两个不同 Numpy 数组中的点之间的最小欧几里德距离，不在 [英] Minimum Euclidean distance between points in two different Numpy arrays, not within

问题描述

我有两个 x-y 坐标数组，我想在一个数组中找到 每个 点之间的最小欧几里德距离all 另一个数组中的点.数组的大小不一定相同.例如:

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]])

我当前的方法循环遍历 xy1 中的每个坐标 xy 并计算该坐标与其他坐标之间的距离.

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()

有没有办法消除 for 循环并以某种方式在两个数组之间进行逐个元素的计算?我设想生成一个距离矩阵，我可以在其中找到每行或每列中的最小元素.

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.

另一种看待问题的方式.假设我将xy1(长度m)和xy2(长度p)连接成xy (length n)，我存储原始数组的长度.从理论上讲，我应该能够从那些坐标中生成一个 n x n 距离矩阵，从中我可以获取一个 m x p 子矩阵.有没有办法有效地生成这个子矩阵?

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?

推荐答案

(几个月后)scipy.spatial.distance.cdist( X, Y )给出所有的距离对，用于 X 和 Y 2 暗、3 暗 ...
它还做了22种不同的规范，详细这里 .

(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 = 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 "
", 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

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