如何找到离线最近的点？

Question

I have a point cloud stored as a numpy array. There is also a line coming through that point cloud specified with two points.

How can I get the closest point to that line in the most optimized version?

I have calculated distance to every point one by one, but it takes to much time to calculate that and it freezes my app....

Please help me to optimize that;(

This is how I have been doing it for every point:

def isectSphere(self, p0, p1, cpt):

        # normalized ray direction
        r_dir = np.subtract(p0, p1)
        r_dir = r_dir / np.linalg.norm(r_dir)

        # nearest point on the ray to the sphere
        p0_cpt = np.subtract(p0, cpt)
        near_pt = np.subtract(p0, r_dir * np.dot(p0_cpt, r_dir))

        # distance to center point
        return np.linalg.norm(np.subtract(near_pt, cpt))

After iterating that code on every point a have been taking the min out of it.

In the point cloud there is around 6 000 000 points.

Answer 1

You can use vectorized calculations:

Assuming p0 and p1 are of shape (m,) and pc is your point cloud array of shape (N, m) , you can use np.cross to calculate vectorized distance:

closest_point = pc[np.argmin(np.linalg.norm(np.cross(p1-p0, p0-pc, axisb=1), axis=1)/np.linalg.norm(p1-p0))]

This on a personal system, takes less than a second for over 6,000,000 points.

Answer 2

import numpy as np


def isectSphere(p0, p1, cloud):
        """
        >>> isectSphere([1, 0], [3, 0], [[0, -4], [2, 3]])
        1
        >>> isectSphere([1, 0, 0], [3, 0, 0], [[0, -4, 0], [2, 3, 0]])
        1
        """
        p0 = np.asarray(p0)
        p1 = np.asarray(p1)
        cloud = np.asarray(cloud)
        product = np.cross(cloud - p0, p1 - p0)
        if product.ndim == 2:
            distances = np.linalg.norm(product, axis=1)
        else:
            distances = np.abs(product)
        return distances.argmin()