I have a numpy array of shape (X,Y,Z). I want to check each of the Z dimension and delete the non-zero dimension really fast.
Detailed explanation:
I would like to check array[:,:,0] if any entry is non-zero.
If yes, ignore and check array[:,:,1].
Else if No, delete dimension array[:,:,0]
Also not 100% sure what your after but I think you want
np.squeeze(array, axis=2)
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.squeeze.html
np.squeeze(your_array) or your_array.squeeze()
array.squeeze(axis=2) syntax though.I'm not certain what you want but this hopefully points in the right direction.
Edit 1st Jan:
Inspired by @J.Warren's use of np.squeeze I think np.compress may be more appropriate.
This does the compression in one line
np.compress((a!=0).sum(axis=(0,1)), a, axis=2) #
To explain the first parameter in np.compress
(a!=0).sum(axis=(0, 1)) # sum across both the 0th and 1st axes.
Out[37]: array([1, 1, 0, 0, 2]) # Keep the slices where the array !=0
My first answer which may no longer be relevant
import numpy as np
a=np.random.randint(2, size=(3,4,5))*np.random.randint(2, size=(3,4,5))*np.random.randint(2, size=(3,4,5))
# Make a an array of mainly zeroes.
a
Out[31]:
array([[[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[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, 0]],
[[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 0, 0],
[1, 0, 0, 0, 0]]])
res=np.zeros(a.shape[2], dtype=np.bool)
for ix in range(a.shape[2]):
res[ix] = (a[...,ix]!=0).any()
res
Out[34]: array([ True, True, False, False, True], dtype=bool)
# res is a boolean array of which slices of 'a' contain nonzero data
a[...,res]
# use this array to index a
# The output contains the nonzero slices
Out[35]:
array([[[0, 0, 0],
[0, 0, 1],
[0, 0, 0],
[0, 0, 0]],
[[0, 1, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 1],
[0, 0, 0],
[1, 0, 0]]])
