analytic_vidhdya_loan_prediction
Amit Kayal
December 1, 2017
Problem Statement
About Company: Dream Housing Finance company deals in all home loans. They have presence across all urban, semi urban and rural areas. Customer first apply for home loan after that company validates the customer eligibility for loan.
Problem: Company wants to automate the loan eligibility process (real time) based on customer detail provided while filling online application form. These details are Gender, Marital Status, Education, Number of Dependents, Income, Loan Amount, Credit History and others. To automate this process, they have given a problem to identify the customers segments, those are eligible for loan amount so that they can specifically target these customers. Here they have provided a partial data set. Probelm link is https://datahack.analyticsvidhya.com/contest/practice-problem-loan-prediction-iii/
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## hurdle and zeroinfl functions by Achim ZeileisWe will first load the data and do basic visualistaion along with descriptive stats Key Observations of variables areare: –ApplicantIncome and CoapplicantIncome variable seems to have outliers –LoanAmount variable has 22 Null Value –Loan_Amount_Term has 14 null values –Credit_History has 50 Null values Data set observation –Data set seems to be skewed towards lo “Loan Approval” value of Yes
my_loan_data <- read.csv(file = 'train_loan.csv')
str(my_loan_data)## 'data.frame': 614 obs. of 13 variables:
## $ Loan_ID : Factor w/ 614 levels "LP001002","LP001003",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ Gender : Factor w/ 3 levels "","Female","Male": 3 3 3 3 3 3 3 3 3 3 ...
## $ Married : Factor w/ 3 levels "","No","Yes": 2 3 3 3 2 3 3 3 3 3 ...
## $ Dependents : Factor w/ 5 levels "","0","1","2",..: 2 3 2 2 2 4 2 5 4 3 ...
## $ Education : Factor w/ 2 levels "Graduate","Not Graduate": 1 1 1 2 1 1 2 1 1 1 ...
## $ Self_Employed : Factor w/ 3 levels "","No","Yes": 2 2 3 2 2 3 2 2 2 2 ...
## $ ApplicantIncome : int 5849 4583 3000 2583 6000 5417 2333 3036 4006 12841 ...
## $ CoapplicantIncome: num 0 1508 0 2358 0 ...
## $ LoanAmount : int NA 128 66 120 141 267 95 158 168 349 ...
## $ Loan_Amount_Term : int 360 360 360 360 360 360 360 360 360 360 ...
## $ Credit_History : int 1 1 1 1 1 1 1 0 1 1 ...
## $ Property_Area : Factor w/ 3 levels "Rural","Semiurban",..: 3 1 3 3 3 3 3 2 3 2 ...
## $ Loan_Status : Factor w/ 2 levels "N","Y": 2 1 2 2 2 2 2 1 2 1 ...summary(my_loan_data)## Loan_ID Gender Married Dependents Education
## LP001002: 1 : 13 : 3 : 15 Graduate :480
## LP001003: 1 Female:112 No :213 0 :345 Not Graduate:134
## LP001005: 1 Male :489 Yes:398 1 :102
## LP001006: 1 2 :101
## LP001008: 1 3+: 51
## LP001011: 1
## (Other) :608
## Self_Employed ApplicantIncome CoapplicantIncome LoanAmount
## : 32 Min. : 150 Min. : 0 Min. : 9.0
## No :500 1st Qu.: 2878 1st Qu.: 0 1st Qu.:100.0
## Yes: 82 Median : 3812 Median : 1188 Median :128.0
## Mean : 5403 Mean : 1621 Mean :146.4
## 3rd Qu.: 5795 3rd Qu.: 2297 3rd Qu.:168.0
## Max. :81000 Max. :41667 Max. :700.0
## NA's :22
## Loan_Amount_Term Credit_History Property_Area Loan_Status
## Min. : 12 Min. :0.0000 Rural :179 N:192
## 1st Qu.:360 1st Qu.:1.0000 Semiurban:233 Y:422
## Median :360 Median :1.0000 Urban :202
## Mean :342 Mean :0.8422
## 3rd Qu.:360 3rd Qu.:1.0000
## Max. :480 Max. :1.0000
## NA's :14 NA's :50table(my_loan_data$Loan_Status)##
## N Y
## 192 422Data Visualistaion Observations are – ApplicantIncome has quite a lot outliers..Data is right skewed if we skip outliers – CoApplicantIncome has quite a lot outliers..Data is right skewed if we skip outliers - It may applicant and co-applicanta are for same location and further analysis will be done this
par(mfrow=c(1, 2)) ### This is added to show multiple plots in same window
boxplot(my_loan_data$ApplicantIncome, outline= TRUE, col = "blue")
boxplot(my_loan_data$ApplicantIncome, outline= FALSE, col = "blue")
#
par(mfrow=c(1, 2)) ### This is added to show multiple plots in same window
boxplot(my_loan_data$CoapplicantIncome, outline= TRUE, col = "blue")
boxplot(my_loan_data$CoapplicantIncome, outline= FALSE, col = "blue")
ggplot(data=my_loan_data, aes(x= my_loan_data$ApplicantIncome)) +
geom_histogram(col="red",fill="green", bins = 15) +
facet_grid(~my_loan_data$Loan_Status)+
theme_bw()
# par(mfrow=c(1, 3)) ### This is added to show multiple plots in same window
# # histogram(my_loan_data$Loan_Status ~ my_loan_data$Property_Area, type = c("count") )
# # histogram(my_loan_data$Loan_Status ~ my_loan_data$Credit_History, type = c("count"))
#par(mfrow=c(1, 3))
# histogram(my_loan_data$Loan_Status ~ my_loan_data$ApplicantIncome , type = c("count") )
# histogram(my_loan_data$Loan_Status ~ my_loan_data$CoapplicantIncome , type = c("count") )Histogram of Property Area shows that: Loan approval is more into Semiurban area than Rural and Urban. Urban area has lowest loan approval. Loan rejection is lowest in Rural area. Semiurban & Urban has same loan rejection.
