Assignment 3
Jason Villa
2026-06-24
- 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
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.3.3library(tidyverse)## Warning: package 'tidyverse' was built under R version 4.3.3## Warning: package 'tibble' was built under R version 4.3.3## Warning: package 'tidyr' was built under R version 4.3.3## Warning: package 'readr' was built under R version 4.3.3## Warning: package 'dplyr' was built under R version 4.3.3## Warning: package 'lubridate' was built under R version 4.3.3## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
## ✔ forcats 1.0.1 ✔ stringr 1.5.1
## ✔ ggplot2 4.0.3 ✔ tibble 3.2.1
## ✔ lubridate 1.9.4 ✔ tidyr 1.3.1
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsdata(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 Downsummary(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
##
##
##
## ggplot(Weekly, aes(x = Today)) +
geom_histogram(bins = 30, fill = "steelblue", color = "white") +
theme_minimal() +
labs(title = "Distribution of Weekly Returns", x = "Today's Return (%)", y = "Count")
ggplot(Weekly, aes(x = Volume)) +
geom_density(fill = "purple", alpha = 0.4) +
theme_minimal() +
labs(title = "Density Plot of Trading Volume", x = "Volume (in billions)", y = "Density")
library(corrplot)## Warning: package 'corrplot' was built under R version 4.3.3## corrplot 0.95 loadedcor_matrix <- cor(Weekly[, -9])
corrplot(cor_matrix, method = "color", type = "upper", tl.col = "black", tl.srt = 45)
We do see some patterns appear. Volume predictor seems to be normally distributed. We can also see from the correlation plot that there is almost no correlation between each Lag predictor and the Today variable.
- 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?
log_model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(log_model)##
## 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: 4Lag2 is the only predictor that 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
library(caret)## Warning: package 'caret' was built under R version 4.3.3## Loading required package: lattice## Warning: package 'lattice' was built under R version 4.3.3##
## Attaching package: 'caret'## The following object is masked from 'package:purrr':
##
## lifttrain <- Weekly[Weekly$Year < 2009, ]
test <- Weekly[Weekly$Year >= 2009, ]
prob <- predict(log_model, newdata = test, type = "response")
pred_class <- ifelse(prob > 0.5, "Up", "Down")
pred_class <- factor(pred_class, levels = c("Down", "Up"))
actual_class <- factor(test$Direction, levels = c("Down", "Up"))
table(Predicted = pred_class, Actual = actual_class)## Actual
## Predicted Down Up
## Down 17 13
## Up 26 48The log model is showing more false positives than false positives. It predicted up when the market was down 26 times compared to opposite which was only 13 times. The model is better at identifying Up days than Down days.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010)
train <- (Weekly$Year < 2009)
test <- Weekly[Weekly$Year >= 2009, ]
log_model1 <- glm(Direction ~ Lag2, data = Weekly, family = binomial)
prob1 <- predict(log_model1, newdata = test, type = "response")
pred_class <- ifelse(prob1 > 0.5, "Up", "Down")
pred_class <- factor(pred_class, levels = c("Down", "Up"))
actual_class <- factor(test$Direction, levels = c("Down", "Up"))
table(Predicted = pred_class, Actual = actual_class)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56- Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## select## The following object is masked from 'package:ISLR2':
##
## BostonDirection.test <- Weekly$Direction[!train]
lda_model <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda_pred <- predict(lda_model, test)
lda.class <- lda_pred$class
table(Predicted = lda.class, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56- Repeat (d) using QDA.
Direction.test <- Weekly$Direction[!train]
qda_model <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda_pred <- predict(qda_model, test)
qda.class <- qda_pred$class
table(Predicted = qda.class, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61- Repeat (d) using KNN with K =1.
library(class)## Warning: package 'class' was built under R version 4.3.3train.index <- (Weekly$Year < 2009)
X.train <- Weekly[train.index, "Lag2", drop = FALSE]
X.test <- Weekly[!train.index, "Lag2", drop = FALSE]
Y.train <- Weekly$Direction[train.index]
Direction.test <- Weekly$Direction[!train.index]
set.seed(1)
knn.pred <- knn(train = X.train, test = X.test, cl = Y.train, k = 1)
table(Predicted = knn.pred, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 21 30
## Up 22 31- Repeat (d) using naive Bayes.
library(e1071)## Warning: package 'e1071' was built under R version 4.3.3##
## Attaching package: 'e1071'## The following object is masked from 'package:ggplot2':
##
## elementtrain <- (Weekly$Year < 2009)
Weekly.test <- Weekly[!train, ]
Direction.test <- Weekly$Direction[!train]
nb.fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb.class <- predict(nb.fit, Weekly.test)
table(Predicted = nb.class, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61- Which of these methods appears to provide the best results on this data?
