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。
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
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