ggplot(data=my_loan_data, aes(my_loan_data$Property_Area)) +
geom_histogram(col="red",fill="green",stat="count" ) +
facet_grid(~my_loan_data$Loan_Status)+
scale_x_discrete()## Warning: Ignoring unknown parameters: binwidth, bins, pad
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## ..- attr(*, "class")= chr [1:2] "element_text" "element"
## $ strip.text.x :List of 11
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## ..$ margin :Classes 'margin', 'unit' atomic [1:4] 5.5 0 5.5 0
## .. .. ..- attr(*, "valid.unit")= int 8
## .. .. ..- attr(*, "unit")= chr "pt"
## ..$ debug : NULL
## ..$ inherit.blank: logi TRUE
## ..- attr(*, "class")= chr [1:2] "element_text" "element"
## $ strip.text.y :List of 11
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## ..$ margin :Classes 'margin', 'unit' atomic [1:4] 0 5.5 0 5.5
## .. .. ..- attr(*, "valid.unit")= int 8
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## $ strip.switch.pad.grid:Class 'unit' atomic [1:1] 0.1
## .. ..- attr(*, "valid.unit")= int 1
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## $ strip.switch.pad.wrap:Class 'unit' atomic [1:1] 0.1
## .. ..- attr(*, "valid.unit")= int 1
## .. ..- attr(*, "unit")= chr "cm"
## - attr(*, "class")= chr [1:2] "theme" "gg"
## - attr(*, "complete")= logi TRUE
## - attr(*, "validate")= logi TRUEHistogram of Applicant Income based on Loan Approval..Histogram shows that low income peoples are mainly applying for loans and number of loan rejection is more in the lowest income segment
ggplot(data=my_loan_data, aes(x= my_loan_data$CoapplicantIncome)) +
geom_histogram(col="red",fill="green", bins = 15) +
facet_grid(~my_loan_data$Loan_Status)+
theme_bw()
Histogram of the Property Area based on loan aproval flag shows that –Loan Rejection rate for graduate is moe than non-graduate –Similarly loan approval rate for graduate is more than non graduate
ggplot(data=my_loan_data, aes(my_loan_data$Education)) +
geom_histogram(col="red",fill="green",stat="count" ) +
facet_grid(~my_loan_data$Loan_Status)+
##scale_x_discrete()
theme_bw()## Warning: Ignoring unknown parameters: binwidth, bins, pad
Histogram of Dependent variable with loan approval shows that –People having no dependents have maximum loan approval and rejection count
ggplot(data=my_loan_data, aes(my_loan_data$Dependents)) +
geom_histogram(col="red",fill="green",stat="count" ) +
facet_grid(~my_loan_data$Loan_Status)+
##scale_x_discrete()
theme_bw()## Warning: Ignoring unknown parameters: binwidth, bins, pad
Histogram based on Gender shows that Male applicant has higher loan approval and rejection count than feamle applicant.. So this looks to be an influencing factor
ggplot(data=my_loan_data, aes(my_loan_data$Gender)) +
geom_histogram(col="red",fill="green",stat="count" , na.rm = TRUE ) +
facet_grid(~my_loan_data$Loan_Status)+
##scale_x_discrete()
theme_bw()## Warning: Ignoring unknown parameters: binwidth, bins, pad
It is essential to identify if there is any corelation among variables…But before that we need to treat the null values and process them. Following are the variables which has NULL values… –LoanAmount –Loan_Amount_Term –Credit_History Following NA treatment says for all the values where value is NA, replace it with the right hand side. You need the na.rm=TRUE piece or else the median function will return NA
my_loan_data$LoanAmount[is.na(my_loan_data$LoanAmount)] <- median(my_loan_data$LoanAmount, na.rm=TRUE)
my_loan_data$Loan_Amount_Term[is.na(my_loan_data$Loan_Amount_Term)] <- median(my_loan_data$Loan_Amount_Term, na.rm=TRUE)
my_loan_data$Credit_History[is.na(my_loan_data$Credit_History)] <- median(my_loan_data$Credit_History, na.rm=TRUE)The data should be splitted into Train and Test data set to ensure we test the decision tree…We will do this through sampling and such smapling helps to remove biasness during split. The column which has indicator based on random variable should be removed from data frame after spliting to ensure that one does not impact the modelling
