Assignment – Part 3 (Case: Auto Finance)
Sameer Mathur
Reading & Explaining the Data
Please download the data using the link
Please review the code below and execute it in your own RMD file.
Reading the Data
# reading the data
df <- read.csv("AutoFinanaceData.csv")
# attach the data frame
attach(df)
# Number of rows and columns
dim(df)## [1] 28906 21Printing the Column Names of the Data
## [1] "Agmt.No" "ContractStatus" "StartDate" "AGE"
## [5] "NOOFDEPE" "MTHINCTH" "SALDATFR" "TENORYR"
## [9] "DWNPMFR" "PROFBUS" "QUALHSC" "QUAL_PG"
## [13] "SEXCODE" "FULLPDC" "FRICODE" "WASHCODE"
## [17] "Region" "Branch" "DefaulterFlag" "DefaulterType"
## [21] "DATASET"List of Data Columns
DEFAULT
1. Defaulter Flag
1: Customer has delayed paying at least once
0: Otherwise
DEMOGRAPHIC VARIABLES
1. Gender
SEXCODE = 1 (Male)
SEXCODE = 0 (Female)
2. Age
3. Education
QUALHSC
QUAL_PG
4. Income
Monthly Income in Thousands (MTHINCTH)
Owns a Fridge (FRICODE)
Owns a Washing Machine (WASHCODE)
5. Profession
- PROFBUS = 1 (BUSINESS)
- PROFBUS = 0 (PROFESSIONAL)
