85 R 银行信用卡风控评分数据分析

1 初始环境准备

读取数据与预览

rm(list=ls())
#setwd("./case")
#install.packages("xlsx")
library(openxlsx)
dat<-read.xlsx("credit.xlsx",1)
View(dat)

2数据预览与处理

数据预览，发现最大值999的异常值，偏离平均值

class(dat)
#describe data
summary(dat)
sum(is.na(dat))

异常值处理，使用na填充

#Outlier filling
dat[,1:6]<-sapply(dat[,1:6],function(x) {x[x==999]<-NA;return(x)} )
nrow(dat)
ncol(dat)
summary(dat[,11])
#Understand data deletion invalid variables
dat<-dat[,-11]

将字符型变量character转化为因子变量factor
#Change string variable type to classification variable

dat1<-dat
sapply(dat1,class)
ch=names(which(sapply(dat1,is.character)))#find the character type variance
dat1[,ch]=as.data.frame(lapply(dat1[,ch], as.factor))

观察数据发现家庭人口数和家庭孩子数密切相关，存在多重共线性，于是产生标识变量代替这二个变量

dat1[,4]<-dat1[,4]-dat1[,3]

table(dat1[,4])
dat1[,4]<-factor(dat1[,4],levels=c(1,2),labels=c("其他","已婚"))

colnames(dat1)<-c("age","income","child","marital","dur_live",
                  "dur_work","housetype","nation","cardtype","loan")
summary(dat1)

3 描述性统计

相关包准备

#install.packages("smbinning")
#install.packages("prettyR")
library(smbinning)
library(prettyR)

library(mvtnorm)
library(kernlab)

盖帽法

异常值可以使用盖帽法处理，使用1%和99%分位数替换异常值

##盖帽法函数 去除异常用99%和1%点分别代替异常值

block<-function(x,lower=T,upper=T){
  if(lower){
    q1<-quantile(x,0.01)
    x[x<=q1]<-q1
  }
  if(upper){
    q99<-quantile(x,0.99)
    x[x>q99]<-q99
  }
  return(x)
}

数据集中1是不违约，0是违约，进行反转设定，使1变为违约，0为不违约

#Odds ratio conversion for later IV calculation
dat1$loan<-as.numeric(!as.logical(dat1$loan))

描述数据分类统计

违约和不违约的人群的区别

#data classification ,discretization of continuous variables
##age
boxplot(age~loan,data=dat1,horizontal=T, frame=F, 
        col="lightgray",main="Distribution")
age<-smbinning(dat1,"loan","age")
age$ivtable

违约与否的年龄箱线图

分箱后的IV图

age<-smbinning(dat1,"loan","age")
age$ivtable

对年龄进行分箱后，查看百分比，weight，good，bad rate，后面的描述性统计大致如此，可以秦楚看出不同年龄层次的区别

par(mfrow=c(2,2))
smbinning.plot(age,option="dist",sub="年龄")
smbinning.plot(age,option="WoE",sub="年龄")
smbinning.plot(age,option="goodrate",sub="年龄")
smbinning.plot(age,option="badrate",sub="年龄"

将IV结果添加到一个向量中

par(mfrow=c(1,1))
age$iv
#Add Iv value to vector
cred_iv<-c("年龄"=age$iv)

关于收入，明显存在异常值，使用盖帽法

##income

boxplot(income~loan,data=dat1,horizontal=T, frame=F, 
        col="lightgray",main="Distribution")
   

盖帽法填充

dat1$income<-block(dat1$income)

填充后明显变正常了

boxplot(income~loan,data=dat1,horizontal=T, frame=F, 
        col="lightgray",main="Distribution")
        ```

IV值测量，同上age

income<-smbinning(dat1,"loan","income")
income$ivtable
smbinning.plot(income,option="WoE",sub="收入")
income$iv
cred_iv<-c(cred_iv,"收入"=income$iv)

child 统计

##child
boxplot(child~loan,data=dat1,horizontal=T, frame=F, 
        col="lightgray",main="Distribution")


child<-smbinning(dat1,"loan","child")
child$ivtable
smbinning.plot(child,option="WoE",sub="孩子数量")
child$iv
cred_iv<-c(cred_iv,"孩子数量"=child$iv)

