Homework 2
Question 3.1
Using the same data set (credit_card_data.txt or credit_card_data-headers.txt) as in Question 2.2, use the
ksvm or kknn function to find a good classifier:
(a) using cross-validation (do this
...
Homework 2
Question 3.1
Using the same data set (credit_card_data.txt or credit_card_data-headers.txt) as in Question 2.2, use the
ksvm or kknn function to find a good classifier:
(a) using cross-validation (do this for the k-nearest-neighbors model; SVM is optional); and
Using leave-one-out crossvalidation with different kernel for classification
data <- read.csv("credit_card_data-headers.txt", header = TRUE, sep = "")
# Splitting data for training (70%) and validating (30%)
number_of_data_points <- nrow(data)
training_sample <- sample(number_of_data_points,
size = round(number_of_data_points * 0.7))
training_data <- data[training_sample,]
validating_data <- data[-training_sample,]
kmax <- 100
model <- train.kknn(R1~., training_data, kmax = kmax, scale = TRUE,
kernel = c("rectangular", "triangular", "epanechnikov",
"gaussian", "rank", "optimal"))
model
##
## Call:
## train.kknn(formula = R1 ~ ., data = training_data, kmax = kmax, kernel = c("rectangular", "triang##
## Type of response variable: continuous
## minimal mean absolute error: 0.1957787
## Minimal mean squared error: 0.107393
## Best kernel: gaussian
## Best k: 41
pred <- predict(model, validating_data[,-11])
accuracy <- sum(as.integer(round(pred) == validating_data[,11])) / nrow(validating_data)
# Accuracy on validation data
cat("Best accuracy for validation data is", accuracy,
" for K value of", model$best.parameters$k, "\n\n")
## Best accuracy for validation data is 0.8469388 for K value of 41
(b) splitting the data into training, validation, and test data sets (pick either KNN or SVM;
the other is optional).
data <- read.csv("credit_card_data-headers.txt", header = TRUE, sep = "")
# spliting data into training: 60%, validating: 20% and testing: 20%
1
number_of_data_points <- nrow(data)
training_sample <- sample(number_of_data_points,
size = round(number_of_data_points * 0.6))
training_data <- data[training_sample,]
non_training_data <- data[-training_sample,]
number_of_non_training_data_points = nrow(non_training_data)
validating_sample <- sample(number_of_non_training_data_points,
size = round(number_of_non_training_data_points * 0.5))
validating_data <- non_training_data[validating_sample,]
testing_data <- non_training_data[-validating_sample,]
# Using kknn for crossvalidation
Ks <- seq(1, 100)
bestK <- 0
bestAcuracy <- 0
bestModel <- NULL
for(k in Ks) {
model <- kknn(R1~., training_data, validating_data, k = k, scale = TRUE)
pred <- round(predict(model))
accuracy <- sum(pred == validating_data[,11]) / nrow(validating_data)
# Keeping the best accuracy data for later use
if(accuracy > bestAcuracy) {
bestAcuracy <- accuracy
bestK <- k
bestModel <- model
}
}
# Best K and it accuracy on validation data
cat("Best K value is", bestK, "with accuracy of", bestAcuracy, "\n\n")
## Best K value is 11 with accuracy of 0.870229
# Running the test data with best K value
model <- kknn(R1~., training_data, testing_data, k = bestK, scale = TRUE)
pred <- round(predict(model))
accuracy <- sum(pred == testing_data[,11]) / nrow(testing_data)
# Accuracy of test data with best K
cat("Acuracy with K value of", bestK, "on test data is", accuracy, "\n\n")
## Acuracy with K value of 11 on test data is 0.8549618
Question 4.1
Describe a situation or problem from your job, everyday life, current events, etc., for which a clustering model
would be appropriate. List some (up to 5) predictors that you might use.
One of the key revenue generator for our e-commerce business is the recommendation based
online sales. In order to make product recommendation, we need to group our online visitors and
returing customers into various groups. Some of the common predictors we use are:
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