R Code for LOGISTIC REGRESSION and C5.0 DECISION TREE
Data Set:- Bank Marketing
Source of Data Set:- UCI Repository (http://archive.ics.uci.edu/ml/datasets/Bank+Marketing)
The Code includes the following:-
1. Data Exploration - Missing Values, Outliers
2. Data Visualisation
3. Correlation Matrix
4. Data Partitioning
5. Logistic Regression
6. C5.0 Decision Tree
6. Confusion Matrix with ROC
## The data has been imported using Import Dataset option in R Environment
## The data set can be obtained from http://archive.ics.uci.edu/ml/datasets/Bank+Marketing
## DATASET UNDERSTANDING
head(bank_full) ## Displays first 6 rows for each variable
str(bank_full) ## Describes each variables
summary(bank_full) ## Provides basic statistical information of each variable
## DATA EXPLORATION - Check for Missing Data
## Option 1
is.na(bank_full) ## Displays True for a missing value
## Since it is a large dataset, graphical display of missing values will prove to be easier
##Option 2
require(Amelia)
missmap(bank_full,main="Missing Data - Bank Subscription", col=c("red","grey"),legend=FALSE)
## No red colour stripes are visible. hence no missing values.
##Option 3
summary(bank_full) ## displays missing values if any under every variable
## DATA VISUALISATION
## Use Box plots (Only for continuous variables)- To Check Ouliers
boxplot(bank_full$age~bank_full$subscribed, main=" AGE",ylab="age of customers",xlab="Subscribed")
boxplot(bank_full$balance~bank_full$subscribed, main=" BALANCE",ylab="Balance of customers",xlab="Subscribed")
boxplot(bank_full$lastday~bank_full$subscribed, main=" LAST DAY",ylab="Last day of contact",xlab="Subscribed")
boxplot(bank_full$lastduration~bank_full$subscribed, main="LAST DURATION",ylab="Last duration of contact",xlab="Subscribed")
boxplot(bank_full$numcontacts~bank_full$subscribed, main="NUM CONTACTS",ylab="number of contacts",xlab="Subscribed")
boxplot(bank_full$pdays~bank_full$subscribed, main=" Previous DAYS",ylab="Previous days of contact",xlab="Subscribed")
boxplot(bank_full$pcontacts~bank_full$subscribed, main=" Previous Contacts",ylab="Previous Contacts with customers",xlab="Subscribed")
## Though some outliers are observed in Previous contacts, NumContacts and LastDuration, they have not bee removed keeping their significance into consideration
## Use Histograms (For both continuous and categorical variables)
## These histograms provide details abpout Skewness, Normal Distribution etc
## Function to create histograms for continuous variables with normal curve
bank_Conthist<-function(VarName,NumBreaks,xlab,main,lengthxfit) ## xlab and main should be mentioned under quotes as they are characters
{
hist(VarName,breaks=NumBreaks,col="yellow",xlab=xlab,main=main)
xfit<-seq(min(VarName),max(VarName),length=lengthxfit)
yfit<-dnorm(xfit,mean=mean(VarName),sd=sd(VarName))
yfit<-yfit*diff(h$mids[1:2])*length(VarName)
lines(xfit,yfit,col="red",lwd=3)
}
bank_Conthist(bank_full$age,10,"age of customers","AGE",30)
bank_Conthist(bank_full$balance,50,"Balance of customers","Balance",100)
## Balance is more skewed towards to Negative or Zero
bank_Conthist(bank_full$lastday,5,"Last Day of contact","LAst Day",10)
bank_Conthist(bank_full$lastduration,100,"LastDuration of COntact","Last Duration",10)
