Python返回数组中的平方Mahalanobis距离函数 - 为什么？

Posted 2021-04-04

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)

但由于某种原因，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]])

看来我的函数只是给了我在第二个参数中输入的2x2矩阵的逆。

答案

假设*运算符正在进行矩阵乘法，你犯了经典的错误。在Python / numpy中不是这样（参见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了点功能

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

另一答案

可以使用scipy的mahalanobis()函数来验证：

import scipy.spatial, numpy as np

scipy.spatial.distance.mahalanobis([2,5], [3,6], np.linalg.inv([[.5,0],[0,2]]))
# 1.5811388300841898
1.5811388300841898**2 # squared Mahalanobis distance
# 2.5000000000000004

def Mahalanobis(x, covariance_matrix, mean):
  x, m, C = np.array(x), np.array(mean), np.array(covariance_matrix)
  return (x-m)@np.linalg.inv(C)@(x-m).T

Mahalanobis([2,5], [[.5,0],[0,2]], [3,6])
# 2.5 

np.isclose(
   scipy.spatial.distance.mahalanobis([2,5], [3,6], np.linalg.inv([[.5,0],[0,2]]))**2, 
   Mahalanobis([2,5], [[.5,0],[0,2]], [3,6])
)
# True