Description

###Project Background:
Liability customers – Majority – Depositors Asset customers – Small – Borrowers Campaign of last year conversion rate of 9.6% [Among the 5000 customers, only 480 (= 9.6%) accepted the personal loan that was offered to them in the earlier campaign.] Goal : use k-NN to predict whether a new customer will accept a loan offer. * Data (rows): 5000 customers *Success class as 1 (loan acceptance) ####Packages used:
library(psych) #for creating dummies
library(caret) #for data partition, normalize data
## Loading required package: ggplot2
##
## Attaching package: ’ggplot2’
## The following objects are masked from ’package:psych’:
##
## %+%, alpha
## Loading required package: lattice
library(FNN) library(class) #for Perfoming knn classification
##
## Attaching package: ’class’
## The following objects are masked from ’package:FNN’:
##
## knn, knn.cv
library(dplyr)
##
## Attaching package: ’dplyr’
## The following objects are masked from ’package:stats’:
##
## filter, lag
## The following objects are masked from ’package:base’:
##
## intersect, setdiff, setequal, union
###importing data
input<- read.csv(“UniversalBank.csv”)
#Eliminating variables [id & zip code] from the dataset df=subset(input, select=-c(ID, ZIP.Code ))
#creating dummies
dummy_Education <- as.data.frame(dummy.code(df$Education)) names(dummy_Education) <- c(“Education_1”, “Education_2″,”Education_3”) df_without_education <- subset(df, select=-c(Education))
UBank_data <- cbind(df_without_education, dummy_Education) #main dataset
#renaming dummy variable
#eliminating education variable
###Data partition
#Partitioning the data into Traning(60%) and Validation(40%) set.seed(1234)
Train_Index = createDataPartition(UBank_data$Age, p= 0.6 , list=FALSE)
Train_Data = UBank_data[Train_Index,] #3001 observations
Validation_Data = UBank_data[-Train_Index,] #1999 observations
###Generating test data
Test_Data <- data.frame(Age=40 , Experience=10, Income = 84, Family = 2, CCAvg = 2, Education_1 = 0, Edu
###Data Normalization
train.norm.df <- Train_Data valid.norm.df <- Validation_Data test.norm.df <- Test_Data maindata.norm.df <- UBank_data head(maindata.norm.df)
## Age Experience Income Family CCAvg Mortgage Personal.Loan Securities.Account
## 1 25 1 49 4 1.6 0 0 1
## 2 45 19 34 3 1.5 0 0 1
## 3 39 15 11 1 1.0 0 0 0
## 4 35 9 100 1 2.7 0 0 0
## 5 35 8 45 4 1.0 0 0 0
## 6 37 13 29 4 0.4 155 0 0
## CD.Account Online CreditCard Education_1 Education_2 Education_3
## 1 0 0 0 1 0 0
## 2 0 0 0 1 0 0
## 3 0 0 0 1 0 0
## 4 0 0 0 0 0 1
## 5 0 0 1 0 0 1
## 6 0 1 0 0 0 1
# use preProcess() from the caret package to normalize . norm.values <- preProcess(Train_Data[,-7], method=c(“center”, “scale”))
train.norm.df[,-7] <- predict(norm.values, Train_Data[,-7]) #Training Data valid.norm.df [,-7]<- predict(norm.values, Validation_Data[,-7])#Validation Data test.norm.df <- predict(norm.values, Test_Data)#Test Data maindata.norm.df[,-7] <- predict(norm.values,UBank_data[,-7]) #Training + Validation data head(maindata.norm.df)
## Age Experience Income Family CCAvg Mortgage
## 1 -1.77136698 -1.6613124 -0.5177762 1.3933091 -0.1845814 -0.5438042
## 2 -0.03145296 -0.0978843 -0.8425723 0.5187388 -0.2419870 -0.5438042
## 3 -0.55342717 -0.4453128 -1.3405930 -1.2304018 -0.5290146 -0.5438042
## 4 -0.90140997 -0.9664555 0.5865306 -1.2304018 0.4468794 -0.5438042
## 5 -0.90140997 -1.0533126 -0.6043885 1.3933091 -0.5290146 -0.5438042
## 6 -0.72741857 -0.6190270 -0.9508377 1.3933091 -0.8734478 1.0035659
## Personal.Loan Securities.Account CD.Account Online CreditCard Education_1
## 1 0 2.9564494 -0.2533042 -1.2038741 -0.6538696 1.1696714
## 2 0 2.9564494 -0.2533042 -1.2038741 -0.6538696 1.1696714
## 3 0 -0.3381309 -0.2533042 -1.2038741 -0.6538696 1.1696714
## 4 0 -0.3381309 -0.2533042 -1.2038741 -0.6538696 -0.8546561
## 5 0 -0.3381309 -0.2533042 -1.2038741 1.5288474 -0.8546561
## 6 0 -0.3381309 -0.2533042 0.8303749 -0.6538696 -0.8546561
## Education_2 Education_3
## 1 -0.6414311 -0.6331615
## 2 -0.6414311 -0.6331615
## 3 -0.6414311 -0.6331615
## 4 -0.6414311 1.5788497
## 5 -0.6414311 1.5788497
## 6 -0.6414311 1.5788497
###Perfoming k-NN classification , using k = 1
set.seed(1234)
prediction <- knn(train = train.norm.df[,-7], test = valid.norm.df[,-7], cl = train.norm.df[,7], k = 1, prob=TRUE)
actual= valid.norm.df$Personal.Loan prediction_prob = attr(prediction,”prob”) table(prediction,actual)
## actual
## prediction 0 1
## 0 1770 68
## 1 25 136
mean(prediction==actual)
## [1] 0.9534767
NROW(train.norm.df)
## [1] 3001
sqrt(3001)
## [1] 54.78138
accuracy.df <- data.frame(k = seq(1, 60, 1), accuracy = rep(0, 60))
# compute knn for different k on validation. for(i in 1:60) { prediction <- knn(train = train.norm.df[,-7], test = valid.norm.df[-7], cl = train.norm.df[,7], k = i, prob=TRUE)
accuracy.df[i,2] <- mean(prediction==actual)
} accuracy.df
## k accuracy
## 1 1 0.9534767
## 2 2 0.9494747
## 3 3 0.9544772
## 4 4 0.9549775
## 5 5 0.9524762
## 6 6 0.9504752
## 7 7 0.9489745
## 8 8 0.9459730
## 9 9 0.9454727
## 10 10 0.9454727
## 11 11 0.9439720
## 12 12 0.9424712
## 13 13 0.9424712
## 14 14 0.9414707
## 15 15 0.9409705
## 16 16 0.9414707
## 17 17 0.9399700
## 18 18 0.9394697
## 19 19 0.9404702
## 20 20 0.9394697
## 21 21 0.9384692
## 22 22 0.9364682
## 23 23 0.9364682
## 24 24 0.9339670
## 25 25 0.9349675
## 26 26 0.9344672
## 27 27 0.9354677
## 28 28 0.9344672
## 29 29 0.9339670
## 30 30 0.9329665
## 31 31 0.9314657
## 32 32 0.9324662
## 33 33 0.9319660
## 34 34 0.9294647
## 35 35 0.9304652
## 36 36 0.9289645
## 37 37 0.9284642
## 38 38 0.9309655
## 39 39 0.9284642
## 40 40 0.9274637
## 41 41 0.9269635
## 42 42 0.9259630
## 43 43 0.9254627
## 44 44 0.9254627
## 45 45 0.9249625
## 46 46 0.9264632
## 47 47 0.9244622
## 48 48 0.9239620
## 49 49 0.9244622
## 50 50 0.9244622
## 51 51 0.9244622
## 52 52 0.9224612
## 53 53 0.9234617
## 54 54 0.9239620
## 55 55 0.9224612
## 56 56 0.9224612
## 57 57 0.9229615
## 58 58 0.9224612
## 59 59 0.9214607
## 60 60 0.9214607
The value of k we choose is 1 as it is given in the question [i.e the choice of k that balances between overfitting and ignoring the predictor information]
####Validation data results using best k value [i.e: k = 1]
set.seed(1234)
prediction <- knn(train = train.norm.df[,-7], test = valid.norm.df[,-7], cl = train.norm.df[,7], k = 1, prob=TRUE)
actual= valid.norm.df$Personal.Loan prediction_prob = attr(prediction,”prob”)
#Answer 3: confusion matrix for the best k value =1 table(prediction,actual)
## actual
## prediction 0 1
## 0 1770 68
## 1 25 136
#accuracy of the best k=1 mean(prediction==actual)
## [1] 0.9534767
prediction_test <- knn(train = maindata.norm.df[,-7], test = Test_Data, cl = maindata.norm.df[,7], k = 1, prob=TRUE)
head(prediction_test)
Classifying the customer using the best k [perfominng k-NN classification on test data]
## [1] 1
## Levels: 0 1
k-NN model predicted that the new customer will accept a loan offer [loan accepted]
5)Repartition the data, this time into training, validation, and test sets (50% : 30% : 20%). Apply the k-NN method with the k chosen above. Compare the confusion matrix of the test set with that of the training and validation sets.
