numpy: extending arrays along a new axis?

Question

Surely there must be a way to do this... I can't work it out.

I have a (9,4) array, and I want to repeat it along a 3rd axis 4096 times... So it becomes simply (9,4,4096), with each value from the 9,4 array simply repeated 4096 times down the new axis.

If my dubious 3D diagram makes sense (the diagonal is a z-axis)

4|   /off to 4096
3|  /
2| /
1|/_ _ _ _ _ _ _ _ _ 
   1 2 3 4 5 6 7 8 9

Cheers

EDIT: Just to clarify, the emphasis here is on the (9,4) array being REPEATED for each of the 4096 'rows' of the new axis. Imagine a cross-section - each original (9,4) array is one of those down the 4096 long cuboid.

Solution 1

Here is one way:

import scipy
X = scipy.rand(9,4,1)
Y = X.repeat(4096,2)

If X is given to you as only (9,4), then

import scipy
X = scipy.rand(9,4)
Y = X.reshape(9,4,1).repeat(4096,2)

Solution 2

You can also rely on the broadcasting rules to repeat-fill a re-sized array:

import numpy
X = numpy.random.rand(9,4)
Y = numpy.resize(X,(4096,9,4))

If you don't like the axes ordered this way, you can then transpose:

Z = Y.transpose(1,2,0)

Solution 3

Question is super old, but here's another option anyway:

import numpy as np
X = np.random.rand(9,4)
Y = np.dstack([X] * 4096)

Solution 4

A simple numpy solution woudl be using numpy.tile

import numpy as np
a = np.random.rand(9, 4)
b = np.tile(a, (4096, 1, 1))

View more solutions

Comments

[abcd]

np.tile is another option. See answers to this question.

[Ajeet Ganga]

Also you have drawn diagram for 4 by 9 matrix. :)

[Duncan Tait]

Thanks again Steve, was getting confused because I didn't realise you had to reshape the array before using the repeat method - it kept telling me I was trying to use an axis that wasn't defined. Makes sense though now.

[blubberdiblub]

Instead of [X for i in range(4096)] you can just do [X] * 4096.