百分比回归-r，python和matlab中的结果不同

Question

I have percentages and need to calculate a regression. According to basic statistics using logistic regression is better than OLS as percentages invalidate the requirement of a continuous and unconstraint value space.

So far, so good. However, I get different results in R, Python, and Matlab. In fact, Matlab even reports significant values where python will not.

My models look like:

R:
summary(glm(foo ~ 1 + bar + baz  , family = "binomial", data = <<data>>))

Python via statsmodels:
smf.logit('foo ~ 1 + bar + baz', <<data>>).fit().summary()

Matlab:
fitglm(<<data>>,'foo ~ 1 + bar + baz','Link','logit')

where Matlab currently produces the best results.

Could there be different initialization values? Different solvers? Different settings for alphas when computing p-values? How can I get the same results at least in similar numeric ranges or same features detected as significant? I do not require exact equal numeric output.

edit

the summary statistics

python:
Dep. Variable:  foo No. Observations:   104
Model:  Logit   Df Residuals:   98
Method: MLE Df Model:   5
Date:   Wed, 28 Aug 2019    Pseudo R-squ.:  inf
Time:   06:48:12    Log-Likelihood: -0.25057
converged:  True    LL-Null:    0.0000
LLR p-value:    1.000
coef    std err z   P>|z|   [0.025  0.975]
Intercept   -16.9863    154.602 -0.110  0.913   -320.001    286.028
bar -0.0278 0.945   -0.029  0.977   -1.880  1.824
baz 18.5550 280.722 0.066   0.947   -531.650    568.760
a   9.9996  153.668 0.065   0.948   -291.184    311.183
b   0.6757  132.542 0.005   0.996   -259.102    260.454
d   0.0005  0.039   0.011   0.991   -0.076  0.077


R:
glm(formula = myformula, family = "binomial", data = r_x)

Deviance Residuals: 
      Min         1Q     Median         3Q        Max  
-0.046466  -0.013282  -0.001017   0.006217   0.104467  

Coefficients:
                                       Estimate Std. Error z value Pr(>|z|)
(Intercept)                          -1.699e+01  1.546e+02  -0.110    0.913
bar                     -2.777e-02  9.449e-01  -0.029    0.977
baz                               1.855e+01  2.807e+02   0.066    0.947
a                       1.000e+01  1.537e+02   0.065    0.948
b                       6.757e-01  1.325e+02   0.005    0.996
d  4.507e-04  3.921e-02   0.011    0.991

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 0.049633  on 103  degrees of freedom
Residual deviance: 0.035684  on  98  degrees of freedom
AIC: 12.486

Matlab:
Estimated Coefficients:
                                            Estimate         SE         tStat        pValue  
                                            _________    __________    ________    __________

    (Intercept)                               -21.044         3.315     -6.3483    6.8027e-09
    bar                        -0.033507      0.022165     -1.5117       0.13383
    d    0.0016149    0.00083173      1.9416      0.055053
    baz                                    21.427        6.0132      3.5632    0.00056774
    a                            14.875        3.7828      3.9322    0.00015712
    b                           -1.2126        2.7535    -0.44038       0.66063


104 observations, 98 error degrees of freedom
Estimated Dispersion: 1.25e-06
F-statistic vs. constant model: 7.4, p-value = 6.37e-06

Answer 1

You are not actually using the binomial distribution in the MATLAB case. You are specifying the link function, but the distribution remains its default value for a normal distribution, which will not give you the expected logistic fit, at least if the sample sizes for the percentages are small. It is also giving you lower p-values, because the normal distribution is less constrained in its variance than the binomial distribution is.

You need to specify the Distribution argument to Binomial :

fitglm(<<data>>, 'foo ~ 1 + bar + baz', 'Distribution', 'binomial ', 'Link', 'logit')

The R and Python code seem to match rather well.