Euclidean distance between points in two different Numpy arrays, not within

Question

I have two arrays of x-y coordinates, and I would like to find the minimum Euclidean distance between each point in one array with all the points in the other array. The arrays are not necessarily the same size. For example:

xy1=numpy.array(
[[  243,  3173],
[  525,  2997]])

xy2=numpy.array(
[[ 682, 2644],
[ 277, 2651],
[ 396, 2640]])

My current method loops through each coordinate xy in xy1 and calculates the distances between that coordinate and the other coordinates.

mindist=numpy.zeros(len(xy1))
minid=numpy.zeros(len(xy1))

for i,xy in enumerate(xy1):
    dists=numpy.sqrt(numpy.sum((xy-xy2)**2,axis=1))
    mindist[i],minid[i]=dists.min(),dists.argmin()

Is there a way to eliminate the for loop and somehow do element-by-element calculations between the two arrays? I envision generating a distance matrix for which I could find the minimum element in each row or column.

Another way to look at the problem. Say I concatenate xy1 (length m) and xy2 (length p) into xy (length n), and I store the lengths of the original arrays. Theoretically, I should then be able to generate a n x n distance matrix from those coordinates from which I can grab an m x p submatrix. Is there a way to efficiently generate this submatrix?

Answer 1

For what you're trying to do:

dists = numpy.sqrt((xy1[:, 0, numpy.newaxis] - xy2[:, 0])**2 + (xy1[:, 1, numpy.newaxis - xy2[:, 1])**2)
mindist = numpy.min(dists, axis=1)
minid = numpy.argmin(dists, axis=1)

Edit: Instead of calling sqrt, doing squares, etc., you can use numpy.hypot:

dists = numpy.hypot(xy1[:, 0, numpy.newaxis]-xy2[:, 0], xy1[:, 1, numpy.newaxis]-xy2[:, 1])

Answer 2

To compute the m by p matrix of distances, this should work:

>>> def distances(xy1, xy2):
...   d0 = numpy.subtract.outer(xy1[:,0], xy2[:,0])
...   d1 = numpy.subtract.outer(xy1[:,1], xy2[:,1])
...   return numpy.hypot(d0, d1)

the .outer calls make two such matrices (of scalar differences along the two axes), the .hypot calls turns those into a same-shape matrix (of scalar euclidean distances).

Answer 3

(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 .