两个不相等的点集之间的最小距离

I want to be able to find the minimum distance between 2 sets of points in the xy-plane. Let's assume the first set of points, set A, has 9 points, and the second set of points, set B, has 3 points. I want to find the minimum total distance that connects each of the points in set A to a points in set B. Obviously there will be some overlap, and maybe even some points in set B that have no links. But all of the points in set A must have 1 and only 1 link coming from it to a point in set B.

我想求出xy平面上两组点之间的最小距离。假设第一组点，集合A有9个点，集合B有3个点。我想求出在集合B中每一个点的最小总距离。很明显，会有一些重叠，甚至在集合B中有一些没有链接的点。但是集合A中的所有点都必须有一个且只有一个从它到集合B中的一个点的链接。

I have found a solution to this problem if both sets have an equal number of points and here is the code for it:

我找到了一个解决这个问题的方法，如果两个集合的点数相等，下面是它的代码:

import random
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import cdist
from scipy.optimize import linear_sum_assignment

points1 = np.array([(x, y) for x in np.linspace(-1,1,3) \
          for y in np.linspace(-1,1,3)])
N = points1.shape[0]
points2 = 2*np.random.rand(N,2)-1

cost12 = cdist(points1, points2)
row_ind12, col_ind12 = linear_sum_assignment(cost12)

plt.plot(points1[:,0], points1[:,1], 'b*')
plt.plot(points2[:,0], points2[:,1], 'rh')
for i in range(N):
    plt.plot([points1[i,0], points2[col_ind12[i],0]], [points1[i,1], 
             points2[col_ind12[i],1]], 'k')
plt.show()

1 个解决方案

#1

The function scipy.cluster.vq.vq does what you want.

scipy.cluster.vq的函数。vq做你想做的。

Here's a modified version of your code that demonstrates vq:

这里有一个修改版本的代码来演示vq:

import numpy as np
from scipy.cluster.vq import vq
import matplotlib.pyplot as plt


# `points1` is the set A described in the question.
points1 = np.array([(x, y) for x in np.linspace(-1,1,3)
                               for y in np.linspace(-1,1,3)])

# `points2` is the set B.  In this example, there are 5 points in B.
N = 5
np.random.seed(1357924)
points2 = 2*np.random.rand(N, 2) - 1

# For each point in points1, find the closest point in points2:
code, dist = vq(points1, points2)


plt.plot(points1[:,0], points1[:,1], 'b*')
plt.plot(points2[:,0], points2[:,1], 'rh')

for i, j in enumerate(code):
    plt.plot([points1[i,0], points2[j,0]],
             [points1[i,1], points2[j,1]], 'k', alpha=0.4)

plt.grid(True, alpha=0.25)
plt.axis('equal')
plt.show()

The script produces the following plot:

剧本产生以下情节:

#1