在Statsmodels中指定常数线性回归? [英] Specifying a Constant in Statsmodels Linear Regression?

问题描述

我想使用statsmodels.regression.linear_model.OLS 包进行预测，但具有指定的常数.

I want to use the statsmodels.regression.linear_model.OLS package to do a prediction, but with a specified constant.

当前，我可以使用参数指定常量的存在:

Currently, I can specify the presence of a constant with an argument:

(来自文档: http://statsmodels.sourceforge. net/devel/generation/statsmodels.regression.linear_model.OLS.html )

class statsmodels.regression.linear_model.OLS(endog，exog = None，missing ='none'，hasconst = None)，其中 hasconst 是布尔值.

class statsmodels.regression.linear_model.OLS(endog, exog=None, missing='none', hasconst=None), where hasconst is a boolean.

我想做的是明确指定一个常数C，然后围绕它拟合线性回归模型.通过使用该OLS，我想生成一个，然后访问所有属性(如resid等).

What I want to do is specify explicitly a constant C, and then fit a linear regression model around it. From using that OLS, I want to generate a and then access all the attributes like resid, etc.

当前的次优解决方法是指定没有常量的OLS，从Y值中减去常量，并创建一个自定义对象，该对象每次都想包装指定的常量和没有常量的OLS进行预测或拟合，首先要从Y变量中减去常数，然后再使用预测.

A current suboptimal work around would be to specify the OLS without a constant, subtract the constant from the Y-values, and create a custom object that wraps both the specified constant and OLS w/o constant, every time I want to do predict or fit, to first subtract the constant from the Y variables, and then use the prediction.

谢谢！

推荐答案

如果将formula API用于statsmodels，则可以将其更简洁地指定为常量截距，作为Patsy设计矩阵规范的一部分.这仍然有点怪异-基本上只是表达您提出的解决方案的一种更简洁的方式-但至少它更短.例如:

If you use the formula API for statsmodels, you can specify a constant intercept more concisely as part of a Patsy design matrix specification. This is still a bit hacky--it's basically just a cleaner way of expressing your proposed solution--but at least it's shorter. E.g.:

>>> import statsmodels.formula.api as smf
>>> import pandas as pd
>>> import numpy as np
>>> c = 3.1416
>>> df = pd.DataFrame(np.random.rand(10, 2), columns=['x', 'y'])
>>> ols = smf.ols('y - c ~ 0 + x', data=df)
>>> result = ols.fit()
>>> print result.summary()
...
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
x              0.7404      0.230      3.220      0.010         0.220     1.261
==============================================================================

如您所见，截距没有系数，x的最佳斜率不是1.

As you can see, there's no coefficient for the intercept, and the best slope for x is not 1.

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