Question

我正在尝试在python中生成逻辑回归，该回归产生与R相同的结果。这似乎很接近，但并不相同。我整理了以下示例，以说明存在差异。数据不真实。

R

# RStudio 1.1.453

d <- data.frame(c(0, 0, 1, 1, 1),
                c(1, 0, 0, 0, 0),
                c(0, 1, 0, 0, 0))

colnames(d) <- c("v1", "v2", "v3")


model <- glm(v1 ~ v2,
         data = d,
         family = "binomial")


summary(model)

R输出

Call:
glm(formula = v1 ~ v2, family = "binomial", data = d)

Deviance Residuals: 
       1         2         3         4         5  
-1.66511  -0.00013   0.75853   0.75853   0.75853  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)
(Intercept)    1.099      1.155   0.951    0.341
v2           -19.665   6522.639  -0.003    0.998

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 6.7301  on 4  degrees of freedom
Residual deviance: 4.4987  on 3  degrees of freedom
AIC: 8.4987

Number of Fisher Scoring iterations: 17

Python

# Python 3.7.1

import pandas as pd # 0.23.4
import statsmodels.api as sm # 0.9.0
import statsmodels.formula.api as smf # 0.9.0

d = pd.DataFrame({"v1" : [0, 0, 1, 1, 1],
                  "v2" : [1, 0, 0, 0, 0],
                  "v3" : [0, 1, 0, 0, 0]})

model = smf.glm(formula = "v1 ~ v2",
               family=sm.families.Binomial(link = sm.genmod.families.links.logit),
               data=d
               ).fit()

model.summary()

Python输出

                 Generalized Linear Model Regression Results                  
==============================================================================
Dep. Variable:                     v1   No. Observations:                    5
Model:                            GLM   Df Residuals:                        3
Model Family:                Binomial   Df Model:                            1
Link Function:                  logit   Scale:                          1.0000
Method:                          IRLS   Log-Likelihood:                -2.2493
Date:                Wed, 07 Nov 2018   Deviance:                       4.4987
Time:                        15:17:52   Pearson chi2:                     4.00
No. Iterations:                    19   Covariance Type:             nonrobust
==============================================================================
                 coef    std err          z      P>|z|      [0.025      0.975]
------------------------------------------------------------------------------
Intercept      1.0986      1.155      0.951      0.341      -1.165       3.362
v2           -21.6647   1.77e+04     -0.001      0.999   -3.48e+04    3.47e+04
==============================================================================

迭代次数有所不同。据我所知，有一些收敛方法可能在两者之间有所不同，但我不明白。我可能还会缺少其他设置吗？

Answer 1

猜测它们在数值稳定性方面有不同的权衡。

v2估计值的方差很大，这可能使他们俩都陷入困境……我要说的是，他们基本上给出了相同的答案，至少在双精度算术可用的极限范围内。 >

R实现允许您传递control参数：

> options(digits=12)
> model <- glm(v1 ~ v2, data=d, family="binomial", control=list(trace=T))
Deviance = 4.67724333758 Iterations - 1
Deviance = 4.5570420311 Iterations - 2
Deviance = 4.51971688994 Iterations - 3
Deviance = 4.50636401333 Iterations - 4
Deviance = 4.50150009179 Iterations - 5
Deviance = 4.49971718523 Iterations - 6
Deviance = 4.49906215541 Iterations - 7
Deviance = 4.49882130019 Iterations - 8
Deviance = 4.4987327103 Iterations - 9
Deviance = 4.49870012203 Iterations - 10
Deviance = 4.49868813377 Iterations - 11
Deviance = 4.49868372357 Iterations - 12
Deviance = 4.49868210116 Iterations - 13
Deviance = 4.4986815043 Iterations - 14
Deviance = 4.49868128473 Iterations - 15
Deviance = 4.49868120396 Iterations - 16
Deviance = 4.49868117424 Iterations - 17

显示了它的收敛性，但是我在Python代码中找不到类似的东西。

看到上面的输出表明他们也可以使用不同的临界值来确定收敛； R使用epsilon = 1e-8

GLM R与Python

1 个答案: