Having a 3-dimensional numpy array A I want to get all the edges (imagine this array as a cuboid).
有一个三维numpy数组A我想得到所有的边(想象这个数组作为一个长方体)。
Well, A[0, 0, :] would give me one edge, A[0, -1, :] second one and A[:, -1, -1] yet another one... so all I'd have to do is get all permutations of 0, -1 and : and use them as indices. Zero and minus one are easy, but how would I do this with colon?
好吧,A [0,0,:]会给我一个边,A [0,-1,:]第二个和A [:, - 1,-1]还有一个......所以我只有要做的是得到0,-1和的所有排列:并将它们用作索引。零和一个很容易,但我怎么用冒号呢?
I can solve it the long way, but it's ugly and I bet there is some neat numpy solution to this. Something like:
我可以解决它很长的路,但它很难看,我打赌这有一个整洁的numpy解决方案。就像是:
for indices in permutations([0, -1, ':']):
edge = A[indices]
...
What I want to do in the end is numpy.any() on the set of all edges to see if all edge-values are zero.
我最后要做的是在所有边的集合上使用numpy.any()来查看所有边值是否为零。
1 个解决方案
#1
1
: is same as slice(None, None, None)
:与slice相同(None,None,None)
A[0, -1, :] is same as
A [0,-1,:]与...相同
obj = (0, -1, slice(None, None, None))
A[obj]
#1
1
: is same as slice(None, None, None)
:与slice相同(None,None,None)
A[0, -1, :] is same as
A [0,-1,:]与...相同
obj = (0, -1, slice(None, None, None))
A[obj]