##marital
xtab(~marital+loan,data=dat1,chisq=T)
marital<-smbinning.factor(dat1,"loan","marital")
marital$ivtable
smbinning.plot(marital,option="WoE",sub="婚姻状态")
marital$iv
cred_iv<-c(cred_iv,"婚姻状态"=marital$iv)

##dur_live
boxplot(dur_live~loan,data=dat1,horizontal=T, 
        frame=F, col="lightgray",main="Distribution")
t.test(dur_live~loan,data=dat1)
dur_live<-smbinning(dat1,"loan","dur_live")
dur_live

观察得到dur_live变量对违约分布区别不大，使用t检验，不能拒绝二者同分布

dur_work变量统计

##dur_work
boxplot(dur_work~loan,data=dat1,horizontal=T, 
        frame=F, col="lightgray",main="Distribution")
t.test(dur_work~loan,data=dat1)
dur_work<-smbinning(dat1,"loan","dur_work")
dur_work$ivtable
smbinning.plot(dur_work,option="WoE",sub="在现工作时间")
dur_work$iv
cred_iv<-c(cred_iv,"在现工作时间"=dur_work$iv)

housetype描述统计

##housetype
xtab(~housetype+loan,data=dat1,chisq=T)
housetype<-smbinning.factor(dat1,"loan","housetype")
housetype$ivtable
smbinning.plot(housetype,option="WoE",sub="住房类型")
housetype$iv
cred_iv<-c(cred_iv,"住房种类"=housetype$iv)

##nation
xtab(~nation+loan,data=dat1,chisq=T)
nation<-smbinning.factor(dat1,"loan","nation")
nation$ivtable
smbinning.plot(nation,option="WoE",sub="国籍")
nation$iv
cred_iv<-c(cred_iv,"国籍"=nation$iv)

cardtype描述统计

##cardtype
xtab(~cardtype+loan,data=dat1,chisq=T)
cardtype<-smbinning.factor(dat1,"loan","cardtype")
cardtype$ivtable
smbinning.plot(cardtype,option="WoE",sub="信用卡类型")
cardtype$iv
cred_iv<-c(cred_iv,"信用卡类型"=cardtype$iv)

总体变量IV值程度

#Drawing shows the amount of information
barplot(cred_iv,main="各变量信息值")

4 属性分箱

#quantity after adding classification
dat2<-dat1
dat2<-smbinning.gen(dat2,age,"glage")
dat2<-smbinning.gen(dat2,income,"glincome")
dat2<-smbinning.gen(dat2,child,"glchild")
dat2<-smbinning.factor.gen(dat2,marital,"glmarital")
dat2<-smbinning.gen(dat2,dur_work,"gldur_work")
dat2<-smbinning.factor.gen(dat2,housetype,"glhousetype")
dat2<-smbinning.factor.gen(dat2,nation,"glnation")
dat2<-smbinning.factor.gen(dat2,cardtype,"glcardtype")
View(dat2)

dat3<-dat2[,c(11:18,10)]

View(dat3)

生成分箱后的数据级，将用户属性转化为区间数据

5 logistic建模

具体打分理论参考链接: link.

模型生成

#Creat logistic regression
cred_mod<-glm(loan~. ,data=dat3,family=binomial())
summary(cred_mod)

6 打分系统

依照打分公式发现，信用最高评分和最低评分分别为797，362分

#Scoring card system
cre_scal<-smbinning.scaling(cred_mod,pdo=45,score=800,odds=50)
cre_scal$logitscaled
cre_scal$minmaxscore

违约过与否的箱线图展示

#Score each item
dat4<-smbinning.scoring.gen(smbscaled=cre_scal, dataset=dat3)
boxplot(Score~loan,data=dat4,horizontal=T, frame=F, 
        col="lightgray",main="Distribution")

生成打分指标

#Standardized scoring table
scaledcard<-cre_scal$logitscaled[[1]][-1,c(1,2,6)]
scaledcard[,1]<-c(rep("年龄",5),rep("收入",3),rep("孩子数量",2),
                  rep("婚否",2),rep("在现工作时间",5),
                  rep("住房类型",3),rep("国籍",8),rep("信用卡类型",7))
scaledcard

7 写入csv文件

ncol(dat4)
dat5=dat4[,10:18]

#write the results
write.table(scaledcard,"card.csv",row.names = F)
write.table(dat4,"card.csv",row.names = F,append = T)
?write.csv

输出文件给业务人员