The purpose of this section is to use tree-based algorithm to predict churn of a customr based on some explanatory variable. The data I use is the Customer Support data from IBM Watson Analytics.

library(survival)
library(ggfortify)
library(rpart)
library(dplyr)
churnAnalysis <- read.csv("WA_Fn-UseC_-Telco-Customer-Churn.csv") 
churnAnalysis <- as.data.frame(churnAnalysis)
#head(churnAnalysis)

   

1. Data pre-processing

Check if there is missing value. There are only 11 missing values in TotalCharges so we can just remove entire row given that the number is small.

sapply(churnAnalysis, function(x) sum(is.na(x)))
##       customerID           gender    SeniorCitizen          Partner 
##                0                0                0                0 
##       Dependents           tenure     PhoneService    MultipleLines 
##                0                0                0                0 
##  InternetService   OnlineSecurity     OnlineBackup DeviceProtection 
##                0                0                0                0 
##      TechSupport      StreamingTV  StreamingMovies         Contract 
##                0                0                0                0 
## PaperlessBilling    PaymentMethod   MonthlyCharges     TotalCharges 
##                0                0                0               11 
##            Churn 
##                0
churnAnalysis <- churnAnalysis[which(!is.na(churnAnalysis[,"TotalCharges"])),]

Then we look at what variables are continuous, and they are: “tenure”, “MonthlyCharges” and “TotalCharges”

2. Kaplan Meier Analysis

library(survival)
library(ggfortify)
churnAnalysis <- churnAnalysis %>% mutate(Churn_flag = ifelse(Churn=="Yes", 1, 0) )
km_churn <- with(churnAnalysis, Surv(tenure, Churn_flag))
km_fit <- survfit(Surv(tenure, Churn_flag) ~ 1, data=churnAnalysis)
autoplot(km_fit)

The figure below shows the conditional churn probability given that the account “survived” the last period. We can see that the churn rate is the highest in the first time unit and gradually reverts to the mean at around 10th time unit.

# Conditional churn probability
s0 <- c(1,km_fit$surv[1:71])
c <- (1-km_fit$surv)/s0
c <- (s0-km_fit$surv)/s0
plot(c, type='l')

km_fit_bill <- survfit(Surv(tenure, Churn_flag) ~ PaperlessBilling, data=churnAnalysis)
autoplot(km_fit_bill)

km_fit_senior <- survfit(Surv(tenure, Churn_flag) ~ SeniorCitizen, data=churnAnalysis)
autoplot(km_fit_senior)

km_fit_pay <- survfit(Surv(tenure, Churn_flag) ~ PaymentMethod, data=churnAnalysis)
autoplot(km_fit_pay)

3. Information value

## 
## Attaching package: 'kableExtra'
## The following object is masked from 'package:dplyr':
## 
##     group_rows
Variable IV
Contract 1.2331890
tenure 0.8347032
OnlineSecurity 0.7152919
TechSupport 0.6971084
InternetService 0.6152530
OnlineBackup 0.5264879
DeviceProtection 0.4976099
PaymentMethod 0.4557558
StreamingMovies 0.3798511
StreamingTV 0.3787158
Variable IV
11 MonthlyCharges 0.3631453
12 TotalCharges 0.3376950
13 PaperlessBilling 0.2020563
14 Dependents 0.1531686
15 Partner 0.1178772
16 SeniorCitizen 0.1050842
17 MultipleLines 0.0081689
18 PhoneService 0.0007129
19 gender 0.0003741
NA NA NA

4. logistic regression with WOEs

Under contruction

5. Decision tree by rpart

To use decision tree, we need to convert the continuous variables to categorical variables be setting some thresold. But first, let’s split the data into training, validation and test sets.

