
## year, data = train_data2)
##
## Prior probabilities of groups:
## 0 1
## 0.4963504 0.5036496
##
## Group means:
## weight displacement horsepower cylinders year
## 0 3641.022 275.2941 130.96324 6.786765 74.39706
## 1 2314.000 114.5290 78.00725 4.188406 77.84058
##
## Coefficients of linear discriminants:
## LD1
## weight -0.001196208
## displacement -0.001452526
## horsepower 0.010098725
## cylinders -0.398890632
## year 0.133073372lda_pred3 <- predict(lda_model3, newdata = test_data2)
lda_class3 <- lda_pred3$class
lda_conf_matrix3 <- table(Predicted = lda_class3, Actual = test_data2$mpg01)
print(lda_conf_matrix)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56test_error <- 1 - sum(diag(lda_conf_matrix3)) / sum(lda_conf_matrix3)
cat("Test Error Rate:", test_error, "\n")## Test Error Rate: 0.1016949qda_model3 <- qda(mpg01 ~ weight + displacement + horsepower + cylinders + year, data = train_data2)
qda_pred3 <- predict(qda_model3, newdata = test_data2)$class
qda_conf_matrix3 <- table(Predicted = qda_pred3, Actual = test_data2$mpg01)
qda_test_error3 <- 1 - sum(diag(qda_conf_matrix3)) / sum(qda_conf_matrix3)
print(qda_conf_matrix3)## Actual
## Predicted 0 1
## 0 53 5
## 1 7 53cat("QDA Test Error:", qda_test_error3, "\n")## QDA Test Error: 0.1016949logit_model3 <- glm(mpg01 ~ weight + displacement + horsepower + cylinders + year,
data = train_data2, family = binomial)
logit_pred_prob2 <- predict(logit_model3, newdata = test_data2, type = "response")
logit_pred2 <- ifelse(logit_pred_prob2 > 0.5, 1, 0)
logit_pred2 <- factor(logit_pred2, levels = c(0, 1))
logit_conf_matrix2 <- table(Predicted = logit_pred2, Actual = test_data2$mpg01)
logit_test_error2 <- 1 - sum(diag(logit_conf_matrix2)) / sum(logit_conf_matrix2)
print(logit_conf_matrix2)## Actual
## Predicted 0 1
## 0 56 9
## 1 4 49cat("Logistic Regression Test Error:", logit_test_error2, "\n")## Logistic Regression Test Error: 0.1101695nb_model3 <- naiveBayes(mpg01 ~ weight + displacement + horsepower + cylinders + year, data = train_data2)
nb_pred3 <- predict(nb_model3, newdata = test_data2)
nb_conf_matrix3 <- table(Predicted = nb_pred3, Actual = test_data2$mpg01)
nb_test_error3 <- 1 - sum(diag(nb_conf_matrix3)) / sum(nb_conf_matrix3)
print(nb_conf_matrix3)## Actual
## Predicted 0 1
## 0 52 4
## 1 8 54cat("Naïve Bayes Test Error:", nb_test_error3, "\n")## Naïve Bayes Test Error: 0.1016949train_knn2 <- scale(train_data2[, c("weight", "displacement", "horsepower", "cylinders", "year")])
test_knn2 <- scale(test_data2[, c("weight", "displacement", "horsepower", "cylinders", "year")],
center = attr(train_knn2, "scaled:center"),
scale = attr(train_knn2, "scaled:scale"))
train_Y2 <- train_data2$mpg01
test_Y2 <- test_data2$mpg01
# Try different values of K
k_values2 <- c(1, 3, 5, 7, 10, 15, 20)
knn_errors2 <- c()
for (k in k_values2) {
knn_pred2 <- knn(train_knn2, test_knn2, train_Y2, k = k)
knn_conf_matrix2 <- table(Predicted = knn_pred2, Actual = test_Y2)
knn_error2 <- 1 - sum(diag(knn_conf_matrix2)) / sum(knn_conf_matrix2)
knn_errors2 <- c(knn_errors2, knn_error2)
cat("K =", k, "Test Error:", knn_error2, "\n")
}## K = 1 Test Error: 0.05932203
## K = 3 Test Error: 0.1186441
## K = 5 Test Error: 0.08474576
## K = 7 Test Error: 0.09322034
## K = 10 Test Error: 0.08474576
## K = 15 Test Error: 0.1016949
## K = 20 Test Error: 0.1016949best_k2 <- k_values2[which.min(knn_errors2)]
cat("Best K:", best_k2, "with Test Error:", min(knn_errors2), "\n")## Best K: 1 with Test Error: 0.05932203- 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 me- dian. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
data("Boston")
crim_median <- median(Boston$crim)