6. No.of Dependents
- NOOFDEPE
7. Region
Structure of the Dataset
## 'data.frame': 28906 obs. of 21 variables:
## $ Agmt.No : chr "AP18100057" "AP18100140" "AP18100198" "AP18100217" ...
## $ ContractStatus: chr "Closed" "Closed" "Closed" "Closed" ...
## $ StartDate : chr "19-01-01" "10-05-01" "05-08-01" "03-09-01" ...
## $ AGE : int 26 28 32 31 36 33 41 47 43 27 ...
## $ NOOFDEPE : int 2 2 2 0 2 2 2 0 0 0 ...
## $ MTHINCTH : num 4.5 5.59 8.8 5 12 ...
## $ SALDATFR : num 1 1 1 1 1 1 1 1 0.97 1 ...
## $ TENORYR : num 1.5 2 1 1 1 2 1 2 1.5 2 ...
## $ DWNPMFR : num 0.27 0.25 0.51 0.66 0.17 0.18 0.37 0.42 0.27 0.47 ...
## $ PROFBUS : int 0 0 0 0 0 0 0 0 0 0 ...
## $ QUALHSC : int 0 0 0 0 0 0 1 0 0 0 ...
## $ QUAL_PG : int 0 0 0 0 0 0 0 0 0 0 ...
## $ SEXCODE : int 1 1 1 1 1 1 1 1 1 1 ...
## $ FULLPDC : int 1 1 1 1 1 0 0 1 1 1 ...
## $ FRICODE : int 0 1 1 1 1 0 0 0 0 0 ...
## $ WASHCODE : int 0 0 1 1 0 0 0 0 0 0 ...
## $ Region : chr "AP2" "AP2" "AP2" "AP2" ...
## $ Branch : chr "Vizag" "Vizag" "Vizag" "Vizag" ...
## $ DefaulterFlag : int 0 0 0 0 0 0 0 0 0 0 ...
## $ DefaulterType : int 0 0 0 0 0 0 0 0 0 0 ...
## $ DATASET : chr " " "BUILD" "BUILD" "BUILD" ...Convert catgorical variables to factor
## 'data.frame': 28906 obs. of 21 variables:
## $ Agmt.No : chr "AP18100057" "AP18100140" "AP18100198" "AP18100217" ...
## $ ContractStatus: chr "Closed" "Closed" "Closed" "Closed" ...
## $ StartDate : chr "19-01-01" "10-05-01" "05-08-01" "03-09-01" ...
## $ AGE : int 26 28 32 31 36 33 41 47 43 27 ...
## $ NOOFDEPE : int 2 2 2 0 2 2 2 0 0 0 ...
## $ MTHINCTH : num 4.5 5.59 8.8 5 12 ...
## $ SALDATFR : num 1 1 1 1 1 1 1 1 0.97 1 ...
## $ TENORYR : num 1.5 2 1 1 1 2 1 2 1.5 2 ...
## $ DWNPMFR : num 0.27 0.25 0.51 0.66 0.17 0.18 0.37 0.42 0.27 0.47 ...
## $ PROFBUS : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
## $ QUALHSC : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 2 1 1 1 ...
## $ QUAL_PG : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
## $ SEXCODE : Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
## $ FULLPDC : Factor w/ 2 levels "0","1": 2 2 2 2 2 1 1 2 2 2 ...
## $ FRICODE : Factor w/ 2 levels "0","1": 1 2 2 2 2 1 1 1 1 1 ...
## $ WASHCODE : Factor w/ 2 levels "0","1": 1 1 2 2 1 1 1 1 1 1 ...
## $ Region : Factor w/ 8 levels "AP1","AP2","Chennai",..: 2 2 2 2 2 2 2 2 2 2 ...
## $ Branch : Factor w/ 14 levels "Bangalore","Chennai",..: 14 14 14 14 14 14 14 14 14 14 ...
## $ DefaulterFlag : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
## $ DefaulterType : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
## $ DATASET : chr " " "BUILD" "BUILD" "BUILD" ...Defaulters based on whether they gave post-dated cheques FULLPDC
attach(df)
# Defaulters based on whether they gave post-dated cheques
round(prop.table(table(DefaulterFlag, FULLPDC), 2)*100, 2)[, 1:2]## FULLPDC
## DefaulterFlag 0 1
## 0 18.17 45.42
## 1 81.83 54.58As you can see above, among the people who are not paying post dated cheques, 81.8% defaulted.
Defaulters based on TENORYR
library(data.table)
# average TENORYR of the borrower by Defaulter Flag
df <- data.table(df)
df[, .(AvgTENORYR = round(mean(TENORYR), 2)),
by=list(DefaulterFlag)]## DefaulterFlag AvgTENORYR
## 1: 0 1.15
## 2: 1 1.34As you can see above, the defaulters have greater average tenor.
Creating Train and Test dataset
Reserve 80% for training and 20% of test
## Warning: package 'caTools' was built under R version 4.0.4Section 1: Logistic Regression
Model1
We Ran Logistics Regression Model using glm() on the training data, the output is as follows.
Model1 <- glm(DefaulterFlag ~ AGE
+ NOOFDEPE
+ MTHINCTH
+ NOOFDEPE
+ SALDATFR
+ TENORYR
+ DWNPMFR
+ PROFBUS
+ QUALHSC
+ QUAL_PG
+ SEXCODE
+ FULLPDC
+ FRICODE
+ WASHCODE
+ Region,data = trainingSet, family = binomial())
summary(Model1)##
## Call:
## glm(formula = DefaulterFlag ~ AGE + NOOFDEPE + MTHINCTH + NOOFDEPE +
## SALDATFR + TENORYR + DWNPMFR + PROFBUS + QUALHSC + QUAL_PG +
## SEXCODE + FULLPDC + FRICODE + WASHCODE + Region, family = binomial(),
## data = trainingSet)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.7470 -1.0215 0.5716 0.7801 2.0874
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 2.2291732 0.1890284 11.793 < 2e-16 ***
## AGE -0.0144968 0.0016680 -8.691 < 2e-16 ***
## NOOFDEPE 0.0566860 0.0107923 5.252 1.50e-07 ***
## MTHINCTH -0.0004025 0.0035613 -0.113 0.910023
## SALDATFR -0.3833870 0.0420223 -9.123 < 2e-16 ***
## TENORYR 0.7727065 0.0456475 16.928 < 2e-16 ***
## DWNPMFR -1.3074501 0.1274734 -10.257 < 2e-16 ***
## PROFBUS1 0.1966576 0.0487903 4.031 5.56e-05 ***
## QUALHSC1 0.1853120 0.0401652 4.614 3.95e-06 ***
## QUAL_PG1 -0.2990904 0.0787907 -3.796 0.000147 ***
## SEXCODE1 0.2339445 0.0600322 3.897 9.74e-05 ***
## FULLPDC1 -1.2365885 0.0368674 -33.541 < 2e-16 ***
## FRICODE1 -0.1761473 0.0377247 -4.669 3.02e-06 ***
## WASHCODE1 -0.2644245 0.0476814 -5.546 2.93e-08 ***
## RegionAP2 -0.5788864 0.1796029 -3.223 0.001268 **
## RegionChennai -1.4136987 0.1408192 -10.039 < 2e-16 ***
## RegionKA1 -0.6529787 0.1411987 -4.625 3.75e-06 ***
## RegionKE2 -0.5753874 0.1450753 -3.966 7.30e-05 ***
## RegionTN1 -0.8084619 0.1362745 -5.933 2.98e-09 ***
## RegionTN2 -0.6142186 0.1458691 -4.211 2.55e-05 ***
## RegionVellore -0.6570233 0.1595604 -4.118 3.83e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 26040 on 21678 degrees of freedom
## Residual deviance: 23025 on 21658 degrees of freedom
## AIC: 23067
##
## Number of Fisher Scoring iterations: 4Section 2: Logistic Regression with Interaction
We want to compare the impact of loan tenor on the probability of load default, for consumers who pay using post dated cheques and remaining consumers who do not pay using post dated cheques, for this purpose we ran a model that gives the following output.
