3d 中点到线段的距离 (Python)

我正在寻找可以计算 3D 中的点 (x_0,y_0,z_0) 到直线的距离的 Python 函数segment由其端点 (x_1,y_1,z_1) 和 (x_2,y_2,z_2) 定义。

我只找到了这个问题的 2D 解决方案。

There are solutions to finding a distance from a point to a line in 3d, but not to a line segment, like here:

（图片取自计算特殊情况下点到线段的距离 https://stackoverflow.com/questions/46079554/calculate-distance-point-to-line-segment-with-special-cases)

这个答案改编自这里：在Python中计算点数组到线段之间的欧几里德距离，无需for循环 https://stackoverflow.com/questions/54442057/calculate-the-euclidian-distance-between-an-array-of-points-to-a-line-segment-in/54442561#54442561.

功能lineseg_dist返回点 p 到线段 [a,b] 的距离。p, a and b是 np.arrays。

import numpy as np

def lineseg_dist(p, a, b):

    # normalized tangent vector
    d = np.divide(b - a, np.linalg.norm(b - a))

    # signed parallel distance components
    s = np.dot(a - p, d)
    t = np.dot(p - b, d)

    # clamped parallel distance
    h = np.maximum.reduce([s, t, 0])

    # perpendicular distance component
    c = np.cross(p - a, d)

    return np.hypot(h, np.linalg.norm(c))