重命名主题，x，y的LS拟合（Mult. Lin. Regr.），
使用intercept_column（转载自我的评论）-好吧，我认为这是因为0平方误差的总和当我们考虑Intersept时，Lin Regr（或残差）-因此，SSE最小化（包括截距的方程）明显适合OLS --从Normal Equationintercept, *coef = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(y)开始，我们使用方程(X.T.dot(X)).dot(β) = (X.T).dot(y)，使L2-norm(xβ-y)最小化（或线性术语(ax-b)）：

import numpy as np
from scipy import linalg as la
from scipy.linalg import solve

# data
f1 = np.array([1., 1.5, 3.5, 4.])
f2 =  np.array([3., 4., 7., 7.25])
# z = np.array([6., 6.5, 8., 9.])
X= np.array([f1, f2]).T
y= np.array([0.5, 1., 1.5, 2.]).T
# adds x0 = 1 to each instance
X = np.column_stack((np.ones(len(y)),X))    # CORRECTION

##################### la.lstsq

x= la.lstsq(X,y)[0]
print("lstsq:", x)

##################### LU-decomposition
from scipy.linalg import lu
p, l, u = lu(X)
# reconstruct
print(p @ l @ u)
print(np.allclose(X - p @ l @ u, np.zeros((4, 3))))

#####################
from scipy.linalg import lu_factor, lu_solve

lu, piv = lu_factor(X.T@X)  # if not cov.matrix (as is square) taken then ValueError: expected square matrix
x = lu_solve((lu, piv), X.T @ y)
print("lu_solve: ", x)

##################### custom ls

# LR: y= ax-b, MLR: beta= (X.T @ X)^(-1)(X.T @ y)
A = X.T @ X # take covariance matrix to make it square for solve !
b = X.T @ y
beta = solve(A, b, overwrite_a=True, overwrite_b=True,
                 check_finite=True)
print("solve: ", beta)

############
# calculate normal equation (NOT STABLE)
theta_best = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(y)
# best values for theta
intercept, *coef = theta_best
print(f"Intercept: {intercept}\n\
Coefficients: {coef}")

##################### plot

# use the optimized parameters to plot the fitted curve in 3D space.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# Create 3D plot of the data points and the fitted curve
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(f1, f2, y, color='blue')
x_range = np.linspace(0, 7, 100)
y_range = np.linspace(0, 7,100)

X, Y = np.meshgrid(x_range, y_range)
Z = beta[0] + beta[1]*X + beta[2]*Y

ax.plot_surface(X, Y, Z, color='red', alpha=0.5)
ax.set_xlabel('feat.1')
ax.set_ylabel('feat.2')
ax.set_zlabel('target')
plt.show()

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N.B.的
不要用矩阵求逆来求解线性方程组专业的算法不会求解矩阵求逆。这很慢，而且会引入不必要的误差。对于小系统来说，这不是一场灾难，但为什么要做一些次优的事情呢？基本上，任何时候你看到数学公式为：x = A^-1 * b，你都需要：x = np.linalg.solve（A，b）
x1c 0d1x的数据