The objective of this case is to get you understand logistic regression (binary classification) and some important ideas such as cross validation, ROC curve, cut-off probability.
We will use a Credit Card Default Data for this lab and illustration. The details of the data can be found at http://archive.ics.uci.edu/ml/datasets/default+of+credit+card+clients. Think about what kind of factors could affect people to fail to pay their credit balance.
We first load the credit scoring data. It is easy to load comma-separated values (CSV).
credit_data <- read.csv(file = "https://xiaoruizhu.github.io/Data-Mining-R/lecture/data/credit_default.csv", header=T)Look at what information do we have.
colnames(credit_data)## [1] "LIMIT_BAL" "SEX"
## [3] "EDUCATION" "MARRIAGE"
## [5] "AGE" "PAY_0"
## [7] "PAY_2" "PAY_3"
## [9] "PAY_4" "PAY_5"
## [11] "PAY_6" "BILL_AMT1"
## [13] "BILL_AMT2" "BILL_AMT3"
## [15] "BILL_AMT4" "BILL_AMT5"
## [17] "BILL_AMT6" "PAY_AMT1"
## [19] "PAY_AMT2" "PAY_AMT3"
## [21] "PAY_AMT4" "PAY_AMT5"
## [23] "PAY_AMT6" "default.payment.next.month"Let’s look at how many people were actually default in this sample.
mean(credit_data$default.payment.next.month)## [1] 0.2193333The name of response variable is too long! I want to make it shorter by renaming. Recall the rename() function.
library(dplyr)
credit_data<- rename(credit_data, default=default.payment.next.month)How about the variable type and summary statistics?
str(credit_data) # structure - see variable type
summary(credit_data) # summary statisticsWe see all variables are int, but we know that SEX, EDUCATION, MARRIAGE are categorical, we convert them to factor.
credit_data$SEX<- as.factor(credit_data$SEX)
credit_data$EDUCATION<- as.factor(credit_data$EDUCATION)
credit_data$MARRIAGE<- as.factor(credit_data$MARRIAGE)We omit other EDA, but you shouldn’t whenever you are doing data analysis.
Randomly split the data to training (80%) and testing (20%) datasets:
index <- sample(nrow(credit_data),nrow(credit_data)*0.80)
credit_train = credit_data[index,]
credit_test = credit_data[-index,]credit_glm0<- glm(default~., family=binomial, data=credit_train)
summary(credit_glm0)##
## Call:
## glm(formula = default ~ ., family = binomial, data = credit_train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.2204 -0.6959 -0.5311 -0.2787 3.3126
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.036e+00 1.549e-01 -6.686 2.29e-11 ***
## LIMIT_BAL -8.374e-07 2.841e-07 -2.947 0.003205 **
## SEX2 -1.410e-01 5.476e-02 -2.575 0.010012 *
## EDUCATION2 -1.576e-01 6.367e-02 -2.476 0.013302 *
## EDUCATION3 -2.250e-01 8.561e-02 -2.628 0.008581 **
## EDUCATION4 -1.216e+00 3.349e-01 -3.630 0.000283 ***
## MARRIAGE2 -1.313e-01 6.209e-02 -2.114 0.034521 *
## MARRIAGE3 -1.607e-01 2.348e-01 -0.684 0.493670
## AGE 7.730e-03 3.357e-03 2.303 0.021292 *
## PAY_0 5.973e-01 3.180e-02 18.783 < 2e-16 ***
## PAY_2 7.976e-02 3.649e-02 2.186 0.028808 *
## PAY_3 5.966e-02 4.021e-02 1.484 0.137892
## PAY_4 5.305e-02 4.489e-02 1.182 0.237349
## PAY_5 4.748e-02 4.831e-02 0.983 0.325690
## PAY_6 1.324e-02 4.006e-02 0.331 0.741000
## BILL_AMT1 -6.889e-06 2.034e-06 -3.388 0.000705 ***
## BILL_AMT2 4.778e-06 2.521e-06 1.895 0.058085 .
## BILL_AMT3 1.826e-06 2.190e-06 0.834 0.404334
## BILL_AMT4 -5.368e-06 2.548e-06 -2.107 0.035120 *
## BILL_AMT5 5.308e-06 2.796e-06 1.899 0.057592 .
## BILL_AMT6 -7.715e-07 2.123e-06 -0.363 0.716315
## PAY_AMT1 -1.097e-05 3.513e-06 -3.123 0.001789 **
## PAY_AMT2 -6.712e-06 3.215e-06 -2.088 0.036838 *
## PAY_AMT3 -3.556e-06 3.418e-06 -1.040 0.298139
## PAY_AMT4 -5.010e-06 3.070e-06 -1.632 0.102657
## PAY_AMT5 -2.095e-06 3.210e-06 -0.653 0.514046
## PAY_AMT6 -3.219e-06 2.152e-06 -1.496 0.134588
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 10071 on 9599 degrees of freedom
## Residual deviance: 8762 on 9573 degrees of freedom
## AIC: 8816
##
## Number of Fisher Scoring iterations: 5You have seen glm() before. In this lab, this is the main function used to build logistic regression model because it is a member of generalized linear model. In glm(), the only thing new is family. It specifies the distribution of your response variable. You may also specify the link function after the name of distribution, for example, family=binomial(logit) (default link is logit). You can also specify family=binomial(link = "probit") to run probit regression. You may also use glm() to build many other generalized linear models.
