Here is vectorized implementation for computing the euclidean distance that is much faster than what you have (even significantly faster than PDIST2 on my machine): D = sqrt( bsxfun(@plus,sum(A.^2,2),sum(B.^2,2)’) – 2*(A*B’) ); It is based on the fact that: ||u-v||^2 = ||u||^2 + ||v||^2 – 2*u.v Consider below a crude comparison between the two methods: … Read more
You can take advantage of the complex type : # build a complex array of your cells z = np.array([complex(c.m_x, c.m_y) for c in cells]) First solution # mesh this array so that you will have all combinations m, n = np.meshgrid(z, z) # get the distance via the norm out = abs(m-n) Second solution … Read more
A distance matrix is a lower triangular matrix in packed format, where the lower triangular is stored as a 1D vector by column. You can check this via str(distMatrix) # Class ‘dist’ atomic [1:10] 1 4 10 15 3 9 14 6 11 5 # … Even if we call dist(vec1, diag = TRUE, upper … Read more
(Months later) scipy.spatial.distance.cdist( X, Y ) gives all pairs of distances, for X and Y 2 dim, 3 dim … It also does 22 different norms, detailed here . # cdist example: (nx,dim) (ny,dim) -> (nx,ny) from __future__ import division import sys import numpy as np from scipy.spatial.distance import cdist #……………………………………………………………………. dim = 10 nx … Read more
The usually given answer here is based on bsxfun (cf. e.g. [1]). My proposed approach is based on matrix multiplication and turns out to be much faster than any comparable algorithm I could find: helpA = zeros(numA,3*d); helpB = zeros(numB,3*d); for idx = 1:d helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*A(:,idx), A(:,idx).^2 ]; helpB(:,3*idx-2:3*idx) = [B(:,idx).^2 , B(:,idx), … Read more
Use numpy.linalg.norm: dist = numpy.linalg.norm(a-b) You can find the theory behind this in Introduction to Data Mining This works because the Euclidean distance is the l2 norm, and the default value of the ord parameter in numpy.linalg.norm is 2.