#Partitioning the data into Traning(50%) ,Validation(30%), Test(20%) set.seed(1234)
Test_Index_1 = createDataPartition(UBank_data$Age, p= 0.2 , list=FALSE) #20% test data
Test_Data_1 = UBank_data [Test_Index_1,]
Rem_DATA = UBank_data[-Test_Index_1,] #80% remaining data [training + validation]
Train_Index_1 = createDataPartition(Rem_DATA$Age, p= 0.5 , list=FALSE)
Train_Data_1 = Rem_DATA[Train_Index_1,] #Training data
Validation_Data_1 = Rem_DATA[-Train_Index_1,] #Validation data
#Data Normalization
# Copy the original data train.norm.df_1 <- Train_Data_1 valid.norm.df_1 <- Validation_Data_1 test.norm.df_1 <- Test_Data_1 rem_data.norm.df_1 <- Rem_DATA
# use preProcess() from the caret package to normalize Sales and Age. norm.values_1 <- preProcess(Train_Data_1[-7], method=c(“center”, “scale”))
train.norm.df_1[-7] <- predict(norm.values_1, Train_Data_1[-7]) #Training Data valid.norm.df_1[-7] <- predict(norm.values_1, Validation_Data_1[-7])#Validation Data
test.norm.df_1[-7] <- predict(norm.values_1, test.norm.df_1[-7]) #Test Data test.norm.df_1[-7] <- predict(norm.values_1, Test_Data_1[-7]) rem_data.norm.df_1[-7] <- predict(norm.values_1,Rem_DATA[-7]) #Training + Validation data head(test.norm.df_1)
## Age Experience Income Family CCAvg Mortgage
## 9 -0.90840439 -0.883582836 0.1435652 0.5333142 -0.780693325 0.4495336
## 28 0.05751618 -0.008054857 1.8189997 -1.2081200 0.234699617 -0.5532869
## 32 -0.46934959 -0.358266049 -0.9878972 -1.2081200 0.009056741 -0.5532869
## 40 -0.64497151 -0.620924443 0.1218063 1.4040313 -0.724282606 2.1948269
## 42 -0.99621536 -0.971135634 -0.3133715 0.5333142 0.178288898 -0.5532869
## 63 -0.29372767 -0.183160453 -1.1402094 -1.2081200 -0.555050449 -0.5532869
## Personal.Loan Securities.Account CD.Account Online CreditCard
## 9 0 -0.3360202 -0.2646808 0.8429167 -0.6350646
## 28 0 -0.3360202 -0.2646808 0.8429167 1.5738557
## 32 0 -0.3360202 -0.2646808 0.8429167 -0.6350646
## 40 0 -0.3360202 -0.2646808 0.8429167 -0.6350646
## 42 0 -0.3360202 -0.2646808 -1.1857637 -0.6350646
## 63 0 -0.3360202 -0.2646808 -1.1857637 -0.6350646
## Education_1 Education_2 Education_3
## 9 -0.827392 -0.6607293 1.566207
## 28 1.208013 -0.6607293 -0.638166
## 32 -0.827392 -0.6607293 1.566207
## 40 -0.827392 1.5127224 -0.638166
## 42 1.208013 -0.6607293 -0.638166
## 63 1.208013 -0.6607293 -0.638166
#Perfoming k-NN classification on Training Data, k = 1 set.seed(1234) prediction_Q5 <- knn(train = train.norm.df_1[,-7], test = valid.norm.df_1[,-7], cl = train.norm.df_1[,7], k = 1, prob=TRUE)
actual= valid.norm.df_1$Personal.Loan prediction_prob = attr(prediction_Q5,”prob”) table(prediction_Q5,actual) #confusion matrix for the best k value =1
## actual
## prediction_Q5 0 1
## 0 1795 69
## 1 16 119
mean(prediction_Q5==actual) #accuracy of the best k=1
## [1] 0.9574787
set.seed(1234)
prediction_Q5 <- knn(train = rem_data.norm.df_1[,-7], test = test.norm.df_1[,-7], cl = rem_data.norm.df_1[,7], k = 1, prob=TRUE)
actual= test.norm.df_1$Personal.Loan prediction_prob = attr(prediction_Q5,”prob”) table(prediction_Q5,actual) #confusion matrix for the best k value =1
## actual
## prediction_Q5 0 1
## 0 907 25
## 1 12 57
mean(prediction_Q5==actual) #accuracy of the best k=1
## [1] 0.963037
The model performed better in the test set, as it got enough data to learn from i.e 80% of the data, Whereas when we were working on training data it only learned from 50% of the data.