set.seed(123)
my_loan_data$spl = sample.split(my_loan_data$Loan_Status,SplitRatio=0.8)
my_loan_data_train <- subset(my_loan_data, spl == "TRUE")
nrow(my_loan_data_train)## [1] 492 ncol(my_loan_data_train)## [1] 14 my_loan_data__test <- subset(my_loan_data, spl == "FALSE")
my_loan_data <- my_loan_data[,-14] ## removing the boolean column which was set for sampling
# mybankdata_test <- mybankdata_test[,-22] ## removing the boolean column which was set for sampling
str(my_loan_data)## 'data.frame': 614 obs. of 13 variables:
## $ Loan_ID : Factor w/ 614 levels "LP001002","LP001003",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ Gender : Factor w/ 3 levels "","Female","Male": 3 3 3 3 3 3 3 3 3 3 ...
## $ Married : Factor w/ 3 levels "","No","Yes": 2 3 3 3 2 3 3 3 3 3 ...
## $ Dependents : Factor w/ 5 levels "","0","1","2",..: 2 3 2 2 2 4 2 5 4 3 ...
## $ Education : Factor w/ 2 levels "Graduate","Not Graduate": 1 1 1 2 1 1 2 1 1 1 ...
## $ Self_Employed : Factor w/ 3 levels "","No","Yes": 2 2 3 2 2 3 2 2 2 2 ...
## $ ApplicantIncome : int 5849 4583 3000 2583 6000 5417 2333 3036 4006 12841 ...
## $ CoapplicantIncome: num 0 1508 0 2358 0 ...
## $ LoanAmount : num 128 128 66 120 141 267 95 158 168 349 ...
## $ Loan_Amount_Term : num 360 360 360 360 360 360 360 360 360 360 ...
## $ Credit_History : num 1 1 1 1 1 1 1 0 1 1 ...
## $ Property_Area : Factor w/ 3 levels "Rural","Semiurban",..: 3 1 3 3 3 3 3 2 3 2 ...
## $ Loan_Status : Factor w/ 2 levels "N","Y": 2 1 2 2 2 2 2 1 2 1 ...Lets now handle categorical variables and convert them into Dummy Varible..We dont want them to do level encoding.. Here we are removing these two columns. First one is Loan Id which will not be required for analysis and Loan Status is our result column.. Then we are creating the dummy columns throug dummies package…
deleteinitially <- c("Loan_ID","Loan_Status") #
##store the loan status value
myloan_revised.loan_status <- my_loan_data_train$Loan_Status
myloan_revised <- my_loan_data_train[, !(names(my_loan_data_train) %in% deleteinitially)]
my_loan_data.conv <- dummy.data.frame(myloan_revised,sep = ".")
deleteafterdummy <- c("Gender.","Married.","Dependents.","Self_Employed.") #
my_loan_data.conv.revised <- my_loan_data.conv[, !(names(my_loan_data.conv) %in% deleteafterdummy)]
my_loan_data.conv.revised <- my_loan_data.conv.revised[,-21]
# ncol(my_loan_data.conv.revised)
# str(my_loan_data.conv.revised)
# my_loan_data.conv.revisedDoing the co-relation here to understand if there are significant co-relation existing. Key Observations from corelation plot are (based on thumb rule of 0.5) Credit History and Education Graduate are highly co-related Credit History and Education Non Graduate are highly co-related Education Graduate and Gender Male are highly corelated Education Graduate and Co-Applicant Income are highly corelated
####
my_loan_data.cormatrix <- cor(my_loan_data.conv.revised)
my_loan_data.cormatrix_rounded<- round(my_loan_data.cormatrix , digits=3)
print(my_loan_data.cormatrix)## Gender.Female Gender.Male Married.No
## Gender.Female 1.000000000 -0.943173399 0.3349658494
## Gender.Male -0.943173399 1.000000000 -0.3210614254
## Married.No 0.334965849 -0.321061425 1.0000000000
## Married.Yes -0.332520983 0.318512491 -0.9955810019
## Dependents.0 0.150231703 -0.143703907 0.3721553876
## Dependents.1 -0.021377612 0.027828347 -0.1453104302
## Dependents.2 -0.126784949 0.128886229 -0.2611541232
## Dependents.3+ -0.081644139 0.054718988 -0.1237135978
## Education.Graduate 0.043706004 -0.065846424 0.0139641576
## Education.Not Graduate -0.043706004 0.065846424 -0.0139641576
## Self_Employed.No -0.018329350 0.022304987 -0.0218301097
## Self_Employed.Yes -0.007374030 -0.006062936 -0.0006760746
## ApplicantIncome -0.051264190 0.007338743 -0.0594898478
## CoapplicantIncome -0.058511447 0.067662732 -0.0385648734
## LoanAmount -0.081773866 0.043111482 -0.1795146719
## Loan_Amount_Term 0.099899635 -0.090071386 0.0992571382
## Credit_History -0.003585075 0.014190689 -0.0544998425
## Property_Area.Rural -0.053317662 0.069582570 0.0259457706
## Property_Area.Semiurban 0.088298839 -0.102461210 -0.0271566237
## Property_Area.Urban -0.039011113 0.037834706 0.0027965779
## Married.Yes Dependents.0 Dependents.1
## Gender.Female -0.3325209830 0.15023170 -0.02137761
## Gender.Male 0.3185124906 -0.14370391 0.02782835
## Married.No -0.9955810019 0.37215539 -0.14531043
## Married.Yes 1.0000000000 -0.36681017 0.14705014
## Dependents.0 -0.3668101703 1.00000000 -0.52145958
## Dependents.1 0.1470501352 -0.52145958 1.00000000
## Dependents.2 0.2627053113 -0.51019087 -0.20143325
## Dependents.3+ 0.1248008566 -0.33720522 -0.13313516
## Education.Graduate -0.0161551581 0.01759206 0.02540320
## Education.Not Graduate 0.0161551581 -0.01759206 -0.02540320
## Self_Employed.No 0.0197960622 0.08726296 -0.06474867
## Self_Employed.Yes 0.0022853500 -0.11482834 0.08813216
## ApplicantIncome 0.0599126895 -0.08488897 0.03117646
## CoapplicantIncome 0.0407763689 0.01788963 -0.04624783
## LoanAmount 0.1786526432 -0.16198422 0.06878314
## Loan_Amount_Term -0.1079793828 0.08043214 -0.05745715
## Credit_History 0.0527181192 0.03863175 0.02769744
## Property_Area.Rural -0.0231851908 0.09966935 -0.07818158
## Property_Area.Semiurban 0.0216121493 -0.04370307 -0.01390199
## Property_Area.Urban 0.0002190612 -0.05131376 0.08957786
## Dependents.2 Dependents.3+ Education.Graduate
## Gender.Female -0.126784949 -0.081644139 0.04370600
## Gender.Male 0.128886229 0.054718988 -0.06584642
## Married.No -0.261154123 -0.123713598 0.01396416
## Married.Yes 0.262705311 0.124800857 -0.01615516
## Dependents.0 -0.510190871 -0.337205219 0.01759206
## Dependents.1 -0.201433250 -0.133135160 0.02540320
## Dependents.2 1.000000000 -0.130258119 -0.02291997
## Dependents.3+ -0.130258119 1.000000000 -0.04915348
## Education.Graduate -0.022919969 -0.049153476 1.00000000
## Education.Not Graduate 0.022919969 0.049153476 -1.00000000
## Self_Employed.No -0.030946147 -0.016849768 -0.01549181
## Self_Employed.Yes 0.029605684 0.024605363 -0.01148449
## ApplicantIncome -0.031485708 0.162552102 0.12880270
## CoapplicantIncome -0.011691628 0.058232409 0.07319029
## LoanAmount 0.021626649 0.180995856 0.18139768
## Loan_Amount_Term -0.007777065 -0.079575148 0.05760322
## Credit_History 0.021461890 -0.098159861 0.06106529
## Property_Area.Rural -0.032611437 -0.006856112 -0.04653699
## Property_Area.Semiurban 0.043906293 0.038820254 0.05188250
## Property_Area.Urban -0.013519958 -0.033134344 -0.00826504
## Education.Not Graduate Self_Employed.No
## Gender.Female -0.04370600 -0.018329350
## Gender.Male 0.06584642 0.022304987
## Married.No -0.01396416 -0.021830110
## Married.Yes 0.01615516 0.019796062
## Dependents.0 -0.01759206 0.087262956
## Dependents.1 -0.02540320 -0.064748668
## Dependents.2 0.02291997 -0.030946147
## Dependents.3+ 0.04915348 -0.016849768
## Education.Graduate -1.00000000 -0.015491805
## Education.Not Graduate 1.00000000 0.015491805
## Self_Employed.No 0.01549181 1.000000000
## Self_Employed.Yes 0.01148449 -0.802515425
## ApplicantIncome -0.12880270 -0.127420686
## CoapplicantIncome -0.07319029 -0.013037661
## LoanAmount -0.18139768 -0.093646803
## Loan_Amount_Term -0.05760322 0.068600264
## Credit_History -0.06106529 -0.085118344
## Property_Area.Rural 0.04653699 -0.030718683
## Property_Area.Semiurban -0.05188250 0.026928256
## Property_Area.Urban 0.00826504 0.002037006
## Self_Employed.Yes ApplicantIncome
## Gender.Female -0.0073740297 -0.051264190
## Gender.Male -0.0060629357 0.007338743
