看来麻烦在于系数计算，你给出的计算系数的公式是标量形式，用于线性回归的最简单情况，即只有一个特征 x。

编辑

现在，在查看了用于系数计算的代码之后，问题变得更加清晰。您不能使用此方程式来计算彼此的每个特征的系数，因为每个系数将取决于所有特征。我建议您在简单情况here和一般情况here下查看此最小二乘优化问题的解的推导。作为通用的提示，只要可以实现，矩阵就会更有效。

However, in this case we have a 10-dimensional feature vector, and so in matrix notation it becomes.

请参见派生here

我怀疑你在这里犯了一些计算错误，因为使用标量公式在 pytn 中实现这个比矩阵等效更乏味和不整洁。

下面是如何实现它的一个例子：

def calc_coefficients(X,Y):
    X=np.mat(X)
    Y = np.mat(Y)
    return np.dot((np.dot(np.transpose(X),X))**(-1),np.transpose(np.dot(Y,X)))
def score_r2(y_pred,y_true):
    ss_tot=np.power(y_true-y_true.mean(),2).sum()
    ss_res = np.power(y_true -y_pred,2).sum()
    return 1 -ss_res/ss_tot
X = np.ones(shape=(506,11))
X[:,1:] = data.values
B=calc_coefficients(X,y)
##### Coeffcients 
B[:]
matrix([[ 2.26053646e+01],
        [-9.64973063e-02],
        [ 5.28108077e-02],
        [ 2.38029890e+00],
        [ 3.94059598e+00],
        [-1.05476566e+00],
        [ 2.82595310e-01],
        [-1.57226536e-02],
        [-7.56519964e-01],
        [ 1.02392192e-02],
        [-5.70698610e-01]])
#### Intercept 
B[0]
matrix([[22.60536463]])
y_pred = np.dot(np.transpose(B),np.transpose(X))
##### First 5 rows predicted
np.array(y_pred)[0][:5]
array([30.42657776, 24.80818347, 30.69339701, 29.35761397, 28.6004966 ])
##### First 5 rows Ground Truth
y[:5]
array([24. , 21.6, 34.7, 33.4, 36.2])
### R^2 score
score_r2(y_pred,y)
0.7278959820021539

完整解决方案-2020-波士顿数据集

正如另一个人所说，要计算线性回归的系数，你必须计算

这给你的系数（所有的 B 为特征 + 截距）。

请确保在 X 中添加一个包含所有 1 的列，以计算截距（代码中有更多内容）

from sklearn.datasets import load_boston
import numpy as np
from CustomLibrary import CustomLinearRegression
from CustomLibrary import CustomMeanSquaredError
boston = load_boston()
X = np.array(boston.data, dtype="f")
Y = np.array(boston.target, dtype="f")
regression = CustomLinearRegression()
regression.fit(X, Y)
print("Projection matrix sk:", regression.coefficients, "\n")
print("bias sk:", regression.intercept, "\n")
Y_pred = regression.predict(X)
loss_sk = CustomMeanSquaredError(Y, Y_pred)
print("Model performance:")
print("--------------------------------------")
print("MSE is {}".format(loss_sk))
print("\n")

CustomLibrary.py

import numpy as np
cl CustomLinearRegression():
    def __init__(self):
        self.coefficients = None
        self.intercept = None
    def fit(self, x , y):
        x = self.add_one_column(x)
        x_T = np.transpose(x)
        inverse = np.linalg.inv(np.dot(x_T, x))
        pseudo_inverse = inverse.dot(x_T)
        coef = pseudo_inverse.dot(y)
        self.intercept = coef[0]
        self.coefficients = coef[1:]
        return coef
    def add_one_column(self, x):
        '''
        the fit metd with x feature return x coefficients ( include the intercept)
        so for have the intercept + x feature coefficients we have to add one column ( in the beginning )
        with all 1ones
        '''
        X = np.ones(shape=(x.shape[0], x.shape[1] +1))
        X[:, 1:] = x
        return X
    def predict(self, x):
        predicted = np.array([])
        for sample in x:
            result = self.intercept
            for idx, feature_value_in_sample in enumerate(sample):
                result += feature_value_in_sample * self.coefficients[idx]
            predicted = np.append(predicted, result)
        return predicted
def CustomMeanSquaredError(Y, Y_pred):
    mse = 0
    for idx,data in enumerate(Y):
        mse += (data - Y_pred[idx])**2
    return  mse * (1 / len(Y))