ISYE - 6501 Homework 2
Due Date: Thursday, September 3rd, 2020
Contents
1 ISYE - 6501 Homework 2 2
2 Homework Analysis 2
2.1 Analysis 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
...
ISYE - 6501 Homework 2
Due Date: Thursday, September 3rd, 2020
Contents
1 ISYE - 6501 Homework 2 2
2 Homework Analysis 2
2.1 Analysis 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.2 Analysis 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 Analysis 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1
1 ISYE - 6501 Homework 2
This document contains my analysis for ISYE - 6501 Homework 2 which is due on Thursday, September 3rd,
2020. Enjoy!
2 Homework Analysis
2.1 Analysis 3.1
Q: 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)
RESULTS
By using cross-validation at 10 folds on a k-nearest-neighbors (KNN) model, at k=15 with a rectangular
kernel, we were able to achieve an accuracy score of roughly 85% (85.47009%). This means that 85 out of
every 100 applicants is predicted correct!
THE CODE:
# needed library
rm(list=ls())
library(kknn)
library(dplyr)
set.seed(12345)
# read data into R
data_path <- "data 3.1/"
data_filename <- "credit_card_data-headers.txt"
credit_data <- read.delim(paste0(data_path, data_filename), header=TRUE)
# train-valid-test-split
sample_split <- sample(1:3, size=nrow(credit_data), prob=c(0.7,0.15,0.15),
replace = TRUE)
train_credit <- credit_data[sample_split==1,]
valid_credit <- credit_data[sample_split==2,]
test_credit <- credit_data[sample_split==3,]
# training our model
train_model <- train.kknn(R1~., train_credit, kmax=100, scale=TRUE, kcv=10,
kernel=c("rectangular", "triangular",
"epanechnikov", "gaussian",
"rank", "optimal"),
kpar=list())
train_model
##
## Call:
2
## train.kknn(formula = R1 ~ ., data = train_credit, kmax = 100, kernel = c("rectangular", "triangul##
## Type of response variable: continuous
## minimal mean absolute error: 0.221968
## Minimal mean squared error: 0.1175795
## Best kernel: rectangular
## Best k: 15
Using cross-validation at 10 folds, we can see that the best kernel for our model is rectangular with a k of
15. Now that we have the best parameters for our model, let’s use them to train our validation data.
# validating our model
valid_model <- train.kknn(R1~., valid_credit, ks=15, kernel="rectangular", scale=TRUE, kpar=list())
valid_pred <- round(predict(valid_model, valid_credit))
accuracy_score <- sum(valid_pred == valid_credit[,11]) / nrow(valid_credit)
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