Model2
Que 1- Write R code for building the Logistics Regression Model with Interaction using glm() on the training data, the expected output should be as follows.
Model2 <- glm(DefaulterFlag ~ AGE
+ NOOFDEPE
+ MTHINCTH
+ NOOFDEPE
+ SALDATFR
+ TENORYR
+ DWNPMFR
+ PROFBUS
+ QUALHSC
+ QUAL_PG
+ SEXCODE
+ FULLPDC
+ FRICODE
+ WASHCODE
+ Region
+ TENORYR*FULLPDC
,data = trainingSet, family = binomial())
summary(Model2)##
## Call:
## glm(formula = DefaulterFlag ~ AGE + NOOFDEPE + MTHINCTH + NOOFDEPE +
## SALDATFR + TENORYR + DWNPMFR + PROFBUS + QUALHSC + QUAL_PG +
## SEXCODE + FULLPDC + FRICODE + WASHCODE + Region + TENORYR *
## FULLPDC, family = binomial(), data = trainingSet)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.5410 -1.0043 0.5721 0.7620 2.1539
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 2.4938627 0.1945079 12.821 < 2e-16 ***
## AGE -0.0144308 0.0016723 -8.629 < 2e-16 ***
## NOOFDEPE 0.0569949 0.0108145 5.270 1.36e-07 ***
## MTHINCTH 0.0006478 0.0035839 0.181 0.856559
## SALDATFR -0.3900738 0.0422177 -9.240 < 2e-16 ***
## TENORYR 0.5723502 0.0557250 10.271 < 2e-16 ***
## DWNPMFR -1.3175535 0.1275771 -10.328 < 2e-16 ***
## PROFBUS1 0.2100458 0.0491541 4.273 1.93e-05 ***
## QUALHSC1 0.1835106 0.0402820 4.556 5.22e-06 ***
## QUAL_PG1 -0.2972520 0.0789155 -3.767 0.000165 ***
## SEXCODE1 0.2352736 0.0600159 3.920 8.85e-05 ***
## FULLPDC1 -1.7550203 0.0952821 -18.419 < 2e-16 ***
## FRICODE1 -0.1719774 0.0378111 -4.548 5.41e-06 ***
## WASHCODE1 -0.2585977 0.0477647 -5.414 6.16e-08 ***
## RegionAP2 -0.5817435 0.1797634 -3.236 0.001211 **
## RegionChennai -1.4436197 0.1414897 -10.203 < 2e-16 ***
## RegionKA1 -0.6604584 0.1417572 -4.659 3.18e-06 ***
## RegionKE2 -0.5973660 0.1458215 -4.097 4.19e-05 ***
## RegionTN1 -0.8355176 0.1369630 -6.100 1.06e-09 ***
## RegionTN2 -0.6171896 0.1464390 -4.215 2.50e-05 ***
## RegionVellore -0.7037441 0.1604099 -4.387 1.15e-05 ***
## TENORYR:FULLPDC1 0.4227544 0.0714356 5.918 3.26e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 26040 on 21678 degrees of freedom
## Residual deviance: 22990 on 21657 degrees of freedom
## AIC: 23034
##
## Number of Fisher Scoring iterations: 4Que 2- Compare the model1 & Model2 based on fit, which model is better and why?
Consider a consumer John, having the following characteristics:
AGE = mean(AGE),
Male,
Education = UG,
MTHINCTH = mean(MTHINCTH)
NoOfDepe = mean(NOOFDEPE),
Owns a Fridge,
Owns a Washing Machine
Working Professional,
SALDATFR = mean(SALDATFR),
Lives in TN1
Down Payment = mean(DWNPMFR) %
Que 3- Using Model2, Calculate probability of John defaulting on the auto finance loan if he does not pay using post dated cheque, when the loan tenor is equal to 1 year?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = 1,
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "0",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred1 <- predict(Model2, Xdata, type = "response")
pred1## 1
## 0.7331136Que 4- Using Model2, Calculate probability of John defaulting on the auto finance loan if he does not pay using post dated cheque, when the loan tenor is equal to 2 years?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = 2,
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "0",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred2 <- predict(Model2, Xdata, type = "response")
pred2## 1
## 0.8296046Que 5- Using Model2, Calculate the change in probability of John defaulting on the auto finance loan if he does not pay using post dated cheque, when the loan tenor increases from 1 year to 2 years duration?