## randomize the data first
churnAnalysis <- churnAnalysis[sample(1:nrow(churnAnalysis)),]
## then split
churnAnalysisTraining <- churnAnalysis[1:(nrow(churnAnalysis)/2),]
churnAnalysisValidation <- churnAnalysis[(nrow(churnAnalysis)/2+1): (3*nrow(churnAnalysis)/4),]
churnAnalysisTest <- churnAnalysis[(3*nrow(churnAnalysis)/4+1),nrow(churnAnalysis),]
hist(churnAnalysisTraining[,"MonthlyCharges"],breaks=40)

MonthlyChargesClass <- ifelse(churnAnalysis[,"MonthlyCharges"]<=30,"<=$30",">30$")

As the distribtion of “MonthlyCharges” does look like normal distribution and there seems to be a spike for monthly charges under 30. Hence, I try to split the data into 2:
1. monthly charges $ 30 $
2. monthly charges $ > 30 $

hist(churnAnalysisTraining[,"TotalCharges"],breaks=40)

TotalChargesClass <- ifelse(churnAnalysis[,"TotalCharges"]<=500,"<=$500",">500$")
hist(churnAnalysisTraining[,"tenure"],breaks=80)

TenureClass <- churnAnalysis[,"tenure"]
TenureClass[which(churnAnalysis[,"tenure"] >=70)] <- ">=70"
TenureClass[which(churnAnalysis[,"tenure"] <=5)] <- "<=5"
TenureClass[which(churnAnalysis[,"tenure"] >5 & churnAnalysis[,"tenure"] <70)] <- "between"

Apply the class transform to all 3 data sets

churnAnalysis["MonthlyCharges"] <- MonthlyChargesClass
churnAnalysis["TotalCharges"] <- TotalChargesClass
churnAnalysis["tenure"] <- TenureClass
churnAnalysisTraining <- churnAnalysis[1:(nrow(churnAnalysis)/2),]
churnAnalysisValidation <- churnAnalysis[(nrow(churnAnalysis)/2+1): floor(3*nrow(churnAnalysis)/4),]
churnAnalysisTest <- churnAnalysis[floor(3*nrow(churnAnalysis)/4+1):nrow(churnAnalysis),]

use rpart to run decision tree from package:rpart

library(rpart)
## xval determine how many data points you want in the validation set.
fitTree <- rpart(Churn~gender+SeniorCitizen+Partner+Dependents+tenure+PhoneService+MultipleLines+InternetService+OnlineSecurity+OnlineBackup+DeviceProtection+TechSupport+StreamingTV+StreamingMovies+Contract+PaperlessBilling+PaymentMethod+MonthlyCharges+TotalCharges, churnAnalysisTraining,xval=0) 
pred_fitTree <- predict(fitTree, churnAnalysisTest,type="class")
table(Predicted = pred_fitTree, Actual = churnAnalysisTest$Churn) ## Confusion Matrix for Decision Tree
##          Actual
## Predicted   No  Yes
##       No  1193  290
##       Yes  108  167
plot(fitTree,uniform=TRUE)
text(fitTree,use.n=T,all=T)

printcp(fitTree) ## cp : complexity parameter
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571

The cp at row one is the (rel error at row 2 - rel error at row 1)/nsplit in row 2. The “rel error” is the relative error = absolute error/root node error

now prune the tree using the validation set

TreeDepth <- nrow(fitTree$cptable)
for(i in 2:TreeDepth)


printcp(fitTree)
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571
## 
## Classification tree:
## rpart(formula = Churn ~ gender + SeniorCitizen + Partner + Dependents + 
##     tenure + PhoneService + MultipleLines + InternetService + 
##     OnlineSecurity + OnlineBackup + DeviceProtection + TechSupport + 
##     StreamingTV + StreamingMovies + Contract + PaperlessBilling + 
##     PaymentMethod + MonthlyCharges + TotalCharges, data = churnAnalysisTraining, 
##     xval = 0)
## 
## Variables actually used in tree construction:
## [1] Contract        InternetService MultipleLines   OnlineSecurity 
## [5] PaymentMethod   tenure         
## 
## Root node error: 939/3516 = 0.26706
## 
## n= 3516 
## 
##         CP nsplit rel error
## 1 0.060170      0   1.00000
## 2 0.047923      2   0.87966
## 3 0.026624      3   0.83174
## 4 0.025559      4   0.80511
## 5 0.013845      5   0.77955
## 6 0.010000      6   0.76571