Question

我有两个带元组（坐标）的列表，例如：

some_pt1 = [(10.76,2.9),(3.24,4.28),(7.98,1.98),(3.21,9.87)]
some_pt2 = [(11.87,6.87), (67.87,8.88), (44.44, 6.78), (9.81, 1.09), (6.91, 0.56), (8.76, 8.97), (8.21, 71.66)]

元组中的每个值都是平面
列表长度不一致

我在两个列表之间找到两个最接近的点。我不知道怎么样，也许可以用距离来做。我希望有一种更有效的方法来做到这一点，因为我需要这个功能尽可能快地工作（它是更大的一部分）。

Answer 1

或者，通过参考Tim Seed的代码。这可以用。

from scipy.spatial import distance
some_pt1 = [(10.76,2.9),(3.24,4.28),(7.98,1.98),(3.21,9.87)]
some_pt2 = [(11.87,6.87), (67.87,8.88), (44.44, 6.78), (9.81, 1.09), (6.91, 0.56), (8.76, 8.97), (8.21, 71.66)]

empthy_dict = {}
for i in range(len(some_pt1)):
    for j in range(len(some_pt2)):
        dist = distance.euclidean(some_pt1[i],some_pt2[j])
        empthy_dict[dist] = [some_pt1[i],some_pt2[j]]

shortest = sorted(empthy_dict.keys())[0]
points = empthy_dict[shortest]
print('Shortest distance is ' ,shortest,' and points are ' ,points)

Answer 2

这个怎么样

from pprint import pprint

some_pt1 = [(10.76,2.9),(3.24,4.28),(7.98,1.98),(3.21,9.87)]
some_pt2 = [(11.87,6.87), (67.87,8.88), (44.44, 6.78), (9.81, 1.09), (6.91, 0.56), (8.76, 8.97), (8.21, 71.66)]


distance = {}
for x in some_pt1:
    for y in some_pt2:
        dist =abs(abs(x[0])-abs(y[0]))+abs(abs(x[1])-abs(y[1]))
        distance[dist]=[x,y]

shortest =sorted(distance.keys())[0]
print("Min Distance is {} Objects are  {} {} ".format(shortest, distance[shortest][0],distance[shortest][0]))

Answer 3

在任何情况下，您都需要进行所有可能的组合，有一些算法可以帮助您以最佳顺序执行或避免重复距离。如果你想快速完成它，你应该使用一个特殊的库来帮助编译或预编译数组，这可以通过Numba或Cython来完成。其他库如scipy有特殊模块，如scipy.spatial.distance。请看这篇文章以获得更多疑问similar cuestion。

示例：

import scipy.spatial.distance as sd
import numpy as np
some_pt1 = [(10.76,2.9),(3.24,4.28),(7.98,1.98),(3.21,9.87)]
some_pt2 = [(11.87,6.87), (67.87,8.88), (44.44, 6.78), (9.81, 1.09), (6.91, 0.56), (8.76, 8.97), (8.21, 71.66)]
np.unravel_index(np.argmin(sd.cdist(some_pt1, some_pt2)), (len(some_pt1), len(some_pt2)))

结果：(2, 4)

此代码将返回第一个列表和第二个列表中的位置。

Answer 4

按欧几里德距离：

>>> some_pt1 = [(10.76,2.9),(3.24,4.28),(7.98,1.98),(3.21,9.87)]
>>> some_pt2 = [(11.87,6.87), (67.87,8.88), (44.44, 6.78), (9.81, 1.09), (6.91, 0.56), (8.76, 8.97), (8.21, 71.66)]
>>> 
>>> def dist_sq(p1_p2):
...     p1, p2 = p1_p2
...     return sum(x*y for x,y in zip(p1, p2))
... 
>>> 
>>> min(((p1, p2) for p1 in some_pt1 for p2 in some_pt2), key=dist_sq)
((3.24, 4.28), (6.91, 0.56))

其中有运行时O（n * m）（其中n，m是列表的长度）。因为你需要查看所有对，所以它不会比这更好。

请注意，比较平方距离就足够了，不需要计算根。

python - 给定两个元组列表找到它们之间最接近的元组（dist

4 个答案: