两个不同的Numpy数组中的点之间的最小欧几里得距离，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，我想找到一个数组中每个点之间的最小欧几里得距离与 all 在另一个数组中的点.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()

有没有一种方法可以消除for循环，并以某种方式在两个数组之间进行逐元素计算?我设想生成一个距离矩阵，为此我可以找到每一行或每一列中的最小元素.

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.

查看问题的另一种方法.假设我将xy1(长度 m )和xy2(长度 p )连接成xy(长度 n )，然后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.有没有一种方法可以有效地生成此子矩阵?

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， 适用于X和Y 2暗，3暗... 它还执行22种不同的规范，详细 此处.

(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 "\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

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