python - sklearn 和 statsmodels OLS 之间摘要的区别

Question

目标是检测并修复为什么我的 sklearn“摘要”实现之间的报告与 OLS statsmodels 的结果不匹配。唯一匹配的是 beta 系数。

import pandas as pd
import numpy as np
from statsmodels.regression.linear_model import OLS
from sklearn import linear_model
from scipy.stats import t


class LinearRegression(linear_model.LinearRegression):
    """
    LinearRegression class after sklearn's, but calculate t-statistics
    and p-values for model coefficients (betas).
    Additional attributes available after .fit()
    are `t` and `p` which are of the shape (y.shape[1], X.shape[1])
    which is (n_features, n_coefs)
    This class sets the intercept to 0 by default, since usually we include it
    in X.
    """

    def __init__(self, *args, **kwargs):
        if not "fit_intercept" in kwargs:
            kwargs['fit_intercept'] = False
        super(LinearRegression, self)\
                .__init__(*args, **kwargs)

    def fit(self, X, y, n_jobs=1):
        self = super(LinearRegression, self).fit(X, y, n_jobs)
       
        # std errors
        uhat = (y-(X@self.coef_).ravel())
        k = np.shape(X)[1]
        s2 = (uhat.T@uhat)/(y.shape[0])
        var = s2*np.linalg.inv(X.T@X)
        self.se = np.sqrt(np.diag(var))

        # T-Stat
        self.t_stats = self.coef_/self.se

        # p-values
        self.df = y.shape[0] - k # -1 degrees of freedom: N minus number of parameters
        self.p_values = 2*(t.sf(abs(self.t_stats),self.df))

        # Rsquared
        tss = (y-np.mean(y)).T@(y-np.mean(y))
        rss = uhat.T@uhat
        self.rsq = 1 - rss/tss

        self.summary = pd.DataFrame({
            "beta":self.coef_.reshape(1,-1).tolist()[0],
            "se":self.se.reshape(1,-1).tolist()[0],
            "t_stats":self.t_stats.reshape(1,-1).tolist()[0],
            "p_values":self.p_values.reshape(1,-1).tolist()[0],
            })

        return self

在玩具数据集中运行函数，我们可以测试结果:

import statsmodels.api as sm
data = sm.datasets.longley.load_pandas()

# Estimate statsmodels OLS
model = OLS(endog=data.endog,exog=data.exog).fit()

# Estimate Sklearn with report like statsmodels OLS
model2 = LinearRegression(fit_intercept=False).fit(data.exog,np.array(data.endog))
model2.summary

我担心某些公式与正确的不匹配。

Answer 1

这里有一个类，您可以使用它来使用 Scikit-learn 获取线性回归模型摘要:

import numpy as np
import pandas as pd
from scipy.stats import t
from sklearn import linear_model

class LinearRegression(linear_model.LinearRegression):

    def __init__(self, *args, **kwargs):
        if not "fit_intercept" in kwargs:
            self.with_intercept = True
            kwargs['fit_intercept'] = True
        else:
            self.with_intercept = False
            kwargs['fit_intercept'] = False
        super(LinearRegression, self).__init__(*args, **kwargs)

    def fit(self, X, y):
        self = super(LinearRegression, self).fit(X, y)

        y_pred = self.predict(X)
        residuals = y - y_pred
        residual_sum_of_squares = residuals.T @ residuals

        if self.with_intercept:
            coefficients = [coef for coef in self.coef_]
            self.coefficients = [self.intercept_] + coefficients

            p = len(X.columns) + 1  # plus one because an intercept is added
            new_X = np.empty(shape=(len(X), p), dtype=float)
            new_X[:, 0] = 1
            new_X[:, 1:p] = X.values
            
        else:
            self.coefficients = self.coef_

            p = len(X.columns)
            new_X = X.values

        # standard errors
        sigma_squared_hat = residual_sum_of_squares / (len(X) - p)
        var_beta_hat = np.linalg.inv(new_X.T @ new_X) * sigma_squared_hat
        self.std_errors = [var_beta_hat[p_, p_] ** 0.5 for p_ in range(p)]

        # t_values
        self.t_values = np.array(self.coefficients)/self.std_errors

        # p values
        freedom_degree = y.shape[0] - X.shape[1]
        self.p_values = 2*(t.sf(abs(self.t_values), freedom_degree))

        # summary
        self.summary = pd.DataFrame()
        self.summary['Coefficients'] = self.coefficients
        self.summary['Standard errors'] = self.std_errors
        self.summary['t values'] = self.t_values
        self.summary['p values'] = self.p_values
    
        return self

这是您可以尝试类(class)并拟合模型的方式:

import statsmodels.api as sm

data = sm.datasets.longley.load_pandas()

X = pd.DataFrame(data.exog)
y = np.array(data.endog)

# in case you want to ignore the intercept include fit_intercept=False as a parameter
model_linear_regression = LinearRegression().fit(X, y)
model_linear_regression.summary

结果:

    Coefficients    Standard errors   t values   p values
0   -3.482259e+06   890420.379        -3.911     0.003
1   1.506190e+01    84.915            0.177      0.863
2   -3.580000e-02   0.033             -1.070     0.310
3   -2.020200e+00   0.488             -4.136     0.002
4   -1.033200e+00   0.214             -4.822     0.001
5   -5.110000e-02   0.226             -0.226     0.826
6   1.829151e+03    455.478           4.016      0.002

您可以训练一个 OLS 模型，在您想要添加截距/常量的情况下进行必要的更改，如下所示:

with_intercept = True
if with_intercept:
    p = len(X.columns) + 1  # plus one because an intercept is added
    new_X = np.empty(shape=(len(X), p), dtype=float)
    new_X[:, 0] = 1
    new_X[:, 1:p] = X.values
    model_statsmodel = sm.OLS(endog=y, exog=new_X).fit()
else:
    model_statsmodel = sm.OLS(endog=y, exog=X).fit()

最后，您可以通过以下方式验证结果是否相同:

assert np.allclose(np.round(model_statsmodel.params, 3), np.round(model_linear_regression.coefficients, 3))
assert np.allclose(np.round(model_statsmodel.bse, 3), np.round(model_linear_regression.std_errors, 3))
assert np.allclose(np.round(model_statsmodel.tvalues, 3), np.round(model_linear_regression.t_values, 3))
# the following assertion might return False, due to a difference in the number of decimals
assert np.allclose(np.round(model_statsmodel.pvalues, 3), np.round(model_linear_regression.p_values, 3))

希望对您有所帮助:)