Question

代码是：

import numpy as np
def Mahalanobis(x, covariance_matrix, mean):
    x = np.array(x)
    mean = np.array(mean)
    covariance_matrix = np.array(covariance_matrix)
    return (x-mean)*np.linalg.inv(covariance_matrix)*(x.transpose()-mean.transpose())

#variables x and mean are 1xd arrays; covariance_matrix is a dxd matrix
#the 1xd array passed to x should be multiplied by the (inverted) dxd array
#that was passed into the second argument
#the resulting 1xd matrix is to be multiplied by a dx1 matrix, the transpose of 
#[x-mean], which should result in a 1x1 array (a number)

但由于某些原因，当我输入参数

时，我得到了输出矩阵

Mahalanobis([2,5], [[.5,0],[0,2]], [3,6])

输出：

out[]: array([[ 2. ,  0. ],
              [ 0. ,  0.5]])

似乎我的函数只是给了我在第二个参数中输入的2x2矩阵的逆。

Answer 1

你犯了一个经典错误，即假设*运算符正在进行矩阵乘法。在Python / numpy中并非如此（请参阅http://www.scipy-lectures.org/intro/numpy/operations.html和https://docs.scipy.org/doc/numpy-dev/user/numpy-for-matlab-users.html）。我将其分解为中间步骤并使用点函数

import numpy as np
def Mahalanobis(x, covariance_matrix, mean):

    x = np.array(x)
    mean = np.array(mean)
    covariance_matrix = np.array(covariance_matrix)

    t1 = (x-mean)
    print(f'Term 1 {t1}')

    icov = np.linalg.inv(covariance_matrix)
    print(f'Inverse covariance {icov}')

    t2 = (x.transpose()-mean.transpose())
    print(f'Term 2 {t2}')

    mahal = t1.dot(icov.dot(t2))

    #return (x-mean)*np.linalg.inv(covariance_matrix).dot(x.transpose()-mean.transpose())
    return mahal

#variables x and mean are 1xd arrays; covariance_matrix is a dxd matrix
#the 1xd array passed to x should be multiplied by the (inverted) dxd array
#that was passed into the second argument
#the resulting 1xd matrix is to be multiplied by a dx1 matrix, the transpose of 
#[x-mean], which should result in a 1x1 array (a number)


Mahalanobis([2,5], [[.5,0],[0,2]], [3,6])

产生

Term 1 [-1 -1]
Inverse covariance [[2.  0. ]
 [0.  0.5]]
Term 2 [-1 -1]
Out[9]:    2.5

Answer 2

一个人可以使用scipy的{{1}}函数进行验证：

mahalanobis()

Python返回数组中的平方Mahalanobis距离函数 - 为什么？

2 个答案: