How to create or fill an numpy array with another

Question

How to create an numpy array with shape [2, 2, 3], where the elements at axis 2 is another array, for example [1, 2, 3]?

So I would like to do something like this invalid code:

a = np.arange(1, 4)
b = np.full((3, 3), a)

Resulting in an array like:

[[[ 1.  2.  3.]
  [ 1.  2.  3.]]
 [[ 1.  2.  3.]
  [ 1.  2.  3.]]]

Could of course make the loop for filling like, but thought there may be a shortcut:

for y in range(b.shape[0]):
    for x in range(b.shape[1]):
        b[y, x, :] = a

Answer 1

Also using np.concatenate or it's wrapper np.vstack

In [26]: a = np.arange(1,4)

In [27]: np.vstack([a[np.newaxis, :]]*4).reshape(2,2, 3)
Out[27]: 
array([[[1, 2, 3],
        [1, 2, 3]],

       [[1, 2, 3],
        [1, 2, 3]]])

In [28]: np.concatenate([a[np.newaxis, :]]*4, axis=0).reshape(2,2, 3)
Out[28]: 
array([[[1, 2, 3],
        [1, 2, 3]],

       [[1, 2, 3],
        [1, 2, 3]]])

Answer 2

If your numpy version is >= 1.10 you can use broadcast_to

a = np.arange(1,4)
a.shape = (1,1,3)
b = np.broadcast_to(a,(2,2,3))

This produces a view rather than copying so will be quicker for large arrays. EDIT this looks to be the result you're asking for with your demo.

Answer 3

Based on Divakar comment, an answer can also be:

import numpy as np
np.full([2, 2, 3], np.arange(1, 4))

Yet another possibility is:

import numpy as np
b = np.empty([2, 2, 3])
b[:] = np.arange(1, 4)

Answer 4

There are multiple ways to achieve this. One is to use np.full in np.full((2,2,3), a) as pointed out by Divakar in the comments. Alternatively, you can use np.tile for this, which allows you to construct an array by repeating an input array a given number of times. To construct your example you could do:

import numpy as np

np.tile(np.arange(1, 4), [2, 2, 1])