我试图使用一个自变量运行逻辑回归，使模型适合数据，然后返回带有随机样本输入的概率输出。
In [153]: df[['Diff1', 'Win']]
Out[153]:
Diff1 Win
0 100 1
1 110 1
2 20 0
3 80 1
4 200 1
5 25 0
In [154]: logit = sm.Logit(df['Win'], df['Diff1'])
In [155]: result=logit.fit()
Optimization terminated successfully.
Current function value: 0.451400
Iterations 6
Logit Regression Results
==============================================================================
Dep. Variable: Win No. Observations: 8
Model: Logit Df Residuals: 7
Method: MLE Df Model: 0
Date: Fri, 11 Dec 2015 Pseudo R-squ.: 0.3177
Time: 13:49:07 Log-Likelihood: -3.6112
converged: True LL-Null: -5.2925
LLR p-value: nan
==============================================================================
coef std err z P>|z| [95.0% Conf. Int.]
------------------------------------------------------------------------------
Diff1 0.0207 0.014 1.435 0.151 -0.008 0.049
==============================================================================
In [158]: result.predict(0)
Out[158]: array([ 0.5])
我显然错误地使用了预测函数，因为在这种情况下输入0不应该产生0.5。该结果用于逻辑模型的非拟合示例。
我会使用简单的OLS回归，但希望我的模型受（0,1）限制。