python中的多元线性回归而不适合原点？

Question

I found this chunk of code on http://rosettacode.org/wiki/Multiple_regression#Python , which does a multiple linear regression in python. Print b in the following code gives you the coefficients of x1, ..., xN. However, this code is fitting the line through the origin (ie the resulting model does not include a constant).

All I'd like to do is the exact same thing except I do not want to fit the line through the origin, I need the constant in my resulting model.

Any idea if it's a small modification to do this? I've searched and found numerous documents on multiple regressions in python, except they are lengthy and overly complicated for what I need. This code works perfect, except I just need a model that fits through the intercept not the origin.

import numpy as np
from numpy.random import random

n=100
k=10
y = np.mat(random((1,n)))
X = np.mat(random((k,n)))

b = y * X.T * np.linalg.inv(X*X.T)
print(b)

Any help would be appreciated. Thanks.

Answer 1

您只需要向X添加全为1的行即可。

Answer 2

Maybe a more stable approach would be to use a least squares algorithm anyway. This can also be done in numpy in a few lines. Read the documentation about numpy.linalg.lstsq .

Here you can find an example implementation:

http://glowingpython.blogspot.de/2012/03/linear-regression-with-numpy.html

Answer 3

What you have written out, b = y * XT * np.linalg.inv(X * XT) , is the solution to the normal equations, which gives the least squares fit with a multi-linear model. swang's response is correct (and EMS's elaboration)---you need to add a row of 1's to X. If you want some idea of why it works theoretically, keep in mind that you are finding b_i such that

y_j = sum_i b_i x_{ij}.

By adding a row of 1's, you are are setting x_{(k+1)j} = 1 for all j , which means that you are finding b_i such that:

y_j = (sum_i b_i x_{ij}) + b_{k+1}

because the k+1 st x_ij term is always equal to one. Thus, b_{k+1} is your intercept term.