Python返回阵列中的平方Mahalanobis距离功能 - 原因

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)

，但由于某种原因，当我输入参数时，我会得到输出的矩阵

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

输出：

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

看来我的功能只是给了我第二个参数中输入的2x2矩阵的倒数。

Answer 1

您犯了一个经典错误，假设 *操作员正在执行矩阵乘法。在python/numpy中是不正确的(请参阅http://www.scipy-lectures.org/intro/numpy/numpy/operations.html和https://docs.scipy.org/doc/doc/doc/numpy-doc/numpy-dev/numpy-dev/numpy-dev/numpy--user/numpy---------------------for-matlab-users.html(。我将其分解为中间步骤，并使用了点函数

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 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