## Married.No -0.0006760746 -0.059489848
## Married.Yes 0.0022853500 0.059912689
## Dependents.0 -0.1148283374 -0.084888965
## Dependents.1 0.0881321619 0.031176462
## Dependents.2 0.0296056835 -0.031485708
## Dependents.3+ 0.0246053632 0.162552102
## Education.Graduate -0.0114844869 0.128802701
## Education.Not Graduate 0.0114844869 -0.128802701
## Self_Employed.No -0.8025154252 -0.127420686
## Self_Employed.Yes 1.0000000000 0.141235228
## ApplicantIncome 0.1412352276 1.000000000
## CoapplicantIncome -0.0474384782 -0.123893746
## LoanAmount 0.1145523029 0.560055535
## Loan_Amount_Term -0.0585719207 -0.074564744
## Credit_History 0.0475355718 -0.037700073
## Property_Area.Rural 0.0249513414 0.009560247
## Property_Area.Semiurban -0.0007726564 -0.015649636
## Property_Area.Urban -0.0232553384 0.006807643
## CoapplicantIncome LoanAmount Loan_Amount_Term
## Gender.Female -0.058511447 -0.081773866 0.099899635
## Gender.Male 0.067662732 0.043111482 -0.090071386
## Married.No -0.038564873 -0.179514672 0.099257138
## Married.Yes 0.040776369 0.178652643 -0.107979383
## Dependents.0 0.017889627 -0.161984218 0.080432144
## Dependents.1 -0.046247826 0.068783142 -0.057457148
## Dependents.2 -0.011691628 0.021626649 -0.007777065
## Dependents.3+ 0.058232409 0.180995856 -0.079575148
## Education.Graduate 0.073190292 0.181397679 0.057603224
## Education.Not Graduate -0.073190292 -0.181397679 -0.057603224
## Self_Employed.No -0.013037661 -0.093646803 0.068600264
## Self_Employed.Yes -0.047438478 0.114552303 -0.058571921
## ApplicantIncome -0.123893746 0.560055535 -0.074564744
## CoapplicantIncome 1.000000000 0.169505266 -0.063887828
## LoanAmount 0.169505266 1.000000000 0.004093816
## Loan_Amount_Term -0.063887828 0.004093816 1.000000000
## Credit_History 0.008421145 -0.026825985 -0.016728696
## Property_Area.Rural 0.001973750 0.045865215 0.044044368
## Property_Area.Semiurban -0.026754977 -0.008016293 0.074324501
## Property_Area.Urban 0.025487932 -0.035995061 -0.118535679
## Credit_History Property_Area.Rural
## Gender.Female -0.003585075 -0.053317662
## Gender.Male 0.014190689 0.069582570
## Married.No -0.054499842 0.025945771
## Married.Yes 0.052718119 -0.023185191
## Dependents.0 0.038631749 0.099669352
## Dependents.1 0.027697441 -0.078181575
## Dependents.2 0.021461890 -0.032611437
## Dependents.3+ -0.098159861 -0.006856112
## Education.Graduate 0.061065288 -0.046536991
## Education.Not Graduate -0.061065288 0.046536991
## Self_Employed.No -0.085118344 -0.030718683
## Self_Employed.Yes 0.047535572 0.024951341
## ApplicantIncome -0.037700073 0.009560247
## CoapplicantIncome 0.008421145 0.001973750
## LoanAmount -0.026825985 0.045865215
## Loan_Amount_Term -0.016728696 0.044044368
## Credit_History 1.000000000 0.002510184
## Property_Area.Rural 0.002510184 1.000000000
## Property_Area.Semiurban 0.008049117 -0.495037307
## Property_Area.Urban -0.010659327 -0.456940571
## Property_Area.Semiurban Property_Area.Urban
## Gender.Female 0.0882988393 -0.0390111126
## Gender.Male -0.1024612102 0.0378347062
## Married.No -0.0271566237 0.0027965779
## Married.Yes 0.0216121493 0.0002190612
## Dependents.0 -0.0437030725 -0.0513137575
## Dependents.1 -0.0139019881 0.0895778646
## Dependents.2 0.0439062934 -0.0135199583
## Dependents.3+ 0.0388202541 -0.0331343443
## Education.Graduate 0.0518824969 -0.0082650405
## Education.Not Graduate -0.0518824969 0.0082650405
## Self_Employed.No 0.0269282557 0.0020370063
## Self_Employed.Yes -0.0007726564 -0.0232553384
## ApplicantIncome -0.0156496365 0.0068076425
## CoapplicantIncome -0.0267549766 0.0254879323
## LoanAmount -0.0080162929 -0.0359950610
## Loan_Amount_Term 0.0743245005 -0.1185356787
## Credit_History 0.0080491170 -0.0106593271
## Property_Area.Rural -0.4950373067 -0.4569405708
## Property_Area.Semiurban 1.0000000000 -0.5466563546
## Property_Area.Urban -0.5466563546 1.0000000000 corrplot(my_loan_data.cormatrix, method="circle", type="full", addCoef.col = "blue", order ="AOE", bg ='grey')
## Try one more corelation way to show in better way
## Try one more ##This plot uses clustering to make it easy to see which variables are closely correlated with each other. ## More close each variable is to each other the higher the relationship while the opposite is true for widely spaced variables.The color of the line represents the direction of the correlation while the line shade and thickness represent the strength of the relationship. The min_cor parameter is the minimum correlation coefficient required to display a line between variables
# my_loan_data.conv %>% correlate() %>% network_plot(min_cor=0.3)
# ## Try another way
# ## Try another way
# ggcorrplot(cor(my_loan_data.conv), p.mat = cor_pmat(my_loan_data.conv), hc.order=TRUE, type='lower')Lets do the first round of logisttic regression…Note that here we did not manage the co-related variables and this will impact the accuracy.. Summary of logistic regression are Credit_History is most statistically significant variable where o value is lowest and suggest strong association of loan approval and credit history.. Positibe coefficient for this predictor suggests that all other variables being equal, loan likely to be approved. Remember that in the logit model the response variable is log odds: ln(odds) = ln(p/(1-p)) = ax1 + bx2 + . + z*xn. S So credit history reduces the log odds by 3.978e+00
Property_Area.Semiurba is 2nd most statistically significant variable
## we first need to
my_loan_data.conv.revised$Loan_Status <- myloan_revised.loan_status
my_loan_data.glm <- glm(formula = my_loan_data.conv.revised$Loan_Status ~., family=binomial(link='logit'), data=my_loan_data.conv.revised )
summary(my_loan_data.glm)##
## Call:
## glm(formula = my_loan_data.conv.revised$Loan_Status ~ ., family = binomial(link = "logit"),
## data = my_loan_data.conv.revised)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4847 -0.3751 0.5261 0.7392 2.5187
##
## Coefficients: (2 not defined because of singularities)
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 9.976e+00 5.354e+02 0.019 0.9851
## Gender.Female 8.689e-01 8.547e-01 1.017 0.3094
## Gender.Male 6.283e-01 8.141e-01 0.772 0.4402
## Married.No -1.299e+01 5.354e+02 -0.024 0.9806
## Married.Yes -1.247e+01 5.354e+02 -0.023 0.9814
## Dependents.0 5.746e-01 1.074e+00 0.535 0.5925
## Dependents.1 2.515e-01 1.090e+00 0.231 0.8176
## Dependents.2 9.797e-01 1.109e+00 0.883 0.3771
## `Dependents.3+` 3.815e-01 1.142e+00 0.334 0.7384
## Education.Graduate 3.253e-01 2.931e-01 1.110 0.2671
## `Education.Not Graduate` NA NA NA NA
## Self_Employed.No 2.125e-01 4.699e-01 0.452 0.6511
## Self_Employed.Yes 2.578e-01 5.684e-01 0.453 0.6502
## ApplicantIncome 4.216e-06 2.629e-05 0.160 0.8726
## CoapplicantIncome -3.568e-05 3.571e-05 -0.999 0.3177
## LoanAmount -1.262e-03 1.857e-03 -0.679 0.4969
## Loan_Amount_Term -4.049e-03 2.236e-03 -1.811 0.0701 .
## Credit_History 3.779e+00 4.623e-01 8.175 2.96e-16 ***
## Property_Area.Rural -2.535e-01 2.839e-01 -0.893 0.3718
## Property_Area.Semiurban 7.678e-01 3.049e-01 2.518 0.0118 *
## Property_Area.Urban NA NA NA NA
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 611.54 on 491 degrees of freedom
## Residual deviance: 452.46 on 473 degrees of freedom
## AIC: 490.46
##
## Number of Fisher Scoring iterations: 12 ##Now we can run the anova() function on the model to analyze the table of devianceAnalyzing the table we can see the drop in deviance when adding each variable one at a time. Again, adding Credit_History,Property_Area.Semiurban and Property_Area.Rural reduces the residual deviance. While no exact equivalent to the R2 of linear regression exists, the McFadden R2 index can be used to assess the model fit.