## Last Duration is more skewed towards 0 to 100 secs.
bank_Conthist(bank_full$numcontacts,30,"Number of Contacts","NUmContacts",20)
## NUmContacts are more skewed towards 1
bank_Conthist(bank_full$pdays,30,"Previous Days of contacts","PDays",20)
## Many were not contacted previously
bank_Conthist(bank_full$pcontacts,20,"Previous Contacts","PContacts",10)
## Since many were not contacted previously, therefore Pcontacts is 0
## Barplots for Categorical Variables
barplot(table(bank_full$job),col="red",main="JOB")
barplot(table(bank_full$marital),col="green",main="Marital")
barplot(table(bank_full$education),col="red",main="Education")
barplot(table(bank_full$creditdefault),col="red",main="Credit Default")
## Since Credit Default is highly skewed towards NO, this shall be removed from further analysis
bank_full[5]<-NULL
str(bank_full)
barplot(table(bank_full$housingloan),col="red",main="Housing Loan")
barplot(table(bank_full$personalloan),col="blue",main="Personal Loan")
barplot(table(bank_full$lastcommtype),col="red",main="Last communication type")
barplot(table(bank_full$lastmonth),col="violet",main="Last Month")
barplot(table(bank_full$poutcome),col="magenta",main="Previous Outcome")
## Correlation Matrix among input (or independent) continuous variables
bank_full.cont<-data.frame(bank_full$age,bank_full$balance,bank_full$lastday,bank_full$lastduration,bank_full$numcontacts,bank_full$pdays,bank_full$pcontacts)
str(bank_full.cont)
cor(bank_full.cont)
## It can be observed that No two variables are highly correlated
## Partitioning Data into Train and Test datasets in 70:30
library(caret)
set.seed(1234567)
train1<-createDataPartition(bank_full$subscribed,p=0.7,list=FALSE)
train<-bank_full[train1,]
test<-bank_full[-train1,]
## Without Any Transformations on the Variables, following are the Classification Models Applied.
## LOGISTIC REGRESSION
train.logistic=glm(subscribed~age+job+marital+education+housingloan+balance+personalloan+lastcommtype+lastday+lastmonth+lastduration+numcontacts+pdays+pcontacts+poutcome,binomial(link="logit"),train)
summary(train.logistic) ## Display the Results. One can remove the variables which are not significant (if P value>0.05) and re-run the model to make it more effective in terms of accurate prediction.
##Deviance was reduced to 22845-18828= 4017 in 40 degrees of freedom (31648-316408)
## Significance Test
1-pchisq(4017,df=40)
## Since its valus is 0, the NUll hypothesis is rejected and the model developed is significant
##Model Diagnostic Plots with train data set
plot(train.logistic)
##Using model for testing the TEST data set
##Initial Probabilities are generated
test.prob<-predict.glm(train.logistic,test[,1:15],type="response")
## PLOT ROC curve to finalise your cut off value
library(pROC)
ROC=roc(subscribed~test.prob,data=test)
plot(ROC) ## Here cut off value = 0.5 has been selected.
## Based on the problem condition, Higher Specificity OR Higher Sensitivity can be selected.In this Scenario, higher sensitivity is preferable
##Now convert the probabilitis to classes/levels with cut off as 0.5
test.prediction<-cut(test.prob,c(-Inf,0.5,Inf),labels=c("No","Yes"))
test.prediction
summary(test.prediction)
##Confusion Matrix
table(test.prediction,test$subscribed)
## 2. C5.0 Decision Tree
library(C50)## Control for C5.0 Model
C5.0Control(subset=TRUE,bands=2,winnow=TRUE,noGlobalPruning=FALSE,CF=0.25,minCases=2,label="C5.0 Outcome")## Train Model using C5.0
train.c50=C5.0(train[,1:15],train$subscribed,trials=10,rules=FALSE,control=C5.0Control(),costs=NULL)
