Minimum 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 nxn distance matrix from those coordinates from which I can grab an mxp submatrix. Is there a way to efficiently generate this submatrix?

Answer 1

(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, detailedhere .

# 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 = 1000
ny = 100
metric = "euclidean"
seed = 1

    # change these params in sh or ipython: run this.py dim=3 ...
for arg in sys.argv[1:]:
    exec( arg )
np.random.seed(seed)
np.set_printoptions( 2, threshold=100, edgeitems=10, suppress=True )

title = "%s  dim %d  nx %d  ny %d  metric %s" % (
        __file__, dim, nx, ny, metric )
print "\n", title

#...............................................................................
X = np.random.uniform( 0, 1, size=(nx,dim) )
Y = np.random.uniform( 0, 1, size=(ny,dim) )
dist = cdist( X, Y, metric=metric )  # -> (nx, ny) distances
#...............................................................................

print "scipy.spatial.distance.cdist: X %s Y %s -> %s" % (
        X.shape, Y.shape, dist.shape )
print "dist average %.3g +- %.2g" % (dist.mean(), dist.std())
print "check: dist[0,3] %.3g == cdist( [X[0]], [Y[3]] ) %.3g" % (
        dist[0,3], cdist( [X[0]], [Y[3]] ))


# (trivia: how do pairwise distances between uniform-random points in the unit cube
# depend on the metric ? With the right scaling, not much at all:
# L1 / dim      ~ .33 +- .2/sqrt dim
# L2 / sqrt dim ~ .4 +- .2/sqrt dim
# Lmax / 2      ~ .4 +- .2/sqrt dim

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

The accepted answer does not fully address the question, which requests to find the minimum distance between the two sets of points, not the distance between every point in the two sets.

Although a straightforward solution to the original question indeed consists of computing the distance between every pair and subsequently finding the minimum one, this is not necessary if one is only interested in the minimum distances. A much faster solution exists for the latter problem.

All the proposed solutions have a running time that scales as m*p = len(xy1)*len(xy2) . This is OK for small datasets, but an optimal solution can be written that scales as m*log(p) , producing huge savings for large xy2 datasets.

This optimal execution time scaling can be achieved using scipy.spatial.KDTree as follows

import numpy as np
from scipy import spatial

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

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

# This solution is optimal when xy2 is very large
tree = spatial.KDTree(xy2)
mindist, minid = tree.query(xy1)
print(mindist)

# This solution by @denis is OK for small xy2
mindist = np.min(spatial.distance.cdist(xy1, xy2), axis=1)
print(mindist)

where mindist is the minimum distance between each point in xy1 and the set of points in xy2

Answer 4

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 5

import numpy as np
P = np.add.outer(np.sum(xy1**2, axis=1), np.sum(xy2**2, axis=1))
N = np.dot(xy1, xy2.T)
dists = np.sqrt(P - 2*N)

Answer 6

I think the following function also works.

import numpy as np
from typing import Optional
def pairwise_dist(X: np.ndarray, Y: Optional[np.ndarray] = None) -> np.ndarray:
    Y = X if Y is None else Y
    xx = (X ** 2).sum(axis = 1)[:, None]
    yy = (Y ** 2).sum(axis = 1)[:, None]
    return xx + yy.T - 2 * (X @ Y.T)

Explanation
Suppose each row of X and Y are coordinates of the two sets of points.
Let their sizes be m X p and p X n respectively.
The result will produce a numpy array of size m X n with the (i, j) -th entry being the distance between the i -th row and the j -th row of X and Y respectively.

Answer 7

I highly recommend using some inbuilt method for calculating squares, and roots for they are customized for optimized way to calculate and very safe against overflows.

@alex answer below is the most safest in terms of overflow and should also be very fast. Also for single points you can use math.hypot which now supports more than 2 dimensions.

>>> 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)

Safety concerns

i, j, k = 1e+200, 1e+200, 1e+200
math.hypot(i, j, k)
# np.hypot for 2d points
# 1.7320508075688773e+200

np.sqrt(np.sum((np.array([i, j, k])) ** 2))
# RuntimeWarning: overflow encountered in square

overflow/underflow/speeds

Answer 8

I think that the most straightforward and efficient solution is to do it like this:

distances = np.linalg.norm(xy1, xy2) # calculate the euclidean distances between the test point and the training features.
min_dist = numpy.min(dists, axis=1) # get the minimum distance 
min_id = np.argmi(distances) # get the index of the class with the minimum distance, i.e., the minimum difference.