Description

library(“dplyr”) library(“tidyr”) library(“ggplot2”) library(“caret”) library(“tidyverse”) library(“SnowballC”) library(‘tinytex’) library(“dplyr”) library(“tidyr”) library(“reshape2”) library(“e1071”)
rm(list=ls())
bank = read.csv(“C:/Users/suraj/Downloads/UniversalBank.csv”) bank$Personal.Loan = as.factor(bank$Personal.Loan) bank$Online = as.factor(bank$Online) bank$CreditCard = as.factor(bank$CreditCard) set.seed(1)
train.index <- sample(row.names(bank), 0.6*dim(bank)[1]) test.index <- setdiff(row.names(bank), train.index) train.df <- bank[train.index, ] test.df <- bank[test.index, ] train <- bank[train.index, ] test = bank[train.index,]
#a. Create a pivot table for the training data with Online as a column variable, CC as a row variable, and Loan as a secondary row variable. The values inside the table should convey the count. In R use functions melt() and cast(), or function table().
melted.bank = melt(train,id=c(“CreditCard”,”Personal.Loan”),variable= “Online”) recast.bank=dcast(melted.bank,CreditCard+Personal.Loan~Online) recast.bank[,c(1:2,14)]
## CreditCard Personal.Loan Online
## 1 0 0 1924 ## 2 0 1 198 ## 3 1 0 801
## 4 1 1 77
#b. Consider the task of classifying a customer who owns a bank credit card and is actively using online banking services. Looking at the pivot table, what is the probability that this customer will accept the loan offer? [This is the probability of loan acceptance (Loan = 1) conditional on having a bank credit card (CC = 1) and being an active user of online banking services (Online = 1)].
melted.bankc1 = melt(train,id=c(“Personal.Loan”),variable = “Online”)
## dropped
melted.bankc2 = melt(train,id=c(“CreditCard”),variable = “Online”)
## dropped
recast.bankc1=dcast(melted.bankc1,Personal.Loan~Online) recast.bankc2=dcast(melted.bankc2,CreditCard~Online)
#c.Create two separate pivot tables for the training data. One will have Loan (rows) as a function of Online
(columns) and the other will have Loan (rows) as a function of CC
Loanline=recast.bankc1[,c(1,13)]
LoanCC = recast.bankc2[,c(1,14)]
Loanline
## Personal.Loan Online
## 1 0 2725
## 2 1 275
LoanCC
## CreditCard Online
## 1 0 2122
## 2 1 878
#d. Compute the following quantities [P (A | B) means “the probability of A given B”]: i. P (CC = 1 | Loan = 1) (the proportion of credit card holders among the loan acceptors) ii. P(Online=1|Loan=1) iii. P (Loan = 1) (the proportion of loan acceptors) iv. P(CC=1|Loan=0) v. P(Online=1|Loan=0) vi. P(Loan=0)
table(train[,c(14,10)])
## Personal.Loan
## CreditCard 0 1
## 0 1924 198 ## 1 801 77
table(train[,c(13,10)])
## Personal.Loan
## Online 0 1 ## 0 1137 109 ## 1 1588 166 table(train[,c(10)])
##
## 0 1
## 2725 275
probability1<-77/(77+198) probability1
## [1] 0.28
probability2<-166/(166+109) probability2
## [1] 0.6036364
probability3<-275/(275+2725) probability3
## [1] 0.09166667
probability4<-801/(801+1924) probability4
## [1] 0.293945
probability5<-1588/(1588+1137) probability5
## [1] 0.5827523
probability6<-2725/(2725+275) probability6
## [1] 0.9083333
#e. Use the quantities computed above to compute the naive Ba1 probability P(Loan = 1 | CC = 1, Online = 1)
(probability1*probability2*probability3)/((probability1*probability2*probability3)+(probability4*probabi ## [1] 0.09055758
#f. Compare this value with the one obtained from the pivot table in (b). Which is a more accurate estimate?
9.05% are very similar to the 9.7% the difference between the exact method and the naive-baise method is the exact method would need the the exact same independent variable classifications to predict, where the naive bayes method does not.
#g. Which of the entries in this table are needed for computing P (Loan = 1 | CC = 1, Online = 1)? In R, run naive Bayes on the data. Examine the model output on training data, and find the entry that corresponds to P (Loan = 1 | CC = 1, Online = 1). Compare this to the number you obtained in (e).
naive.train = train.df[,c(10,13:14)] naive.test = test.df[,c(10,13:14)]
naivebayes = naiveBayes(Personal.Loan~.,data=naive.train) naivebayes
##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1 ## 0.90833333 0.09166667
##
## Conditional probabilities:
## Online
## Y 0 1 ## 0 0.4172477 0.5827523
## 1 0.3963636 0.6036364
##
## CreditCard
## Y 0 1
## 0 0.706055 0.293945
## 1 0.720000 0.280000
the naive bayes is the exact same output we recieved in the previous methods. (.280)(.603)(.09)/(.280.603.09+.29.58.908) = .09 which is the same response provided as above.