sklearn Standardscaler Ridge Pipeline

Question

我正在尝试标准化特征，然后运行岭回归。

正如所提供的，这两个答案是不同的。

当我设置 ridge=0 时，答案是一样的。当我删除 StandardScaler 和 Dn 时，答案也是一样的。

我不知道如何协调这两个版本（原始版本和使用 sklearn）。

感谢您的帮助

from sklearn.linear_model import Ridge
from sklearn.linear_model import LinearRegression
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import StandardScaler
import numpy as np
np.random.seed(0)

x = np.random.randn(100, 3)
y = np.random.randn(100, 2)

xx = x.T @ x
xy = x.T @ y
Dn = np.diag(1 / np.sqrt(np.diag(xx)))

ridge = 1

xx = Dn @ xx @ Dn
xy = Dn @ xy
beta_raw = Dn @ np.linalg.solve(xx + np.eye(len(xx)) * ridge, xy)
f_raw = x @ beta_raw

model = Pipeline([("scaler", StandardScaler(with_mean=False)), ("regression", Ridge(ridge, fit_intercept=False))])
trained_model = model.fit(x, y)
f_ml = trained_model.predict(x)

print(f_ml[:3] / f_raw[:3])

Answer 1

您正在按不同的值进行缩放，检查：

np.diag(Dn)
array([0.09699826, 0.10123938, 0.1016412 ])

model.steps[0][1].scale_
array([1.02603414, 0.98202661, 0.97598415])

您的标准差是协方差矩阵对角线的平方。即使您没有使矩阵居中，您仍然需要减去均值以获得协方差。见this post for more information

所以如果我们做对了：

x_m = x.mean(axis=0)
x_cov = np.dot((x - x_m).T, x - x_m) / (x.shape[0])
Dn = np.diag(1 / np.sqrt(np.diag(x_cov)))

xx = x.T @ x
xy = x.T @ y

ridge = 1

xx = Dn @ xx @ Dn
xy = Dn @ xy
beta_raw = Dn @ np.linalg.solve(xx + np.eye(len(xx)) * ridge, xy)
f_raw = x @ beta_raw

model = Pipeline([("scaler", StandardScaler(with_mean=False)), ("regression", Ridge(ridge, fit_intercept=False))])
trained_model = model.fit(x, y)
f_ml = trained_model.predict(x)

print(f_ml[:3] / f_raw[:3])

[[1. 1.]
 [1. 1.]
 [1. 1.]]