如果 2D 中的 4-connectivity 就足够了,您可以使用最近邻树在 n log n 时间内获得也是前景的相邻像素。
然后是构建图并找到连接组件(也是 n log n,IIRC)。
#!/usr/bin/env python
"""
https://stackoverflow.com/questions/66724201/connected-component-labling-for-arrays-quasi-images-with-many-dimension
"""
import numpy as np
import networkx as nx
from scipy.spatial import cKDTree
def get_components(boolean_array):
# find neighbours
coordinates = list(zip(*np.where(boolean_array)))
tree = cKDTree(coordinates)
neighbours_by_pixel = tree.query_ball_tree(tree,r=1,p=1) # p=1 -> Manhatten distance; r=1 -> what would be 4-connectivity in 2D
# create graph and find components
G = nx.Graph()
for ii,neighbours in enumerate(neighbours_by_pixel):
if len(neighbours) > 1:
G.add_edges_from([(ii,jj) for jj in neighbours[1:]]) # skip first neighbour as that is a self-loop
components = nx.connected_components(G)
# create output image
output = np.zeros_like(data,dtype=np.int)
for ii,component in enumerate(components):
for idx in component:
output[coordinates[idx]] = ii+1
return output
if __name__ == '__main__':
shape = (5,2,3,6,10)
D = len(shape)
data = np.random.rand(*shape) < 0.1
output = get_components(data)
对于形状为 (50,50,50) 的数组,我在笔记本电脑上得到以下时间:
In [48]: %timeit output = get_components(data)
5.85 s ± 279 ms per loop (mean ± std. dev. of 7 runs,1 loop each)
,
scipy.ndimage.label
直接做你想做的事:
In [1]: import numpy as np
In [2]: arr = np.random.random((5,10)) > 0.5
In [3]: from scipy import ndimage as ndi
In [4]: labeled,n = ndi.label(arr)
In [5]: n
Out[5]: 11
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