布尔二维数组的子域

Question

假设我有一个二维布尔数组。我想得到一个切片列表，其中每个切片表示数组中包含真值的最小（大小）子区域，而其边框包含所有假值

我可以为每一行和每一列循环，并在满足这样的条件时存储索引，但我想知道您是否知道另一种方法或库可以有效地做到这一点？可以假定原始布尔数组的边界始终为False/0

例1

例2

编辑！添加了具有正确解决方案的新示例。很抱歉给你带来了困惑

Answer 1

您可以从图形的角度来处理这个问题，其中的坐标是图形元素，8路连接-然后您只需在图形中找到连接的组件。如果数据是稀疏的，这应该比在可能的正方形大小中循环要快得多。这是它如何工作的一个例子：

from itertools import combinations

def find_squares(a):
    # Find ones
    ones = [(i, j) for i, row in enumerate(a) for j, val in enumerate(row) if val]
    # Make graph of connected ones
    graph = {a: [] for a in ones}
    for a, b in combinations(ones, 2):
        if abs(a[0] - b[0]) <= 1 and abs(a[1] - b[1]) <= 1:
            graph[a].append(b)
            graph[b].append(a)
    # Find connected components in graph
    components = []
    for a, a_neigh in graph.items():
        if any(a in c for c in components):
            continue
        component = {a, *a_neigh}
        pending = [*a_neigh]
        visited = {a}
        while pending:
            b = pending.pop()
            if b in visited:
                continue
            visited.add(b)
            component.add(b)
            b_neigh = graph[b]
            component.update(b_neigh)
            pending.extend(c for c in b_neigh if c not in visited)
        components.append(component)
    # Find bounds for each component
    bounds = [((min(a[0] for a in c), min(a[1] for a in c)),
               (max(a[0] for a in c), max(a[1] for a in c)))
              for c in components]
    return bounds

a = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 0, 0, 0, 1, 0, 1, 0, 0],
     [0, 0, 0, 1, 0, 0, 1, 0, 1, 0],
     [0, 0, 0, 0, 0, 0, 0, 1, 0, 0],
     [0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
     [0, 1, 0, 0, 1, 0, 0, 0, 0, 0],
     [0, 0, 0, 0, 1, 1, 0, 0, 1, 0],
     [0, 0, 0, 0, 1, 0, 1, 0, 0, 0],
     [0, 0, 0, 1, 0, 0, 0, 0, 1, 0],
     [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
square_bounds = find_squares(a)
print(*square_bounds, sep='\n')
# ((1, 5), (3, 8))
# ((2, 3), (2, 3))
# ((4, 1), (5, 1))
# ((5, 3), (8, 6))
# ((6, 8), (6, 8))
# ((8, 8), (8, 8))

Answer 2

这就是连通成分分析，它已经被asked and answered before。根据您的需要调整公认的答案，一个可能的解决方案非常简短：

import numpy as np
from scipy.ndimage.measurements import label


def analysis(array):
    labeled, _ = label(array, np.ones((3, 3), dtype=np.int))
    for i in np.arange(1, np.max(labeled)+1):
        pixels = np.array(np.where(labeled == i))
        x1 = np.min(pixels[1, :])
        x2 = np.max(pixels[1, :])
        y1 = np.min(pixels[0, :])
        y2 = np.max(pixels[0, :])
        print(str(i) + ' | slice: array[' + str(y1) + ':' + str(y2) + ', ' + str(x1) + ':' + str(x2) + ']')


example1 = np.array([
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 1, 0, 1, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 1, 1, 0],
    [0, 0, 0, 0, 0, 0, 1, 1, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]).astype(bool)

example2 = np.array([
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 1, 0, 1, 0, 0],
    [0, 0, 0, 1, 0, 0, 1, 0, 1, 0],
    [0, 0, 0, 0, 0, 0, 0, 1, 0, 0],
    [0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 1, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 0, 0, 1, 0],
    [0, 0, 0, 0, 1, 0, 1, 0, 0, 0],
    [0, 0, 0, 1, 0, 0, 0, 0, 1, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]).astype(bool)

for a in [example1, example2]:
    print(a, '\n')
    analysis(a)
    print('\n')

这就是输出（没有示例）：

[[...]] 

1 | slice: array[1:2, 3:5]
2 | slice: array[4:6, 6:8]
3 | slice: array[8:8, 2:2]

[[...]] 

1 | slice: array[1:3, 5:8]
2 | slice: array[2:2, 3:3]
3 | slice: array[4:5, 1:1]
4 | slice: array[5:8, 3:6]
5 | slice: array[6:6, 8:8]
6 | slice: array[8:8, 8:8]

希望有帮助

         
System information
         
Python:  3.8.1
SciPy:   1.4.1