计算给定线 start(x,y) end(x,y) 上的投影点位置 (x,y)

Question

如果我有三个点 P1、P2、P3 及其坐标（x，y）

P1(x,y) 和 P3(x,y) 是线（起点，终点）的坐标，P3 是需要投影的点。

我怎样才能找到点 r(x,y) 的坐标，它是 P3 在 P1 和 P2 上的投影

Answer 1

适用于 2D 和 3D 的 Numpy 代码（基于https://gamedev.stackexchange.com/questions/72528/how-can-i-project-a-3d-point-onto-a-3d-line）：

import numpy as np

def point_on_line(a, b, p):
    ap = p - a
    ab = b - a
    result = a + np.dot(ap, ab) / np.dot(ab, ab) * ab
    return result

A = np.array([2, 0])
B = np.array([4, 4])
P = np.array([1, 3])
projected = point_on_line(A, B, P)
print(projected)

更新

阴谋：

A = np.array([ 10.5, 15.6 ])
B = np.array([ 2, 6 ])
P = np.array([ 18.561, -19.451])

projected = point_on_line(A, B, P) 
print(projected)
# [-3.35411076 -0.04699568]

plt.xlim(-20, 20)
plt.ylim(-20, 20)
plt.axis('equal')

x_values = [A[0], B[0]]
y_values = [A[1], B[1]]

plt.plot(B[0], B[1], 'ro')
plt.plot(A[0], A[1], 'ro')
plt.plot(P[0], P[1], 'ro')
plt.plot(x_values, y_values, 'b-')
plt.plot(projected[0], projected[1], 'rx')

更新2

如果需要点属于线段，则需要做一个小修改

def point_on_line(a, b, p):
    ap = p - a
    ab = b - a
    t = np.dot(ap, ab) / np.dot(ab, ab)
    # if you need the the closest point belonging to the segment
    t = max(0, min(1, t))
    result = a + t * ab
    return result

Answer 2

此解决方案扩展到具有任何几何尺寸（2D、3D、4D 等）的点。它假设所有点都是一维 numpy 数组（或一维形状为 1 的二维）。我不确定您是否需要投影落在线段或线段的延伸上，所以我将两者都包括在内。您可以选择最适合您的问题的：

#distance between p1 and p2
l2 = np.sum((p1-p2)**2)
if l2 == 0:
  print('p1 and p2 are the same points')

#The line extending the segment is parameterized as p1 + t (p2 - p1).
#The projection falls where t = [(p3-p1) . (p2-p1)] / |p2-p1|^2

#if you need the point to project on line extention connecting p1 and p2
t = np.sum((p3 - p1) * (p2 - p1)) / l2

#if you need to ignore if p3 does not project onto line segment
if t > 1 or t < 0:
  print('p3 does not project onto p1-p2 line segment')

#if you need the point to project on line segment between p1 and p2 or closest point of the line segment
t = max(0, min(1, np.sum((p3 - p1) * (p2 - p1)) / l2))

projection = p1 + t * (p2 - p1)