3

假设我有一个这样的数组：

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 1s、3s、4s 和右 1s。

我将如何获得每个区域边界的路径？区域顶点的坐标，按顺序排列。

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

（我最终想得到类似的东西start at 0, 0. Right 2. Down 1. Right -1. Down 1. Right -1. Down -2.，但它很容易转换，因为它只是相邻顶点之间的差异）

我可以将其拆分为以下区域scipy.ndimage.label：

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)]

看起来更像这样：

{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
 ]
}

那么我如何将其转换为路径？

4

1 回答 1

0

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..

尝试可视化所有可能的坐标相邻单元组合，这将使您对可能的流动有一个直观的了解。另请注意，使用此方法应该不可能遍历所有 4 个邻居都为 False 的坐标。

于 2018-09-26T11:33:56.267 回答