在二进制numpy数组中提取1的区域

Question

我正在寻找一种方法来提取二进制numpy数组中的1的区域，以放置不同的值，例如，用于以下数组:

x=[[0,1,1,0,0,0],
[0,1,1,0,0,0],
[0,1,0,0,0,0],
[0,0,0,1,1,0],
[0,0,1,1,1,0],
[0,0,0,0,0,0]]

预期结果:

x=[[0,2,2,0,0,0],
[0,2,2,0,0,0],
[0,2,0,0,0,0],
[0,0,0,3,3,0],
[0,0,3,3,3,0],
[0,0,0,0,0,0]]

Answer 1

使用scipy.ndimage.label:

x=[[0,1,1,0,0,0],
[0,1,1,0,0,0],
[0,1,0,0,0,0],
[0,0,0,1,1,0],
[0,0,1,1,1,0],
[0,0,0,0,0,0]]
a = np.array(x)
from scipy.ndimage import label
b = label(a)[0]

输出:

# b
array([[0, 1, 1, 0, 0, 0],
[0, 1, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 2, 2, 0],
[0, 0, 2, 2, 2, 0],
[0, 0, 0, 0, 0, 0]], dtype=int32)

从2:

开始标记

b = (label(a)[0]+1)*a

输出:

array([[0, 2, 2, 0, 0, 0],
[0, 2, 2, 0, 0, 0],
[0, 2, 0, 0, 0, 0],
[0, 0, 0, 3, 3, 0],
[0, 0, 3, 3, 3, 0],
[0, 0, 0, 0, 0, 0]])