Repeating a Numpy array to match dimension of another array

Question

I'd like to repeat an array along multiple dimensions, such that it matches the dimensions of another array.

For example, lets put:

import numpy as np

a = np.arange(300)
b = np.zeros((300, 10, 20, 40, 50))

I'd like to expand a such that it matches the dimensions of b , considering than be can have an arbitrary number of dimensions of arbitrary length.

For now, the only thing I can do is a loop over the dimensions like:

c = np.copy(a)
for axis, length in enumerate(b.shape[1:]):
  c = np.repeat(c[..., None], length, axis + 1)

But it is rather inefficient for a large number of dimensions...

Answer 1

reshape and expand_dims can make an array that will broadcast like a b shaped array

tile can expand that. Check the code, but I think it does repeated repeats as you do:

d=np.tile(np.expand_dims(a,[1,2,3,4]),(1,)+b.shape[1:])

Another way to expand the array to full size, but without the full memory use is:

w = np.broadcast_to(np.expand_dims(a,[1,2,3,4]),b.shape)

w should have the same shape, but strides will be (8, 0, 0, 0, 0) , compared to (3200000, 320000, 16000, 400, 8) for b or c .

But making w.copy() will take nearly as long making c or d , because it has to make the full Gb array.

Answer 2

One option is to reshape a so it can be broadcasted against b , and then assign values to target array inplace:

c = np.zeros(b.shape)
c[:] = a.reshape(b.shape[:1] + (1,) * (len(b.shape) - 1))

Or use np.expand_dims to reshape:

c[:] = np.expand_dims(a, axis=tuple(range(1, len(b.shape))))

Answer 3

You could use np.empty_like together with np.copyto

c = np.emtpy_like(b)
np.copyto(c.T,a)

#check:
(c == a[:,None,None,None,None]).all()
# True

Or, if you want to retain a 's dtype:

c = np.empty(b.shape,a.dtype)
# etc.