anova(my_loan_data.glm, test="Chisq")## Analysis of Deviance Table
##
## Model: binomial, link: logit
##
## Response: my_loan_data.conv.revised$Loan_Status
##
## Terms added sequentially (first to last)
##
##
## Df Deviance Resid. Df Resid. Dev Pr(>Chi)
## NULL 491 611.54
## Gender.Female 1 0.015 490 611.53 0.90347
## Gender.Male 1 0.684 489 610.84 0.40810
## Married.No 1 7.007 488 603.84 0.00812 **
## Married.Yes 1 0.648 487 603.19 0.42077
## Dependents.0 1 0.530 486 602.66 0.46643
## Dependents.1 1 0.063 485 602.60 0.80183
## Dependents.2 1 5.245 484 597.35 0.02200 *
## `Dependents.3+` 1 0.482 483 596.87 0.48752
## Education.Graduate 1 2.191 482 594.68 0.13883
## `Education.Not Graduate` 0 0.000 482 594.68
## Self_Employed.No 1 0.949 481 593.73 0.32985
## Self_Employed.Yes 1 0.105 480 593.62 0.74596
## ApplicantIncome 1 0.223 479 593.40 0.63686
## CoapplicantIncome 1 1.247 478 592.15 0.26409
## LoanAmount 1 1.162 477 590.99 0.28110
## Loan_Amount_Term 1 2.863 476 588.13 0.09062 .
## Credit_History 1 122.974 475 465.15 < 2e-16 ***
## Property_Area.Rural 1 6.169 474 458.98 0.01300 *
## Property_Area.Semiurban 1 6.528 473 452.46 0.01062 *
## Property_Area.Urban 0 0.000 473 452.46
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 pR2(my_loan_data.glm)## llh llhNull G2 McFadden r2ML
## -226.2280529 -305.7713151 159.0865243 0.2601397 0.2762770
## r2CU
## 0.3883160Now we wll reun the logistic with our selective variables and compare the MfFadden R2 value
## we first need to
my_loan_data.glmrev <- glm(formula = my_loan_data.conv.revised$Loan_Status ~ my_loan_data.conv.revised$Property_Area.Semiurban + my_loan_data.conv.revised$Credit_History + my_loan_data.conv.revised$Education.Graduate + my_loan_data.conv.revised$CoapplicantIncome, family=binomial(link='logit'), data=my_loan_data.conv.revised)
summary(my_loan_data.glmrev)##
## Call:
## glm(formula = my_loan_data.conv.revised$Loan_Status ~ my_loan_data.conv.revised$Property_Area.Semiurban +
## my_loan_data.conv.revised$Credit_History + my_loan_data.conv.revised$Education.Graduate +
## my_loan_data.conv.revised$CoapplicantIncome, family = binomial(link = "logit"),
## data = my_loan_data.conv.revised)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0669 -0.3710 0.5227 0.7646 2.3630
##
## Coefficients:
## Estimate Std. Error
## (Intercept) -2.876e+00 5.040e-01
## my_loan_data.conv.revised$Property_Area.Semiurban 8.833e-01 2.626e-01
## my_loan_data.conv.revised$Credit_History 3.768e+00 4.541e-01
## my_loan_data.conv.revised$Education.Graduate 2.342e-01 2.791e-01
## my_loan_data.conv.revised$CoapplicantIncome -3.534e-05 3.292e-05
## z value Pr(>|z|)
## (Intercept) -5.705 1.16e-08 ***
## my_loan_data.conv.revised$Property_Area.Semiurban 3.364 0.000769 ***
## my_loan_data.conv.revised$Credit_History 8.299 < 2e-16 ***
## my_loan_data.conv.revised$Education.Graduate 0.839 0.401475
## my_loan_data.conv.revised$CoapplicantIncome -1.074 0.282953
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 611.54 on 491 degrees of freedom
## Residual deviance: 466.73 on 487 degrees of freedom
## AIC: 476.73
##
## Number of Fisher Scoring iterations: 5 anova(my_loan_data.glmrev, test="Chisq")## Analysis of Deviance Table
##
## Model: binomial, link: logit
##
## Response: my_loan_data.conv.revised$Loan_Status
##
## Terms added sequentially (first to last)
##
##
## Df Deviance Resid. Df
## NULL 491
## my_loan_data.conv.revised$Property_Area.Semiurban 1 9.706 490
## my_loan_data.conv.revised$Credit_History 1 133.417 489
## my_loan_data.conv.revised$Education.Graduate 1 0.556 488
## my_loan_data.conv.revised$CoapplicantIncome 1 1.133 487
## Resid. Dev Pr(>Chi)
## NULL 611.54
## my_loan_data.conv.revised$Property_Area.Semiurban 601.84 0.001836 **
## my_loan_data.conv.revised$Credit_History 468.42 < 2.2e-16 ***
## my_loan_data.conv.revised$Education.Graduate 467.86 0.455907
## my_loan_data.conv.revised$CoapplicantIncome 466.73 0.287221
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 pR2(my_loan_data.glmrev)## llh llhNull G2 McFadden r2ML