## Variable Importance Measure
C5imp(train.c50,metric="usage",pct=TRUE)## Lower % Usage variables can be dicarded and the Train function is re-run to get better accuracy
summary(train.c50) ## Check the percentage of accuracy of the model## Predict on the TEST data using trained Model
test.c50=predict(object=train.c50,newdata=test[,1:15],trials=1,type="class")
table(test.c50,test$subscribed) ## CONFUSION MATRIX from C5.0
Following is the ROC curve of Logistic Regression for the above solution
Data Set:- Bank Marketing
Source of Data Set:- UCI Repository (http://archive.ics.uci.edu/ml/datasets/Bank+Marketing)
The Code includes the following:-
1. Data Exploration - Missing Values, Outliers
2. Data Visualisation
3. Correlation Matrix
4. Data Partitioning
5. Logistic Regression
6. C5.0 Decision Tree
6. Confusion Matrix with ROC
## The data has been imported using Import Dataset option in R Environment
## The data set can be obtained from http://archive.ics.uci.edu/ml/datasets/Bank+Marketing
## DATASET UNDERSTANDING
head(bank_full) ## Displays first 6 rows for each variable
str(bank_full) ## Describes each variables
summary(bank_full) ## Provides basic statistical information of each variable
## DATA EXPLORATION - Check for Missing Data
## Option 1
is.na(bank_full) ## Displays True for a missing value
## Since it is a large dataset, graphical display of missing values will prove to be easier
##Option 2
require(Amelia)
missmap(bank_full,main="Missing Data - Bank Subscription", col=c("red","grey"),legend=FALSE)
## No red colour stripes are visible. hence no missing values.
##Option 3
summary(bank_full) ## displays missing values if any under every variable
## DATA VISUALISATION
## Use Box plots (Only for continuous variables)- To Check Ouliers
boxplot(bank_full$age~bank_full$subscribed, main=" AGE",ylab="age of customers",xlab="Subscribed")
boxplot(bank_full$balance~bank_full$subscribed, main=" BALANCE",ylab="Balance of customers",xlab="Subscribed")
boxplot(bank_full$lastday~bank_full$subscribed, main=" LAST DAY",ylab="Last day of contact",xlab="Subscribed")
boxplot(bank_full$lastduration~bank_full$subscribed, main="LAST DURATION",ylab="Last duration of contact",xlab="Subscribed")
boxplot(bank_full$numcontacts~bank_full$subscribed, main="NUM CONTACTS",ylab="number of contacts",xlab="Subscribed")
boxplot(bank_full$pdays~bank_full$subscribed, main=" Previous DAYS",ylab="Previous days of contact",xlab="Subscribed")
boxplot(bank_full$pcontacts~bank_full$subscribed, main=" Previous Contacts",ylab="Previous Contacts with customers",xlab="Subscribed")
## Though some outliers are observed in Previous contacts, NumContacts and LastDuration, they have not bee removed keeping their significance into consideration
## Use Histograms (For both continuous and categorical variables)
## These histograms provide details abpout Skewness, Normal Distribution etc
## Function to create histograms for continuous variables with normal curve
bank_Conthist<-function(VarName,NumBreaks,xlab,main,lengthxfit) ## xlab and main should be mentioned under quotes as they are characters
{
hist(VarName,breaks=NumBreaks,col="yellow",xlab=xlab,main=main)
xfit<-seq(min(VarName),max(VarName),length=lengthxfit)
yfit<-dnorm(xfit,mean=mean(VarName),sd=sd(VarName))
yfit<-yfit*diff(h$mids[1:2])*length(VarName)
lines(xfit,yfit,col="red",lwd=3)
}
bank_Conthist(bank_full$age,10,"age of customers","AGE",30)
bank_Conthist(bank_full$balance,50,"Balance of customers","Balance",100)
## Balance is more skewed towards to Negative or Zero
bank_Conthist(bank_full$lastday,5,"Last Day of contact","LAst Day",10)
bank_Conthist(bank_full$lastduration,100,"LastDuration of COntact","Last Duration",10)
