I pretend to remove slices from the third dimension of a 4d numpy array if it’s contains only zeros.
I have a 4d numpy array of dimensions [256,256,336,6]
and I need to delete the slices in the third dimension that only contains zeros. So the result would have a shape like this , e.g. [256,256,300,6]
if 36 slices are fully zeros. I have tried multiple approaches including for loops, np.delete
and all()
, any()
functions without success.
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Answer
I’m not an afficionado with numpy, but does this do what you want?
I take the following small example matrix with 4 dimensions all full of 1s and then I set some slices to zero:
import numpy as np a=np.ones((4,4,5,2))
The shape of a
is:
>>> a.shape (4, 4, 5, 2)
I will artificially set some of the slices in dimension 3 to zero:
a[:,:,0,:]=0 a[:,:,3,:]=0
I can find the indices of the slices with not all zeros by calculating sums (not very efficient, perhaps!)
indices = [i for i in range(a.shape[2]) if a[:,:,i,:].sum() != 0] >>> indices [1, 2, 4]
So, in your general case you could do this:
indices = [i for i in range(a.shape[2]) if a[:,:,i,:].sum() != 0] a_new = a[:, :, indices, :].copy()
Then the shape of a_new is:
>>> anew.shape (4, 4, 3, 2)