Mahalanobis distance inverting the covariance matrix

user_1110675 0
Viewed 5442

I am writing a function to take the Mahalanobis distance between two vectors. I understand that this is achieved using the equation a'*C^-1*b, where a and b are vectors and C is the covariance matrix. My question is, is there an efficient way to find the inverse of the matrix without using Gauss-Jordan elimination, or is there no way around this? I'm looking for a way to do this myself, not with any predefined functions.

I know that C is a Hermitian, positive definite matrix, so is there any way that I can algorithmically take advantage of this fact? Or is there some clever way compute the Mahalanobis distance without calculating the inverse of the covariance at all? Any help would be appreciated.

***Edit: The Mahalanobis distance equation above is incorrect. It should be x'*C^-1*x where x = (b-a), and b and a are the two vectors whose distance we are trying to find (thanks LRPurser). The solution posited in the selected answer is therefore as follows:

d=x'*b, where b = C^-1*x C*b = x, so solve for b using LU factorization or LDL' factorization.

linear-algebra matrix c algorithm
3 Answers
# Mahalanobis Distance Matrix

import numpy as np
from scipy.spatial import distance
from scipy.spatial.distance import mahalanobis
from scipy.spatial.distance import pdist
from scipy.spatial.distance import squareform

# Example
Data = np.array([[1,2],[3,2],[4,3]])
Cov = np.cov(np.transpose([[1,2],[3,2],[4,3]]))
invCov = np.linalg.inv(Cov)

Y = pdist(Data, 'mahalanobis', invCov)
MD = squareform(Y)
user_177082150
Related