Python，成对的“距离”，需要一种快速的方法来实现

您可以使用numpy的矢量化功能来加快计算速度。我的版本一次计算距离矩阵的所有元素，然后将对角线和下三角设置为零。

def pairwise_distance2(s):    # we need this because we're gonna divide by zero    old_settings = np.seterr(all="ignore")    N = N_segments # just shorter, could also use len(s)    # we repeat p0 and p1 along all columns    p0 = np.repeat(s[:,0:3].reshape((N, 1, 3)), N, axis=1)    p1 = np.repeat(s[:,3:6].reshape((N, 1, 3)), N, axis=1)    # and q0, q1 along all rows    q0 = np.repeat(s[:,0:3].reshape((1, N, 3)), N, axis=0)    q1 = np.repeat(s[:,3:6].reshape((1, N, 3)), N, axis=0)    # element-wise dot product over the last dimension,    # while keeping the number of dimensions at 3    # (so we can use them together with the p* and q*)    a = np.sum((p1 - p0) * (p1 - p0), axis=-1).reshape((N, N, 1))    b = np.sum((p1 - p0) * (q1 - q0), axis=-1).reshape((N, N, 1))    c = np.sum((q1 - q0) * (q1 - q0), axis=-1).reshape((N, N, 1))    d = np.sum((p1 - p0) * (p0 - q0), axis=-1).reshape((N, N, 1))    e = np.sum((q1 - q0) * (p0 - q0), axis=-1).reshape((N, N, 1))    # same as above    s = (b*e-c*d)/(a*c-b*b)    t = (a*e-b*d)/(a*c-b*b)    # almost same as above    pairwise = np.sqrt(np.sum( (p0 + (p1 - p0) * s - ( q0 + (q1 - q0) * t))**2, axis=-1))    # turn the error reporting back on    np.seterr(**old_settings)    # set everything at or below the diagonal to 0    pairwise[np.tril_indices(N)] = 0.0    return pairwise

现在让我们旋转一下。以您的示例

N = 1000

%timeit pairwise_distance(List_of_segments)1 loops, best of 3: 10.5 s per loop%timeit pairwise_distance2(List_of_segments)1 loops, best of 3: 398 ms per loop

当然，结果是一样的：

(pairwise_distance2(List_of_segments) == pairwise_distance(List_of_segments)).all()

返回

True

顺便说一句：我尝试过首先简单地使用numba而不成功。虽然不知道为什么。