无法使用 statsmodels.api 计算 y 截距以进行多元线性回归

Question

我的自变量数据集如下：

>>> reg_data_pd
             a         b         c
0     0.794527  0.033651  0.352414
1     0.794914  0.001086  0.093222
2     0.794476  0.004711  0.027977
3     0.776916  0.035780  0.023156
4     0.773526  0.002273  0.035269
5     0.797933  0.001838  0.131261
6     0.806997  0.011498  0.180022
7     0.780709  0.000766  0.522399
8     0.779954  0.001397  0.036386
9     0.756837  0.010448  0.035893
10    0.775064  0.029471  0.036798
11    0.787213  0.013467  0.081323
12    0.757511  0.016465  0.021611
13    0.794530  0.004141  0.157539
14    0.783696  0.019909  0.021765
15    0.793892  0.003597  0.063312
16    0.762702  0.003547  0.052479
17    0.780336  0.004958  0.084910
18    0.787005  0.006372  0.048153
19    0.824416  0.000513  0.045102
20    0.790552  0.009652  0.581571
21    0.773064  0.000889  0.263941
22    0.772039  0.021499  0.260455
23    0.780298  0.022814  0.061621
24    0.794924  0.020585  0.020638
25    0.772452  0.085798  0.215673
26    0.784202  0.000013  0.233638
27    0.822010  0.082684  0.028724
28    0.772587  0.027979  0.118953
29    0.765530  0.006655  0.018605
...        ...       ...       ...
4771  0.968364  0.227303  0.153739
4772  0.968401  0.159052  0.132388
4773  0.959733  0.278948  0.132163
4774  0.957354  0.315088  0.136973
4775  0.954627  0.447764  0.139494
4776  0.952442  0.305559  0.206204
4777  0.948925  0.235244  0.116273
4778  0.953192  0.228221  0.247231
4779  0.952769  0.327529  0.229617
4780  0.954471  0.396722  0.210942
4781  0.955292  0.336075  0.179493
4782  0.950516  0.320840  0.289505
4783  0.950454  0.316647  0.200065
4784  0.947313  0.291446  0.155215
4785  0.945677  0.292084  0.585302
4786  0.951083  0.285946  0.536361
4787  0.943909  0.346754  0.457234
4788  0.941971  0.276125  0.207159
4789  0.945111  0.440802  0.222561
4790  0.951011  0.407192  0.167613
4791  0.925485  0.464954  0.237568
4792  0.926332  0.252929  0.190035
4793  0.931606  0.020075  0.179730
4794  0.929963  0.426511  0.134418
4795  0.941986  0.640994  0.123444
4796  0.943526  0.232498  0.139800
4797  0.945268  0.460201  0.106471
4798  0.953572  0.398044  0.151489
4799  0.947673  0.479376  0.174330
4800  0.952663  0.532027  0.409197

[4801 rows x 3 columns]

而因变量的数据集是：

>>> yu_pd
             y
0     0.290740
1     0.295920
2     0.295920
3     0.192100
4     0.266000
5     0.284700
6     0.284700
7     0.272300
8     0.282680
9     0.243260
10    0.243260
11    0.273150
12    0.273150
13    0.282850
14    0.300325
15    0.192525
16    0.192525
17    0.269620
18    0.286825
19    0.207700
20    0.207700
21    0.292380
22    0.292380
23    0.282600
24    0.278212
25    0.243512
26    0.243512
27    0.309025
28    0.361740
29    0.249520
...        ...
4771  0.251480
4772  0.287500
4773  0.287500
4774  0.282071
4775  0.313343
4776  0.287463
4777  0.287463
4778  0.298700
4779  0.272920
4780  0.272920
4781  0.371314
4782  0.388429
4783  0.305200
4784  0.305200
4785  0.296725
4786  0.287920
4787  0.271580
4788  0.305486
4789  0.318571
4790  0.337975
4791  0.337975
4792  0.319988
4793  0.192360
4794  0.312871
4795  0.323000
4796  0.347088
4797  0.347088
4798  0.324986
4799  0.184320
4800  0.352100

[4801 rows x 1 columns]

我计算多线回归的代码如下：

>>> import statsmodels.api as sm
>>> model = sm.OLS(yu_pd,reg_data_pd)
>>> results = model.fit()
>>> results.summary()
<class 'statsmodels.iolib.summary.Summary'>
"""
                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      y   R-squared:                       0.896
Model:                            OLS   Adj. R-squared:                  0.896
Method:                 Least Squares   F-statistic:                 1.379e+04
Date:                Thu, 28 Jan 2016   Prob (F-statistic):               0.00
Time:                        16:45:03   Log-Likelihood:                 6693.6
No. Observations:                4801   AIC:                        -1.338e+04
Df Residuals:                    4798   BIC:                        -1.336e+04
Df Model:                           3                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
a              0.1933      0.002     78.058      0.000         0.188     0.198
b              0.0135      0.005      2.796      0.005         0.004     0.023
c             -0.0221      0.006     -3.984      0.000        -0.033    -0.011
==============================================================================
Omnibus:                      151.028   Durbin-Watson:                   0.452
Prob(Omnibus):                  0.000   Jarque-Bera (JB):              166.568
Skew:                           0.430   Prob(JB):                     6.77e-37
Kurtosis:                       3.306   Cond. No.                         6.75
==============================================================================

Warnings:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
"""

我得到了所有系数“a”、“b”和“c”，但我没有得到 Y 截距的值。

Answer 1

如果您有一个包含 3 个自变量的回归模型，为什么您希望数据摘要中的 Y 截距有一个唯一定义？我认为这是一个适用于单个自变量的单个函数的概念。

考虑以下几点：

import numpy as np
from numpy import random
import statsmodels.api as sm

x = np.linspace(0,1,100)
X = 0.45*x
X1 = 0.45*x
x3 = np.zeros(200).reshape((100,2))
x3[:,0] = X[:]
x3[:,1] = X1[:]

y = 0.45*x

model = sm.OLS(y, x3)
results = model.fit()
print(results.summary())
                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      y   R-squared:                         1.000
Model:                            OLS   Adj. R-squared:                  1.000
Method:                 Least Squares   F-statistic:                 5.710e+33
Date:                Thu, 28 Jan 2016   Prob (F-statistic):               0.00
Time:                        15:11:53   Log-Likelihood:                 3649.3
No. Observations:                 100   AIC:                            -7297.
Df Residuals:                      99   BIC:                            -7294.
Df Model:                           1                                         
Covariance Type:            nonrobust                                         
             coef    std err          t      P>|t|      [95.0% Conf. Int.]
x1             0.5000   6.62e-18   7.56e+16      0.000         0.500     0.500
x2             0.5000   6.62e-18   7.56e+16      0.000         0.500     0.500

现在这告诉我我的系数确实是两条线的斜率。因此，在您的情况下，值 a、b 和 c 是自变量输入的三行的斜率，与 Y 变量输出相匹配。

如果这些线中的任何一条具有 Y 截距，例如 Y = a0 + a*x，其中 x 是自变量，您可以通过将 x 和 Y 设置为从 deta 获得的值来求解斜率；例如，如果点x = Y = 1出现在同一行，则设置x = Y = 1，得到Y-int：a0 = 1-a。你可以对你适合的其他两条线做同样的事情。我相信这现在是您问题的完整答案。