## -233.3655783 -305.7713151 144.8114736 0.2367970 0.2549711
## r2CU
## 0.3583698 ##Now we can run the anova() function on the model to analyze the table of devianceWe will try to make thus model simple and will only take credit history as significant variable
my_loan_data.glmrevk <- glm(formula = my_loan_data.conv.revised$Loan_Status ~ my_loan_data.conv.revised$Credit_History , family=binomial(link='logit'), data=my_loan_data.conv.revised)
summary(my_loan_data.glmrevk)##
## Call:
## glm(formula = my_loan_data.conv.revised$Loan_Status ~ my_loan_data.conv.revised$Credit_History,
## family = binomial(link = "logit"), data = my_loan_data.conv.revised)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.7530 -0.4265 0.6960 0.6960 2.2101
##
## Coefficients:
## Estimate Std. Error z value
## (Intercept) -2.3514 0.4271 -5.505
## my_loan_data.conv.revised$Credit_History 3.6457 0.4432 8.226
## Pr(>|z|)
## (Intercept) 3.69e-08 ***
## my_loan_data.conv.revised$Credit_History < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 611.54 on 491 degrees of freedom
## Residual deviance: 481.26 on 490 degrees of freedom
## AIC: 485.26
##
## Number of Fisher Scoring iterations: 4 pR2(my_loan_data.glmrevk)## llh llhNull G2 McFadden r2ML
## -240.6307112 -305.7713151 130.2812077 0.2130370 0.2326400
## r2CU
## 0.3269827 myglm <- glm(formula = Loan_Status ~ Credit_History , family=binomial(link='logit'), data=my_loan_data_train)
summary(myglm)##
## Call:
## glm(formula = Loan_Status ~ Credit_History, family = binomial(link = "logit"),
## data = my_loan_data_train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.7530 -0.4265 0.6960 0.6960 2.2101
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.3514 0.4271 -5.505 3.69e-08 ***
## Credit_History 3.6457 0.4432 8.226 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 611.54 on 491 degrees of freedom
## Residual deviance: 481.26 on 490 degrees of freedom
## AIC: 485.26
##
## Number of Fisher Scoring iterations: 4 pR2(myglm)## llh llhNull G2 McFadden r2ML
## -240.6307112 -305.7713151 130.2812077 0.2130370 0.2326400
## r2CU
## 0.3269827In the steps above, we briefly evaluated the fitting of the model, now we would like to see how the model is doing when predicting y on a new set of data. Our model prediction is 83% which is excellent
nrow(my_loan_data__test)## [1] 122 nrow(my_loan_data_train)## [1] 492 my_loan_data__test <- my_loan_data__test[,-14]
# newdata=subset(my_loan_data__test, select = c("Credit_History"))
### clar categorical variable from train model
my_loan_data__test.loan_status <- my_loan_data__test$Loan_Status
my_loan_data__test_rev <- my_loan_data__test[, !(names(my_loan_data__test) %in% deleteinitially)]
my_loan_data__test.conv <- dummy.data.frame(my_loan_data__test_rev,sep = ".")
deleteafterdummy <- c("Gender.","Married.","Dependents.","Self_Employed.") #
my_loan_data__test.revised <- my_loan_data__test.conv[, !(names(my_loan_data__test.conv) %in% deleteafterdummy)]
my_loan_data__test.revised <- my_loan_data__test.revised[,-21]
# ncol(my_loan_data.conv.revised)
my_loan_data__test.revised$Loan_Status <- my_loan_data__test.loan_status
my_loan_data__test$Loan_Status_Predicted <- predict(myglm,newdata=my_loan_data__test, type = 'response')
my_loan_data__test$Loan_Status_Predicted <- ifelse(my_loan_data__test$Loan_Status_Predicted > 0.2,"Y","N")
myloan <- table(actualclass=my_loan_data__test.revised$Loan_Status, predictedclass=my_loan_data__test$Loan_Status_Predicted ) ## generting the table format
confusionMatrix(myloan) # generating the confusion matrix## Confusion Matrix and Statistics
##
## predictedclass
## actualclass N Y
## N 19 19
## Y 1 83
##
## Accuracy : 0.8361
## 95% CI : (0.7582, 0.8969)
## No Information Rate : 0.8361
## P-Value [Acc > NIR] : 0.5593301
##
## Kappa : 0.5608
## Mcnemar's Test P-Value : 0.0001439
##
## Sensitivity : 0.9500
## Specificity : 0.8137
## Pos Pred Value : 0.5000
## Neg Pred Value : 0.9881
## Prevalence : 0.1639
## Detection Rate : 0.1557
## Detection Prevalence : 0.3115
## Balanced Accuracy : 0.8819
##
## 'Positive' Class : N
##