As we talked in the lecture, people may be more interested in the classification results. But we have to define a cut-off probability first.
These tables illustrate the impact of choosing different cut-off probability. Choosing a large cut-off probability will result in few cases being predicted as 1, and choosing a small cut-off probability will result in many cases being predicted as 1.
pred_glm0_train <- predict(credit_glm0, type="response")
table(credit_train$default, (pred_glm0_train > 0.9)*1, dnn=c("Truth","Predicted"))## Predicted
## Truth 0 1
## 0 7498 8
## 1 2080 14table(credit_train$default, (pred_glm0_train > 0.5)*1, dnn=c("Truth","Predicted"))## Predicted
## Truth 0 1
## 0 7299 207
## 1 1546 548table(credit_train$default, (pred_glm0_train > 0.2)*1, dnn=c("Truth","Predicted"))## Predicted
## Truth 0 1
## 0 4613 2893
## 1 615 1479table(credit_train$default, (pred_glm0_train > 0.0001)*1, dnn=c("Truth","Predicted"))## Predicted
## Truth 0 1
## 0 2 7504
## 1 0 2094Therefore, determine the optimal cut-off probability is crucial. The simplest way to determine the cut-off is to use the proportion of “1” in the original data. We will intriduce a more appropriate way to determine the optimal p-cut.
In the case of giving loan to someone, the cost function can indicate the trade off between the risk of giving loan to someone who cannot pay (predict 0, truth 1), and risk of rejecting someone who qualifies (predict 1, truth 0). Given different business situation, one may need to have asymmetric costs for false positive and false negative. Meanwhile, when you want a binary classification decision rule, you need to choose different cut-off probability. Choosing a large cut-off probability will result in few cases being predicted as 1, and choosing a small cut-off probability will result in many cases being predicted as 1.
#Symmetric cost
cost1 <- function(r, pi, pcut){
mean(((r==0)&(pi>pcut)) | ((r==1)&(pi<pcut)))
}
#Asymmetric cost
cost2 <- function(r, pi, pcut){
weight1 <- 5
weight0 <- 1
c1 <- (r==1)&(pi<pcut) #logical vector - true if actual 1 but predict 0
c0 <-(r==0)&(pi>pcut) #logical vector - true if actual 0 but predict 1
return(mean(weight1*c1+weight0*c0))
}pcut <- 1/(5+1)
#Symmetric cost
cost1(r = credit_train$default, pi = pred_glm0_train, pcut)## [1] 0.46375#Asymmetric cost
cost2(r = credit_train$default, pi = pred_glm0_train, pcut)## [1] 0.6704167pcuts <- costs1 <- costs2 <- matrix(c(0.9,0.5,0.2,0.0001), nrow = 4, ncol = 1)
costs1[] <- vapply(X = pcuts, FUN = cost1, FUN.VALUE = numeric(1),
r = credit_train$default, pi = pred_glm0_train)
costs2[] <- vapply(X = pcuts, FUN = cost2, FUN.VALUE = numeric(1),
r = credit_train$default, pi = pred_glm0_train)
All_costs <- as.matrix(cbind(costs1, costs2))
dimnames(All_costs) <- list(c(0.9,0.5,0.2,0.0001), c("Symm cost", "Asym cost"))
All_costs## Symm cost Asym cost
## 0.9 0.2175000 1.0841667
## 0.5 0.1826042 0.8267708
## 0.2 0.3654167 0.6216667
## 1e-04 0.7816667 0.7816667Know how to use glm() to build logistic regression;
Know how to get ROC and AUC based on predicted probability;
Know how to get PR curve and AUC based on predicted probability;
Know how to do binary classification, and calculation of MR, FPR, FNR, and cost;
Know how to use LASSO for logistic regression
EDA
Train logistic model
Prediction (ROC, AUC; PR, AUC)
Model comparison using AUC
Find optimal cut-off based on training sample
Classification – Obtain the confusion matrix and calculate the MR, asymmetric cost, FPR, Precision, and Recall.
Build new models by Variable Selection
Calculate all criteria
Comprehensive comparison