在 2D numpy 数组中找到给定角度的最近项目

Question

给定一个 numpy 二维数组，在给定角度的情况下，从指定坐标（'X' 所在的位置）获取最近项（对于本例“1”）的最佳方法是什么。

例如，假设我们有 'X' 位于二维数组中的 (1,25)，如下所示。假设角度为 225 度，假设 0 度向右笔直，90 度笔直向上。如何获得位于该矢量方向的“1”的最近坐标？

[
0000000000000000000000000000
0000000000000000000000000X00
0000000000000000000000000000
1110000000000000000000000000
1111100000000000000000000000
1111110000000000000000000000
1111111000000000000000000000
1111111110000000000000000000
1111111111100000000000000000
]

Answer 1

我假设朝着那个方向你的意思是像在那条射线上一样。那么255°无解所以我冒昧的改成了195°

然后您可以对其进行暴力破解：

import numpy as np

a = """
0000000000000000000000000000
0000000000000000000000000X00
0000000000000000000000000000
1110000000000000000000000000
1111100000000000000000000000
1111110000000000000000000000
1111111000000000000000000000
1111111110000000000000000000
1111111111100000000000000000
"""

a = np.array([[int(i) for i in row] for row in a.strip().replace('X', '2').split()], dtype=np.uint8)

x = np.argwhere(a==2)[0]
y = np.argwhere(a==1)
d = y-x

phi = 195 # 255 has no solutions

on_ray = np.abs(d@(np.sin(np.radians(-phi-90)), np.cos(np.radians(-phi-90))))<np.sqrt(0.5)

show_ray = np.zeros_like(a)
show_ray[tuple(y[on_ray].T)] = 1
print(show_ray)

ymin=y[on_ray][np.argmin(np.einsum('ij,ij->i', d[on_ray], d[on_ray]))]
print(ymin)

输出：

# [[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
#  [1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
# [6 6]

在 2D numpy 数组中找到给定角度的最近项目

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