## 1
## 0.09649103Que 6- Using Model2, Calculate probability of John defaulting on the auto finance loan if he is paying using post dated cheque, when the loan tenor is equal to 1 year?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = 1,
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "1",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred3 <- predict(Model2, Xdata, type = "response")
pred3## 1
## 0.42024Que 7- Using Model2, Calculate probability of John defaulting on the auto finance loan if he is paying using post dated cheque, when the loan tenor is equal to 2 years?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = 2,
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "1",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred4 <- predict(Model2, Xdata, type = "response")
pred4## 1
## 0.6622454Que 8- Using Model2, Calculate the change in probability of John defaulting on the auto finance loan if he is paying using post dated cheque, when the loan tenor increases from 1 year to 2 years duration?
## 1
## 0.2420054Que 9- What problem would you face if you try to solve Que 5 & Que 8 using Model1 (without interaction), please explain in detail.
Que 10- Using Model1, Write R code to predict the probabilities of default of the consumers in the testing data
Que 11- Using Model1, Write R Code to generate the Confusion Matrix, shown below, assuming threshold probability of 50%
## Loading required package: lattice## Loading required package: ggplot2# confusion matrix using caret package
yPred <- ifelse(predProbTestLogit > 0.5, "Yes", "No")
predY <- as.factor(yPred)
levels(testSet$DefaulterFlag) <- c("No", "Yes")
confusionMatrix(data = predY, reference = testSet$DefaulterFlag, positive = "Yes")## Confusion Matrix and Statistics
##
## Reference
## Prediction No Yes
## No 635 417
## Yes 1448 4727
##
## Accuracy : 0.7419
## 95% CI : (0.7317, 0.752)
## No Information Rate : 0.7118
## P-Value [Acc > NIR] : 5.682e-09
##
## Kappa : 0.2624
##
## Mcnemar's Test P-Value : < 2.2e-16
##
## Sensitivity : 0.9189
## Specificity : 0.3048
## Pos Pred Value : 0.7655
## Neg Pred Value : 0.6036
## Prevalence : 0.7118
## Detection Rate : 0.6541
## Detection Prevalence : 0.8544
## Balanced Accuracy : 0.6119
##
## 'Positive' Class : Yes
## Que 12- Using Model1, Write R code to draw this ROC Plot based on testing data
## Warning: package 'ROCR' was built under R version 4.0.4# predicted probabilities
predProbTest <- predict(Model1, testSet, type = "response")
lgPredObj <- prediction(predProbTest,testSet$DefaulterFlag)
lgPerfObjlogit <- performance(lgPredObj, "tpr","fpr")
plot(lgPerfObjlogit,main = "ROC Curve",col = 2,lwd = 2)
abline(a = 0,b = 1,lwd = 2,lty = 3,col = "black")
Que 13- Using Model1, Write R code to to calculate area under curve (AUC)
# auc for logit
aucLogit <- performance(lgPredObj, measure = "auc")
aucLogit <- aucLogit@y.values[[1]]
aucLogit## [1] 0.7263962Section 3: Decision Tree
Que 14- Write R code to model decision tree using gini on training dataset, also write R code to visulise the decision tree as shown below.
set.seed(2345)
dTree <- train(DefaulterFlag ~ AGE
+ NOOFDEPE
+ MTHINCTH
+ SALDATFR
+ TENORYR
+ DWNPMFR
+ PROFBUS
+ QUALHSC
+ QUAL_PG
+ SEXCODE
+ FULLPDC
+ FRICODE
+ WASHCODE
+ Region,
data = trainingSet,
method = "rpart",
parms = list(split = "gini"),
trControl = trainControl(method = "cv"))
dTree## CART
##
## 21679 samples
## 14 predictor
## 2 classes: '0', '1'
##
## No pre-processing
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 19511, 19511, 19512, 19511, 19511, 19511, ...
