我想对二进制结果变量（0和1）进行逻辑回归。数据摘要如下所示：
Classes ‘data.table’ and 'data.frame': 1044 obs. of 16 variables:
$ age : int 18 17 15 15 16 16 16 17 15 15 ...
$ Medu : Factor w/ 5 levels "0","1","2","3",..: 5 2 2 5 4 5 3 5 4 4 ...
$ Fedu : Factor w/ 5 levels "0","1","2","3",..: 5 2 2 3 4 4 3 5 3 5 ...
$ got_parent : Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ...
$ traveltime : Factor w/ 4 levels "1","2","3","4": 2 1 1 1 1 1 1 2 1 1 ...
$ studytime : Factor w/ 4 levels "1","2","3","4": 2 2 2 3 2 2 2 2 2 2 ...
$ failures : int 0 0 3 0 0 0 0 0 0 0 ...
$ famrel : Factor w/ 5 levels "1","2","3","4",..: 4 5 4 3 4 5 4 4 4 5 ...
$ freetime : Factor w/ 5 levels "1","2","3","4",..: 3 3 3 2 3 4 4 1 2 5 ...
$ goout : Factor w/ 5 levels "1","2","3","4",..: 4 3 2 2 2 2 4 4 2 1 ...
$ Dalc : Factor w/ 5 levels "1","2","3","4",..: 1 1 2 1 1 1 1 1 1 1 ...
$ Walc : Factor w/ 5 levels "1","2","3","4",..: 1 1 3 1 2 2 1 1 1 1 ...
$ health : Factor w/ 5 levels "1","2","3","4",..: 3 3 3 5 5 5 3 1 1 5 ...
$ absences : int 6 4 10 2 4 10 0 6 0 0 ...
$ final_grade: int 6 6 10 15 10 15 11 6 19 15 ...
$ binge_drink: Factor w/ 2 levels "0","1": 1 1 2 1 1 1 1 1 1 1 ...
- attr(*, ".internal.selfref")=<externalptr>
其中binge_drink是我要预测的Y变量。我将数据分为70％的分割率的训练和测试集，但是当我按如下方式运行代码时，得到警告消息：
m1 <- glm(binge_drink ~., data = data_train, family = "binomial")
Warning messages:
1: glm.fit: algorithm did not converge
2: glm.fit: fitted probabilities numerically 0 or 1 occurred
摘要输出如下：
> summary(m1)
Call:
glm(formula = binge_drink ~ ., family = "binomial", data = data_train)
Deviance Residuals:
Min 1Q Median 3Q Max
-6.443e-06 -2.908e-06 -2.315e-06 2.110e-08 1.441e-05
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.975e+01 5.800e+05 0.000 1.000
age 2.793e-02 1.018e+04 0.000 1.000
Medu1 2.662e+00 5.230e+05 0.000 1.000
Medu2 2.049e+00 5.229e+05 0.000 1.000
Medu3 2.335e+00 5.235e+05 0.000 1.000
Medu4 2.500e+00 5.233e+05 0.000 1.000
Fedu1 2.100e-01 1.442e+05 0.000 1.000
Fedu2 4.332e-01 1.457e+05 0.000 1.000
Fedu3 4.615e-01 1.469e+05 0.000 1.000
Fedu4 3.043e-01 1.477e+05 0.000 1.000
got_parent1 -4.496e-01 4.591e+04 0.000 1.000
traveltime2 1.706e-01 2.492e+04 0.000 1.000
traveltime3 5.142e-01 4.216e+04 0.000 1.000
traveltime4 3.188e-01 7.544e+04 0.000 1.000
studytime2 -3.050e-02 2.572e+04 0.000 1.000
studytime3 2.261e-02 3.546e+04 0.000 1.000
studytime4 -4.446e-01 6.649e+04 0.000 1.000
failures -1.537e-03 1.925e+04 0.000 1.000
famrel2 5.388e-01 8.514e+04 0.000 1.000
famrel3 3.928e-01 6.791e+04 0.000 1.000
famrel4 1.007e-01 6.622e+04 0.000 1.000
famrel5 1.544e-01 6.772e+04 0.000 1.000
freetime2 3.437e-01 5.248e+04 0.000 1.000
freetime3 3.095e-01 4.963e+04 0.000 1.000
freetime4 -7.933e-02 5.250e+04 0.000 1.000
freetime5 -3.947e-01 6.455e+04 0.000 1.000
goout2 -7.329e-02 4.856e+04 0.000 1.000
goout3 -1.290e-01 4.948e+04 0.000 1.000
goout4 8.735e-02 5.056e+04 0.000 1.000
goout5 3.687e-01 5.613e+04 0.000 1.000
Dalc2 5.186e+01 3.300e+04 0.002 0.999
Dalc3 5.050e+01 5.089e+04 0.001 0.999
Dalc4 5.030e+01 8.331e+04 0.001 1.000
Dalc5 4.882e+01 8.648e+04 0.001 1.000
Walc2 2.194e-01 3.208e+04 0.000 1.000
Walc3 6.827e-01 3.267e+04 0.000 1.000
Walc4 5.085e+01 3.812e+04 0.001 0.999
Walc5 4.912e+01 6.719e+04 0.001 0.999
health2 2.159e-01 4.387e+04 0.000 1.000
health3 6.133e-02 4.077e+04 0.000 1.000
health4 -2.094e-01 4.418e+04 0.000 1.000
health5 2.766e-01 3.684e+04 0.000 1.000
absences -4.326e-03 1.951e+03 0.000 1.000
final_grade 1.050e-02 3.322e+03 0.000 1.000
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 9.4833e+02 on 729 degrees of freedom
Residual deviance: 7.4180e-09 on 686 degrees of freedom
AIC: 88
Number of Fisher Scoring iterations: 25
我尝试为分类预测变量创建虚拟变量，但输出相似。有人知道为什么吗？
谢谢。