我喜欢用它来方便地计算几种数组格式的距离...它不是一行程序，但它很好用。关于它的实现细节可以在堆栈的其他地方找到，方法是搜索'einsum'和numpy作为关键字。需要将numpy作为np导入，这只是def，你需要2个数组

import numpy as np

def e_dist(a, b, metric='euclidean'):

"""Distance calculation for 1D, 2D and 3D points using einsum
: a, b   - list, tuple, array in 1,2 or 3D form
: metric - euclidean ('e','eu'...), sqeuclidean ('s','sq'...),
:-----------------------------------------------------------------------
"""
a = np.asarray(a)
b = np.atleast_2d(b)
a_dim = a.ndim
b_dim = b.ndim
if a_dim == 1:
    a = a.reshape(1, 1, a.shape[0])
if a_dim >= 2:
    a = a.reshape(np.prod(a.shape[:-1]), 1, a.shape[-1])
if b_dim > 2:
    b = b.reshape(np.prod(b.shape[:-1]), b.shape[-1])
diff = a - b
dist_arr = np.einsum('ijk,ijk->ij', diff, diff)
if metric[:1] == 'e':
    dist_arr = np.sqrt(dist_arr)
dist_arr = np.squeeze(dist_arr)
return dist_arr

然后您可以根据需要对结果进行排序，例如

a = np.random.randint(0, 10, size=(10,2))

orig = np.array([0,0])

    e_dist(a, orig)

    array([  4.12,   9.9 ,   7.07,   6.08,   3.16,  10.63,   8.54,   7.28,   7.21,
             6.08])
    np.sort(e_dist(a, orig))

    array([  3.16,   4.12,   6.08,   6.08,   7.07,   7.21,   7.28,   8.54,   9.9 ,
            10.63])

附录
我应该补充说，您可以使用argsort获得排序后的值，如下所示

np.argsort(e_dist(a, orig))
array([4, 0, 3, 9, 2, 8, 7, 6, 1, 5], dtype=int64)

idx = np.argsort(art.e_dist(a, orig))

closest = a[idx]
array([[3, 1],
       [1, 4],
       [1, 6],
       [6, 1],
       [5, 5],
       [4, 6],
       [2, 7],
       [8, 3],
       [7, 7],
       [7, 8]])

def distance_squared(x1,y1,x2,y2):
    return (x1-x2)**2 + (y1-y2)**2
target_point = 0,0
sorted(a,key=lambda point:distance_squared(target_point[0],target_point[1],*point[0]))

你可以使用scikit-learn内置的pairwise-distance函数。

from sklearn.metrics.pairwise import pairwise_distances
d = pairwise_distances(X, Y, metric="euclidean")