I have a 2D numpy array, say array1 with values. array1 is of dimensions 2x4. I want to create a 4D numpy array array2 with dimensions 20x20x2x4 and I wish to replicate the array array1 to get this array.
That is, if array1 was
[[1, 2, 3, 4],
[5, 6, 7, 8]]
I want
array2[0, 0] = array1
array2[0, 1] = array1
array2[0, 2] = array1
array2[0, 3] = array1
# etc.
How can I do this?
One approach with initialization -
array2 = np.empty((20,20) + array1.shape,dtype=array1.dtype)
array2[:] = array1
Runtime test -
In [400]: array1 = np.arange(1,9).reshape(2,4)
In [401]: array1
Out[401]:
array([[1, 2, 3, 4],
[5, 6, 7, 8]])
# @MSeifert's soln
In [402]: %timeit np.tile(array1, (20, 20, 1, 1))
100000 loops, best of 3: 8.01 µs per loop
# Proposed soln in this post
In [403]: %timeit initialization_based(array1)
100000 loops, best of 3: 4.11 µs per loop
# @MSeifert's soln for READONLY-view
In [406]: %timeit np.broadcast_to(array1, (20, 20, 2, 4))
100000 loops, best of 3: 2.78 µs per loop
There are two easy ways:
array2 = np.broadcast_to(array1, (20, 20, 2, 4)) # array2 is a READONLY-view
and np.tile:
array2 = np.tile(array1, (20, 20, 1, 1)) # array2 is a normal numpy array
If you don't want to modify your array2 then np.broadcast_to should be really fast and simple. Otherwise np.tile or assigning to a new allocated array (see Divakars answer) should be preferred.
i got the answer.
array2[:, :, :, :] = array1.copy()
this should work fine
array2, you don't need that copy().
:, :, :, :, one simple : would also work (by broadcasting).
array1.copy() is un-necessary. You are assigned values into array2, which could directly be fed from array1 itself, rather than making a copy of array1 and then feeding from it. Thus, you could simply do : array2[:, :, :, :] = array1.