Python返回数组中的平方马氏距离函数为什么?

2024-09-28 01:32:23 发布

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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,我得到了一个输出矩阵

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输出:

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

似乎我的函数只是给我在第二个参数中输入的2x2矩阵的倒数。在


Tags: thetobyisnpbemeanarray
2条回答

可以使用scipymahalanobis()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

你犯了一个经典的错误,假设*运算符是做矩阵乘法的。这在Python/numpy中是不正确的(请参见http://www.scipy-lectures.org/intro/numpy/operations.htmlhttps://docs.scipy.org/doc/numpy-dev/user/numpy-for-matlab-users.html)。我把它分解成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])

生产

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