3D tiling of a numpy array

Question

Provided a 1D array as a:

a=np.arange(8)

I would like it to be reproduced in a 3D scheme in order to have such shape (n1, len(a), n3). Is there any working way to obtain this via np.tile? It seems trivial, but trying:

np.shape(  np.tile(a, (n1,1,n3))  )

or

np.shape(  np.tile( np.tile(a, (n1,1)), (1,1,n2) )  )

I never obtain what I need, being the resulting shapes (n1, 1, len(a)*n3) or (1, n1, len(a)*n3). Maybe it is me not understanding how tile works ...

Answer 1

What's happening is that a is being made a 1x1x8 array before the tiling is applied. You'll need to make a a 1x8x1 array and then call tile.

As the documentation for tile notes:

If A.ndim < d, A is promoted to be d-dimensional by prepending new axes. So a shape (3,) array is promoted to (1, 3) for 2-D replication, or shape (1, 1, 3) for 3-D replication. If this is not the desired behavior, promote A to d-dimensions manually before calling this function.

The easiest way to get the result you're after is to slice a with None (or equivalently, np.newaxis) to make it the correct shape.

As a quick example:

import numpy as np

a = np.arange(8)
result = np.tile(a[None, :, None], (4, 1, 5))
print result.shape