获取二维数组中相邻区域边界的路径

Question

假设我有这样一个数组：

import numpy as np

arr = np.array([
   [1, 1, 3, 3, 1],
   [1, 3, 3, 1, 1],
   [4, 4, 3, 1, 1],
   [4, 4, 1, 1, 1]
])

有4个不同的区域：左上角的1s、3s、4s和右侧的1s。

如何获取每个区域边界的路径？按顺序列出区域的顶点坐标。

例如，对于左上角的1s，它是(0, 0), (0, 2), (1, 2), (1, 1), (2, 1), (2, 0)。

（最终我想得到类似于"从0,0开始。向右2。向下1。向右-1。向下1。向右-1。向下-2"这样的东西，但很容易转换，因为它只是相邻顶点之间的差异）

我可以使用scipy.ndimage.label将其分成区域：

from scipy.ndimage import label

regions = {}
# region_value is the number in the region
for region_value in np.unique(arr):
    labeled, n_regions = label(arr == region_value)
    regions[region_value] = [labeled == i for i in range(1, n_regions + 1)]

这看起来更像是这样：

{1: [
    array([
        [ True,  True, False, False, False],
        [ True, False, False, False, False],
        [False, False, False, False, False],
        [False, False, False, False, False]
    ], dtype=bool),  # Top left 1s region
    array([
        [False, False, False, False,  True],
        [False, False, False,  True,  True],
        [False, False, False,  True,  True],
        [False, False,  True,  True,  True]
    ], dtype=bool)  # Right 1s region
 ],
 3: [
    array([
        [False, False,  True,  True, False],
        [False,  True,  True, False, False],
        [False, False,  True, False, False],
        [False, False, False, False, False]
    ], dtype=bool)  # 3s region
 ],
 4: [
    array([
        [False, False, False, False, False],
        [False, False, False, False, False],
        [ True,  True, False, False, False],
        [ True,  True, False, False, False]
    ], dtype=bool)  # 4s region
 ]
}

那么我该如何将其转换为路径？

- Artyer

---

只是一个随机的想法。使用现有的轮廓线代码: https://dev59.com/7WMl5IYBdhLWcg3wkXlx#18309914 - Paul Panzer

Answer 1

一个伪代码的想法是执行以下操作：

scan multi-dim array horizontally and then vertically until you find True value (for second array it is (0,4))
output that as a start coord
since you have been scanning as determined above your first move will be to go right.
repeat until you come back:
    move one block in the direction you are facing.
    you are now at coord x,y
    check values of ul=(x-1, y-1), ur=(x-1, y), ll=(x, y-1), lr=(x,y)
    # if any of above is out of bounds, set it as False
    if ul is the only True:
         if previous move right:
             next move is up
         else:
             next move is left
         output previous move
         move by one
   ..similarly for other single True cells..
   elif ul and ur only True or ul and ll only True or ll and lr only True or ur and lr only True:
        repeat previous move
   elif ul and lr only True:
        if previous move left:
            next move down
        elif previous move right:
            next move up
        elif preivous move down:
            next move left:
        else:
            next move right
        output previous move
        move one
   elif ul, ur, ll only Trues:
        if previous move left:
            next move down
        else:
            next move right
        output previous move, move by one
   ...similarly for other 3 True combos...

对于第二个数组，它将执行以下操作：

finds True val at 0,4
start at 0,4
only lower-right cell is True, so moves right to 0,5 (previous move is None, so no output)
now only lower-left cell is True, so moves down to 1,5 (previous move right 1 is output)
now both left cells are True, so repeat move (moves down to 2,5)
..repeat until hit 4,5..
only upper-left cell is True, so move left (output down 4)
both upper cells are true, repeat move (move left to 3,4)
both upper cells are true, repeat move (move left to 2,4)
upper right cell only true, so move up (output right -3)
..keep going until back at 0,4..

尝试将所有可能的相邻单元格组合可视化，这将为您提供可能的流动的视觉想法。 还要注意，使用此方法应该不可能遍历具有4个邻居都为False的坐标。

- Landar