如何解决在 Python 中手动线性回归函数
import numpy as np
import pandas as pd
def linear_regress(y,X):
X_trp = np.transpose(X)
b = np.linalg.inv(X_trp@X)@X_trp@y
e = y - X@b
e_trp = np.transpose(e)
sigma_sqr = (e_trp@e) / (X.shape[0] - X.shape[1])
var_b = sigma_sqr@(np.linalg.inv(X_trp@X))
SE = np.sqrt(var_b)
z_score = 1.96
upper = b+SE*z_score
lower = b-SE*z_score
CI = [lower,upper]
results = {"Regression Coefficients": b,"Standard Error (SE)": SE,"95% Confidence Interval": CI}
return results
当我定义数组并运行函数时它返回:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-17-8307755b9bbc> in <module>
----> 1 linear_regress(indt,dept)
<ipython-input-16-a6336ec1997c> in linear_regress(y,X)
8 e_trp = np.transpose(e)
9 sigma_sqr = (e_trp@e) / (X.shape[0] - X.shape[1])
---> 10 var_b = sigma_sqr@(np.linalg.inv(X_trp@X))
11 SE = np.sqrt(var_b)
12 z_score = 1.96
ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0,with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 2 is different from 1)
你能帮我找出问题所在吗?
感谢您的帮助!
解决方法
如果不正确了解数据,就很难判断自己做错了什么。 但我猜你正在尝试实施正态方程来获得体重。 如果是这样的话,这似乎太复杂了,这里有一个更简单的方法。
def LinearRegression(X,y):
X_t = X.T
theta = np.dot(np.linalg.inv(np.dot(X.T,X)),np.dot(y,X))
mean_sqr_error = np.mean((y - np.dot(theta,X.T))**2)
return theta,mean_sqr_error
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