R语言逻辑回归深入讲解

逻辑回归

> ###############逻辑回归
> setwd("/users/yaozhilin/downloads/r_edu/data")
> accepts<-read.csv("accepts.csv")
> names(accepts)
 [1] "application_id" "account_number" "bad_ind"    "vehicle_year"  "vehicle_make" 
 [6] "bankruptcy_ind" "tot_derog"   "tot_tr"     "age_oldest_tr" "tot_open_tr"  
[11] "tot_rev_tr"   "tot_rev_debt"  "tot_rev_line"  "rev_util"    "fico_score"  
[16] "purch_price"  "msrp"      "down_pyt"    "loan_term"   "loan_amt"   
[21] "ltv"      "tot_income"   "veh_mileage"  "used_ind"   
> accepts<-accepts[complete.cases(accepts),]
> select<-sample(1:nrow(accepts),length(accepts$application_id)*0.7)
> train<-accepts[select,]###70%用于建模
> test<-accepts[-select,]###30%用于检测
> attach(train)
> ###用glm(y~x,family=binomial(link="logit"))
> gl<-glm(bad_ind~fico_score,family=binomial(link = "logit"))
> summary(gl)

call:
glm(formula = bad_ind ~ fico_score, family = binomial(link = "logit"))

deviance residuals: 
  min    1q  median    3q   max 
-2.0794 -0.6790 -0.4937 -0.3073  2.6028 

coefficients:
       estimate std. error z value pr(>|z|)  
(intercept) 9.049667  0.629120  14.38  <2e-16 ***
fico_score -0.015407  0.000938 -16.43  <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: 2989.2 on 3046 degrees of freedom
residual deviance: 2665.9 on 3045 degrees of freedom
aic: 2669.9

number of fisher scoring iterations: 5

多元逻辑回归

> ###多元逻辑回归
> gls<-glm(bad_ind~fico_score+bankruptcy_ind+age_oldest_tr+
+      tot_derog+rev_util+veh_mileage,family = binomial(link = "logit"))
> summary(gls)

call:
glm(formula = bad_ind ~ fico_score + bankruptcy_ind + age_oldest_tr + 
  tot_derog + rev_util + veh_mileage, family = binomial(link = "logit"))

deviance residuals: 
  min    1q  median    3q   max 
-2.2646 -0.6743 -0.4647 -0.2630  2.8177 

coefficients:
         estimate std. error z value pr(>|z|)  
(intercept)   8.205e+00 7.433e-01 11.039 < 2e-16 ***
fico_score   -1.338e-02 1.092e-03 -12.260 < 2e-16 ***
bankruptcy_indy -3.771e-01 1.855e-01 -2.033  0.0421 * 
age_oldest_tr  -4.458e-03 6.375e-04 -6.994 2.68e-12 ***
tot_derog    3.012e-02 1.552e-02  1.941  0.0523 . 
rev_util     3.763e-04 5.252e-04  0.717  0.4737  
veh_mileage   2.466e-06 1.381e-06  1.786  0.0741 . 
---
signif. codes: 0 ‘***' 0.001 ‘**' 0.01 ‘*' 0.05 ‘.' 0.1 ‘ ' 1

(dispersion parameter for binomial family taken to be 1)

  null deviance: 2989.2 on 3046 degrees of freedom
residual deviance: 2601.4 on 3040 degrees of freedom
aic: 2615.4

number of fisher scoring iterations: 5

> glss<-step(gls,direction = "both")
start: aic=2615.35
bad_ind ~ fico_score + bankruptcy_ind + age_oldest_tr + tot_derog + 
  rev_util + veh_mileage

         df deviance  aic
- rev_util    1  2601.9 2613.9
<none>        2601.3 2615.3
- veh_mileage   1  2604.4 2616.4
- tot_derog    1  2605.1 2617.1
- bankruptcy_ind 1  2605.7 2617.7
- age_oldest_tr  1  2655.9 2667.9
- fico_score   1  2763.8 2775.8

step: aic=2613.88
bad_ind ~ fico_score + bankruptcy_ind + age_oldest_tr + tot_derog + 
  veh_mileage

         df deviance  aic
<none>        2601.9 2613.9
- veh_mileage   1  2604.9 2614.9
+ rev_util    1  2601.3 2615.3
- tot_derog    1  2605.7 2615.7
- bankruptcy_ind 1  2606.1 2616.1
- age_oldest_tr  1  2656.9 2666.9
- fico_score   1  2773.2 2783.2

> #出来的数据是logit，我们需要转换
> train$pre<-predict(glss,train)
> #出来的数据是logit，我们需要转换
> train$pre<-predict(glss,train)
> summary(train$pre)
  min. 1st qu. median  mean 3rd qu.  max. 
 -4.868 -2.421 -1.671 -1.713 -1.011  2.497 
> train$pre_p<-1/(1+exp(-1*train$pre))
> summary(train$pre_p)
  min. 1st qu. median  mean 3rd qu.  max. 
0.00763 0.08157 0.15823 0.19298 0.26677 0.92395

 #逻辑回归不需要检测扰动项，但需要检测共线性
 > library(car)
 > vif(glss)
 > fico_score bankruptcy_ind age_oldest_tr   tot_derog  veh_mileage 
 >1.271283    1.144846    1.075603    1.423850    1.003616 

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