## Last Duration is more skewed towards 0 to 100 secs.
bank_Conthist(bank_full$numcontacts,30,"Number of Contacts","NUmContacts",20)
## NUmContacts are more skewed towards 1
bank_Conthist(bank_full$pdays,30,"Previous Days of contacts","PDays",20)
## Many were not contacted previously
bank_Conthist(bank_full$pcontacts,20,"Previous Contacts","PContacts",10)
## Since many were not contacted previously, therefore Pcontacts is 0
## Barplots for Categorical Variables
barplot(table(bank_full$job),col="red",main="JOB")
barplot(table(bank_full$marital),col="green",main="Marital")
barplot(table(bank_full$education),col="red",main="Education")
barplot(table(bank_full$creditdefault),col="red",main="Credit Default")
## Since Credit Default is highly skewed towards NO, this shall be removed from further analysis
bank_full[5]<-NULL
str(bank_full)
barplot(table(bank_full$housingloan),col="red",main="Housing Loan")
barplot(table(bank_full$personalloan),col="blue",main="Personal Loan")
barplot(table(bank_full$lastcommtype),col="red",main="Last communication type")
barplot(table(bank_full$lastmonth),col="violet",main="Last Month")
barplot(table(bank_full$poutcome),col="magenta",main="Previous Outcome")
## Correlation Matrix among input (or independent) continuous variables
bank_full.cont<-data.frame(bank_full$age,bank_full$balance,bank_full$lastday,bank_full$lastduration,bank_full$numcontacts,bank_full$pdays,bank_full$pcontacts)
str(bank_full.cont)
cor(bank_full.cont)
## It can be observed that No two variables are highly correlated
## Partitioning Data into Train and Test datasets in 70:30
library(caret)
set.seed(1234567)
train1<-createDataPartition(bank_full$subscribed,p=0.7,list=FALSE)
train<-bank_full[train1,]
test<-bank_full[-train1,]
## Without Any Transformations on the Variables, following are the Classification Models Applied.
## LOGISTIC REGRESSION
train.logistic=glm(subscribed~age+job+marital+education+housingloan+balance+personalloan+lastcommtype+lastday+lastmonth+lastduration+numcontacts+pdays+pcontacts+poutcome,binomial(link="logit"),train)
summary(train.logistic) ## Display the Results. One can remove the variables which are not significant (if P value>0.05) and re-run the model to make it more effective in terms of accurate prediction.
##Deviance was reduced to 22845-18828= 4017 in 40 degrees of freedom (31648-316408)
## Significance Test
1-pchisq(4017,df=40)
## Since its valus is 0, the NUll hypothesis is rejected and the model developed is significant
##Model Diagnostic Plots with train data set
plot(train.logistic)
##Using model for testing the TEST data set
##Initial Probabilities are generated
test.prob<-predict.glm(train.logistic,test[,1:15],type="response")
## PLOT ROC curve to finalise your cut off value
library(pROC)
ROC=roc(subscribed~test.prob,data=test)
plot(ROC) ## Here cut off value = 0.5 has been selected.
## Based on the problem condition, Higher Specificity OR Higher Sensitivity can be selected.In this Scenario, higher sensitivity is preferable
##Now convert the probabilitis to classes/levels with cut off as 0.5
test.prediction<-cut(test.prob,c(-Inf,0.5,Inf),labels=c("No","Yes"))
test.prediction
summary(test.prediction)
##Confusion Matrix
table(test.prediction,test$subscribed)
## 2. C5.0 Decision Tree
library(C50)## Control for C5.0 Model
C5.0Control(subset=TRUE,bands=2,winnow=TRUE,noGlobalPruning=FALSE,CF=0.25,minCases=2,label="C5.0 Outcome")## Train Model using C5.0
train.c50=C5.0(train[,1:15],train$subscribed,trials=10,rules=FALSE,control=C5.0Control(),costs=NULL)
## Variable Importance Measure
C5imp(train.c50,metric="usage",pct=TRUE)## Lower % Usage variables can be dicarded and the Train function is re-run to get better accuracy
summary(train.c50) ## Check the percentage of accuracy of the model## Predict on the TEST data using trained Model
test.c50=predict(object=train.c50,newdata=test[,1:15],trials=1,type="class")
table(test.c50,test$subscribed) ## CONFUSION MATRIX from C5.0
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