3d空间中点到线的最短距离 [英] Shortest distance between a point and a line in 3 d space

问题描述

我正在尝试使用 numpy 或 python 中的任何东西找到从点 (x0,y0,z0) 到由 (x1,y1,z1) 和 (x2,y2,z2) 连接的线的最小距离.不幸的是，我在网上能找到的所有内容都与 2d 空间有关，而且我对 python 还很陌生.任何帮助将不胜感激.提前致谢！

I am trying to find the minimum distance from a point (x0,y0,z0) to a line joined by (x1,y1,z1) and (x2,y2,z2) using numpy or anything in python. Unfortunately, all i can find on the net is related to 2d spaces and i am fairly new to python. Any help will be appreciated. Thanks in advance!

推荐答案

StackOverflow 不支持 Latex，所以我将掩盖一些数学问题.一种解决方案来自这样一种想法:如果您的线跨越点 p 和 q，那么该线上的每个点都可以表示为 t*(pq)+q 用于一些实值 t.然后，您希望最小化给定点 r 与该线上任何点之间的距离，距离很方便是单个变量 t 的函数，因此标准微积分技巧有效美好的.考虑以下示例，该示例计算 r 与 p 和 q 跨越的线之间的最小距离.手工，我们知道答案应该是1.

StackOverflow doesn't support Latex, so I'm going to gloss over some of the math. One solution comes from the idea that if your line spans the points p and q, then every point on that line can be represented as t*(p-q)+q for some real-valued t. You then want to minimize the distance between your given point r and any point on that line, and distance is conveniently a function of the single variable t, so standard calculus tricks work fine. Consider the following example, which calculates the minimum distance between r and the line spanned by p and q. By hand, we know the answer should be 1.

import numpy as np

p = np.array([0, 0, 0])
q = np.array([0, 0, 1])
r = np.array([0, 1, 1])

def t(p, q, r):
    x = p-q
    return np.dot(r-q, x)/np.dot(x, x)

def d(p, q, r):
    return np.linalg.norm(t(p, q, r)*(p-q)+q-r)

print(d(p, q, r))
# Prints 1.0

这适用于任意数量的维度，包括 2、3 和 10 亿.唯一真正的限制是 p 和 q 必须是不同的点，以便它们之间有一条唯一的线.

This works fine in any number of dimensions, including 2, 3, and a billion. The only real constraint is that p and q have to be distinct points so that there is a unique line between them.

我在上面的例子中分解了代码，以展示我在数学上思考它的方式(找到 t 然后计算距离)产生的两个不同的步骤.这不一定是最有效的方法，如果您想知道各种点和同一条线的最小距离，则肯定不是最有效的方法——如果维数很小，则加倍如此.如需更有效的方法，请考虑以下事项:

I broke the code down in the above example in order to show the two distinct steps arising from the way I thought about it mathematically (finding t and then computing the distance). That isn't necessarily the most efficient approach, and it certainly isn't if you want to know the minimum distance for a wide variety of points and the same line -- doubly so if the number of dimensions is small. For a more efficient approach, consider the following:

import numpy as np

p = np.array([0, 0, 0])  # p and q can have shape (n,) for any
q = np.array([0, 0, 1])  # n>0, and rs can have shape (m,n)
rs = np.array([          # for any m,n>0.
    [0, 1, 1],
    [1, 0, 1],
    [1, 1, 1],
    [0, 2, 1],
])

def d(p, q, rs):
    x = p-q
    return np.linalg.norm(
        np.outer(np.dot(rs-q, x)/np.dot(x, x), x)+q-rs,
        axis=1)

print(d(p, q, rs))
# Prints array([1.        , 1.        , 1.41421356, 2.        ])

可能有一些我遗漏的简化或其他可以加快速度的事情，但至少应该是一个好的开始.

There may well be some simplifications I'm missing or other things that could speed that up, but it should be a good start at least.

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