Boston$crim01 <- ifelse(Boston$crim > crim_median, 1, 0)
Boston$crim01 <- as.factor(Boston$crim01)
str(Boston)## 'data.frame': 506 obs. of 15 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
## $ crim01 : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...cor_matrix <- cor(Boston[, sapply(Boston, is.numeric)])
print(cor_matrix["crim",]) ## crim zn indus chas nox rm
## 1.00000000 -0.20046922 0.40658341 -0.05589158 0.42097171 -0.21924670
## age dis rad tax ptratio black
## 0.35273425 -0.37967009 0.62550515 0.58276431 0.28994558 -0.38506394
## lstat medv
## 0.45562148 -0.38830461set.seed(123)
train_indices3 <- sample(1:nrow(Boston), size = 0.7 * nrow(Boston))
train_data3 <- Boston[train_indices3, ]
test_data3 <- Boston[-train_indices3, ]Logistic Regression:
logit_model4 <- glm(crim01 ~ lstat + dis + rm + age + indus + nox,
data = train_data3, family = binomial)
logit_pred_prob3 <- predict(logit_model4, newdata = test_data3, type = "response")
logit_pred3 <- ifelse(logit_pred_prob3 > 0.5, 1, 0)
logit_pred3 <- factor(logit_pred3, levels = c(0, 1))
logit_conf_matrix3 <- table(Predicted = logit_pred3, Actual = test_data3$crim01)
logit_test_error3 <- 1 - sum(diag(logit_conf_matrix3)) / sum(logit_conf_matrix3)
print(logit_conf_matrix3)## Actual
## Predicted 0 1
## 0 60 9
## 1 15 68cat("Logistic Regression Test Error:", logit_test_error3, "\n")## Logistic Regression Test Error: 0.157894715.8% is moderate accuracy, with 9 high crime areas and 15 misclassified low crime areas
LDA:
lda_model4 <- lda(crim01 ~ lstat + dis + rm + age + indus + nox, data = train_data3)
lda_pred3 <- predict(lda_model4, newdata = test_data3)$class
lda_conf_matrix3 <- table(Predicted = lda_pred3, Actual = test_data3$crim01)
lda_test_error3 <- 1 - sum(diag(lda_conf_matrix3)) / sum(lda_conf_matrix3)
print(lda_conf_matrix3)## Actual
## Predicted 0 1
## 0 65 11
## 1 10 66cat("LDA Test Error:", lda_test_error3, "\n")## LDA Test Error: 0.138157913.8% test error is better than logistic regression. Misclassification is slightly lower.
nb_model4 <- naiveBayes(crim01 ~ lstat + dis + rm + age + indus + nox, data = train_data3)
nb_pred3 <- predict(nb_model4, newdata = test_data3)
nb_conf_matrix3 <- table(Predicted = nb_pred3, Actual = test_data3$crim01)
nb_test_error3 <- 1 - sum(diag(nb_conf_matrix3)) / sum(nb_conf_matrix3)
print(nb_conf_matrix3)## Actual
## Predicted 0 1
## 0 64 12
## 1 11 65cat("Naïve Bayes Test Error:", nb_test_error3, "\n")## Naïve Bayes Test Error: 0.1513158Error rate is similar to logistic regression.
train_knn3 <- scale(train_data3[, c("lstat", "dis", "rm", "age", "indus", "nox")])
test_knn3 <- scale(test_data3[, c("lstat", "dis", "rm", "age", "indus", "nox")],
center = attr(train_knn3, "scaled:center"),
scale = attr(train_knn3, "scaled:scale"))
train_Y3 <- train_data3$crim01
test_Y3 <- test_data3$crim01
k_values3 <- c(1, 3, 5, 7, 10, 15, 20)
knn_errors3 <- c()
for (k in k_values3) {
knn_pred3 <- knn(train_knn3, test_knn3, train_Y3, k = k)
knn_conf_matrix3 <- table(Predicted = knn_pred3, Actual = test_Y3)
knn_error3 <- 1 - sum(diag(knn_conf_matrix3)) / sum(knn_conf_matrix3)
knn_errors3 <- c(knn_errors3, knn_error3)
cat("K =", k, "Test Error:", knn_error3, "\n")
}## K = 1 Test Error: 0.1052632
## K = 3 Test Error: 0.08552632
## K = 5 Test Error: 0.09868421
## K = 7 Test Error: 0.1118421
## K = 10 Test Error: 0.1315789
## K = 15 Test Error: 0.1578947
## K = 20 Test Error: 0.1513158best_k3 <- k_values3[which.min(knn_errors3)]
cat("Best K:", best_k3, "with Test Error:", min(knn_errors3), "\n")## Best K: 3 with Test Error: 0.08552632The best performing value of K is 3, provides the lowest test error rate at 8.6% compared to all other models above.