Log regression and LDA tie for the best results.
- Experiment with different combinations of predictors, includ ing possible transformations and interactions, for each of the methods. Report the variables, method, and associated confu sion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
train <- (Weekly$Year < 2009)
Weekly.test <- Weekly[!train, ]
Direction.test <- Weekly$Direction[!train]
best.logit <- glm(Direction ~ Lag1 + Lag2,
data = Weekly,
family = binomial,
subset = train)
best.probs <- predict(best.logit, Weekly.test, type = "response")
best.pred <- rep("Down", length(best.probs))
best.pred[best.probs > 0.5] <- "Up"
table(Predicted = best.pred, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 7 8
## Up 36 53X.train <- Weekly[train, "Lag2", drop = FALSE]
X.test <- Weekly[!train, "Lag2", drop = FALSE]
Y.train <- Weekly$Direction[train]
set.seed(1)
knn.pred.9 <- knn(train = X.train, test = X.test, cl = Y.train, k = 9)
table(Predicted = knn.pred.9, Actual = Direction.test)## Actual
## Predicted Down Up
## Down 17 20
## Up 26 41- 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.
mpg_median <- median(Auto$mpg)
mpg01 <- as.integer(Auto$mpg > mpg_median)
Auto_new <- data.frame(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
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500- 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.
boxplot(cylinders ~ mpg01, data = Auto_new, main = "Cylinders vs mpg01", xlab = "mpg01", ylab = "Cylinders", col = "lightblue")
boxplot(displacement ~ mpg01, data = Auto_new, main = "Displacement vs mpg01", xlab = "mpg01", ylab = "Displacement", col = "lightgreen")
boxplot(horsepower ~ mpg01, data = Auto_new, main = "Horsepower vs mpg01", xlab = "mpg01", ylab = "Horsepower", col = "lightpink")
boxplot(weight ~ mpg01, data = Auto_new, main = "Weight vs mpg01", xlab = "mpg01", ylab = "Weight", col = "lightyellow")
boxplot(acceleration ~ mpg01, data = Auto_new, main = "Acceleration vs mpg01", xlab = "mpg01", ylab = "Acceleration", col = "lavender")
boxplot(year ~ mpg01, data = Auto_new, main = "Year vs mpg01", xlab = "mpg01", ylab = "Year", col = "wheat")
pairs(Auto_new[, c("mpg01", "cylinders", "displacement", "horsepower", "weight", "acceleration", "year")])
- Split the data into a training set and a test set.
library(caret)
set.seed(42)
train_size <- floor(0.80 * nrow(Auto_new))
train_indices <- sample(seq_len(nrow(Auto_new)), size = train_size)
train_set <- Auto_new[train_indices, ]
test_set <- Auto_new[-train_indices, ]- 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_automodel <- lda(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set)
print(lda_automodel)## Call:
## lda(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set)
##
## Prior probabilities of groups:
## 0 1
## 0.5271565 0.4728435
##
## Group means:
## weight displacement horsepower cylinders
## 0 3597.970 272.5394 129.38788 6.751515
## 1 2345.905 117.0912 78.66892 4.202703
##
## Coefficients of linear discriminants:
## LD1
## weight -0.0009597286
## displacement -0.0016031063
## horsepower 0.0013528155
## cylinders -0.3798340772lda_autopred <- predict(lda_automodel, newdata = test_set)
lda_autoclass <- lda_autopred$class
test_error <- mean(lda_autoclass != test_set$mpg01)
print(test_error)## [1] 0.06329114Test error is .0633
- 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_automodel <- qda(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set)
print(qda_automodel)## Call:
## qda(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set)
##
## Prior probabilities of groups:
## 0 1
## 0.5271565 0.4728435
##
## Group means:
## weight displacement horsepower cylinders
## 0 3597.970 272.5394 129.38788 6.751515
## 1 2345.905 117.0912 78.66892 4.202703qda_autopred <- predict(qda_automodel, newdata = test_set)
qda_autoclass <- qda_autopred$class
test_error_qda <- mean(qda_autoclass != test_set$mpg01)