## Resampling results across tuning parameters:
##
## cp Accuracy Kappa
## 0.01024164 0.7292763 0.2129099
## 0.01152184 0.7269239 0.2189725
## 0.01408225 0.7186214 0.1326211
##
## Accuracy was used to select the optimal model using the largest value.
## The final value used for the model was cp = 0.01024164.## Loading required package: rpart
Recall the the profile of John,
AGE = mean(AGE),
Male,
Education = UG,
MTHINCTH = mean(MTHINCTH)
NoOfDepe = mean(NOOFDEPE),
Owns a Fridge,
Owns a Washing Machine
Working Professional,
SALDATFR = mean(SALDATFR),
Lives in TN1
Down Payment = mean(DWNPMFR) %
TENORYR = mean(TENORYR)
Que 15- On the basis of John’s profile, would he default or not, assuming that he does not pay using post dated cheque?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = mean(TENORYR),
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "0",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred1 <- predict(dTree, Xdata, type = "raw")
pred1## [1] 1
## Levels: 0 1Que 16- On the basis of John’s profile, would he default or not, assuming that he payes using post dated cheque?
Xdata <- data.frame(
AGE = mean(AGE),
NOOFDEPE = mean(NOOFDEPE),
SALDATFR = mean(SALDATFR),
TENORYR = mean(TENORYR),
DWNPMFR = mean(DWNPMFR),
PROFBUS = "0",
QUALHSC = "0",
QUAL_PG = "0",
SEXCODE = "1",
FULLPDC = "1",
FRICODE = "1",
WASHCODE = "1",
Region = "TN1",
MTHINCTH = mean(MTHINCTH))
pred1 <- predict(dTree, Xdata, type = "raw")
pred1## [1] 1
## Levels: 0 1Que 17- Using Decision Tree, Write R code to predict the probabilities of default of the consumers in the testing data.
Que 18- Using Decision Tree, Write R Code to generate the Confusion Matrix, shown below, assuming threshold probability of 50%.
library(caret)
# confusion matrix using caret package
yPred <- ifelse(predProbTestTree[2] > 0.5, "Yes", "No")
predY <- as.factor(yPred)
levels(testSet$DefaulterFlag) <- c("No", "Yes")
confusionMatrix(data = predY, reference = testSet$DefaulterFlag, positive = "Yes")## Confusion Matrix and Statistics
##
## Reference
## Prediction No Yes
## No 434 297
## Yes 1649 4847
##
## Accuracy : 0.7307
## 95% CI : (0.7203, 0.7409)
## No Information Rate : 0.7118
## P-Value [Acc > NIR] : 0.0001805
##
## Kappa : 0.1867
##
## Mcnemar's Test P-Value : < 2.2e-16
##
## Sensitivity : 0.9423
## Specificity : 0.2084
## Pos Pred Value : 0.7462
## Neg Pred Value : 0.5937
## Prevalence : 0.7118
## Detection Rate : 0.6707
## Detection Prevalence : 0.8989
## Balanced Accuracy : 0.5753
##
## 'Positive' Class : Yes
## Que 19- Which Machine Leraning technique (Decision Tree of Logistic Regression) is better based on sensitivity? Explain your reasoning
Que 20- Which Machine Leraning technique (Decision Tree of Logistic Regression) is better based on Specificity? Explain your reasoning
Que 21- Write R code to draw ROC curves for Logistic Regression and Decision Tree on same graph using test dataset.
library(ROCR)
# predicted probabilities
predProbTest <- predict(dTree, testSet, type = "prob")
lgPredObj <- prediction(predProbTest[2],testSet$DefaulterFlag)
lgPerfObjTree <- performance(lgPredObj, "tpr","fpr")
# List of predictions
predList <- list(predProbTestLogit, predProbTestTree[2])
# List of actual values (same for all)
m <- length(predList)
# ROC curves (logit and tree)
plot(lgPerfObjlogit, col = "red", lwd = 2,
main = "ROC Curves \n (Logit and Tree)")
plot(lgPerfObjTree, add = TRUE, col = "blue", lwd = 3,
main = "ROC Curve for CC Default \n (Logit and Tree)")
legend(x = "bottomright",
legend = c("Tree", "Logit"),
fill = 1:m)
Que 22- Using Decision Tree, Write R code to to calculate area under curve (AUC)
# auc for logit
aucTree <- performance(lgPredObj, measure = "auc")
aucTree <- aucTree@y.values[[1]]
aucTree## [1] 0.6829464