print(test_error_qda)## [1] 0.07594937Test error is .0760
- Perform logistic regression on the training data in order to pre dict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
log_automodel <- glm(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set, family = binomial)
summary(log_automodel)##
## Call:
## glm(formula = mpg01 ~ weight + displacement + horsepower + cylinders,
## family = binomial, data = train_set)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 11.4938653 1.8908145 6.079 1.21e-09 ***
## weight -0.0018412 0.0007409 -2.485 0.0130 *
## displacement -0.0109249 0.0088261 -1.238 0.2158
## horsepower -0.0471302 0.0148478 -3.174 0.0015 **
## cylinders -0.0045738 0.3720303 -0.012 0.9902
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 432.99 on 312 degrees of freedom
## Residual deviance: 174.26 on 308 degrees of freedom
## AIC: 184.26
##
## Number of Fisher Scoring iterations: 7prob_autolog <- predict(log_automodel, newdata = test_set, type = "response")
pred_autolog <- ifelse(prob_autolog > 0.5, 1, 0)
test_error_log <- mean(pred_autolog != test_set$mpg01)- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
nb_automodel <- naiveBayes(mpg01 ~ weight + displacement + horsepower + cylinders, data = train_set)
print(nb_automodel)##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5271565 0.4728435
##
## Conditional probabilities:
## weight
## Y [,1] [,2]
## 0 3597.970 670.9549
## 1 2345.905 409.4496
##
## displacement
## Y [,1] [,2]
## 0 272.5394 89.45761
## 1 117.0912 41.22017
##
## horsepower
## Y [,1] [,2]
## 0 129.38788 36.80148
## 1 78.66892 16.11328
##
## cylinders
## Y [,1] [,2]
## 0 6.751515 1.4542036
## 1 4.202703 0.7186482nb_autopred <- predict(nb_automodel, newdata = test_set)
test_error_nb <- mean(nb_autopred != test_set$mpg01)
print(test_error_nb)## [1] 0.06329114- 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?
predictors <- c("weight", "displacement", "horsepower", "cylinders")
scaled_predictors <- scale(Auto_new[, predictors])
train_X <- scaled_predictors[train_indices, ]
test_X <- scaled_predictors[-train_indices, ]
train_y <- Auto_new$mpg01[train_indices]
test_y <- Auto_new$mpg01[-train_indices]
k_values <- c(1, 3, 5, 7, 10, 15, 20, 50)
knn_errors <- numeric(length(k_values))
for (i in seq_along(k_values)) {
set.seed(42)
knn_pred <- knn(train = train_X, test = test_X, cl = train_y, k = k_values[i])
knn_errors[i] <- mean(knn_pred != test_y)
}
knn_results <- data.frame(K = k_values, Test_Error = round(knn_errors, 4))
print(knn_results)## K Test_Error
## 1 1 0.1266
## 2 3 0.0886
## 3 5 0.0759
## 4 7 0.0633
## 5 10 0.0506
## 6 15 0.0633
## 7 20 0.0633
## 8 50 0.0633The best K-value would 10 followe by 15, 20, and 50.
- 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.
crime_median <- median(Boston$crim)
crim01 <- as.integer(Boston$crim > crime_median)
Boston_new <- data.frame(crim01, Boston)
Boston_new$crim <- NULL
cor(Boston_new)[, "crim01"]## crim01 zn indus chas nox rm
## 1.00000000 -0.43615103 0.60326017 0.07009677 0.72323480 -0.15637178
## age dis rad tax ptratio black
## 0.61393992 -0.61634164 0.61978625 0.60874128 0.25356836 -0.35121093
## lstat medv
## 0.45326273 -0.26301673boxplot(rad ~ crim01, data = Boston_new, main = "rad vs crim01", xlab = "crim01", ylab = "rad", col = "lightblue")
boxplot(indus ~ crim01, data = Boston_new, main = "indus vs crim01", xlab = "crim01", ylab = "indus", col = "lightgreen")
boxplot(nox ~ crim01, data = Boston_new, main = "nox vs crim01", xlab = "crim01", ylab = "nox", col = "lightpink")
boxplot(age ~ crim01, data = Boston_new, main = "age vs crim01", xlab = "crim01", ylab = "age", col = "lightyellow")
boxplot(dis ~ crim01, data = Boston_new, main = "dis vs crim01", xlab = "crim01", ylab = "dis", col = "lavender")
boxplot(tax ~ crim01, data = Boston_new, main = "Year vs crim01", xlab = "crim01", ylab = "tax", col = "wheat")
set.seed(123)
train_indices <- sample(1:nrow(Boston_new), 0.8 * nrow(Boston_new))
bostontrain_set <- Boston_new[train_indices, ]
bostontest_set <- Boston_new[-train_indices, ]boston_log <- glm(crim01 ~ nox + rad + tax + dis, data = bostontrain_set, family = binomial)
summary(boston_log)##
## Call:
## glm(formula = crim01 ~ nox + rad + tax + dis, family = binomial,
## data = bostontrain_set)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -21.432095 3.566290 -6.010 1.86e-09 ***
## nox 39.659745 6.067602 6.536 6.31e-11 ***
## rad 0.552769 0.123378 4.480 7.45e-06 ***
## tax -0.009133 0.002500 -3.654 0.000258 ***
## dis 0.077851 0.156504 0.497 0.618880
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 560.02 on 403 degrees of freedom
## Residual deviance: 202.12 on 399 degrees of freedom
## AIC: 212.12
##
## Number of Fisher Scoring iterations: 8bostonlog_prob <- predict(boston_log, newdata = bostontest_set, type = "response")
bostonlog_pred <- ifelse(bostonlog_prob > 0.5, 1, 0)
bostonlog_error <- mean(bostonlog_pred != test_set$crim01)
print(bostonlog_error)## [1] NaNboston_lda <- lda(crim01 ~ nox + rad + tax + dis, data = bostontrain_set)
print(boston_lda)## Call:
## lda(crim01 ~ nox + rad + tax + dis, data = bostontrain_set)
##
## Prior probabilities of groups:
## 0 1
## 0.5049505 0.4950495
##
## Group means:
## nox rad tax dis
## 0 0.4699054 4.22549 309.6765 5.147072
## 1 0.6361000 14.79000 508.6700 2.502883
##
## Coefficients of linear discriminants:
## LD1
## nox 8.311749661
## rad 0.088010880
## tax -0.001745742
## dis -0.141699629bostonpred_lda <- predict(boston_lda, newdata = bostontest_set)$class
lda_bostonerror <- mean(bostonpred_lda != bostontest_set$crim01)
print(lda_bostonerror)## [1] 0.1568627boston_nb <- naiveBayes(crim01 ~ nox + rad + tax + dis, data = bostontrain_set)
print(boston_nb)##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5049505 0.4950495
##
## Conditional probabilities:
## nox
## Y [,1] [,2]
## 0 0.4699054 0.05497320
## 1 0.6361000 0.09875222
##
## rad
## Y [,1] [,2]
## 0 4.22549 1.618288
## 1 14.79000 9.563115
##
## tax
## Y [,1] [,2]
## 0 309.6765 91.01661
## 1 508.6700 168.01885
##
## dis
## Y [,1] [,2]
## 0 5.147072 2.09731
## 1 2.502883 1.07322boston_nbpred <- predict(boston_nb, newdata = test_set)## Warning in predict.naiveBayes(boston_nb, newdata = test_set): Type mismatch
## between training and new data for variable 'nox'. Did you use factors with
## numeric labels for training, and numeric values for new data?## Warning in predict.naiveBayes(boston_nb, newdata = test_set): Type mismatch
## between training and new data for variable 'rad'. Did you use factors with
## numeric labels for training, and numeric values for new data?## Warning in predict.naiveBayes(boston_nb, newdata = test_set): Type mismatch
## between training and new data for variable 'tax'. Did you use factors with
## numeric labels for training, and numeric values for new data?## Warning in predict.naiveBayes(boston_nb, newdata = test_set): Type mismatch
## between training and new data for variable 'dis'. Did you use factors with
## numeric labels for training, and numeric values for new data?boston_nberror <- mean(boston_nbpred != bostontest_set$crim01)## Warning in `!=.default`(boston_nbpred, bostontest_set$crim01): longer object
## length is not a multiple of shorter object length## Warning in is.na(e1) | is.na(e2): longer object length is not a multiple of
## shorter object lengthprint(boston_nberror)## [1] 0.5196078bostonscaled_features <- scale(Boston_new[, -1])
train_boston <- bostonscaled_features[train_indices, c("nox", "rad", "tax", "dis")]
test_boston <- bostonscaled_features[-train_indices, c("nox", "rad", "tax", "dis")]
train_y <- Boston_new$crim01[train_indices]
test_y <- Boston_new$crim01[-train_indices]
set.seed(123)
knn_pred <- knn(train_boston, test_boston, train_y, k = 5)
err_knn <- mean(knn_pred != test_y)
print(err_knn)## [1] 0.06862745