Numpy N-Dimensional Array Unfolding

Question

I am basically trying to achieve this, but need the unfolding to be done in a different fashion. i want all samples of the N-1th dimension to be concatenated. For example, if my unfolding were to be applied to an RGB image of (100,100,3) the new array would basically become a (100,300) where the 3 colour channel images are now side by side in the new array.

All my attempts to use a neat built in numpy function like flatten and concatenate yielded no results. (flatten, because the end goal is to apply this unfolding until it is a 1D array)

Can't even think of a slicing way of doing it in a loop since the starting number of dimensions isn't constant (array = array[:,...,:,0]+...+array[:,...,:,0])

EDIT

I just came up with this way of achieving what I want, but would still welcome better, more pure, numpy solutions.

shape = numpy.random.randint(100, size=numpy.random.randint(100))
array = numpy.random.uniform(size=shape)

array = array.T
for i in range(0, len(shape)-1, -1):
    array = numpy.concatenate(array)

As mentioned, the image example is just an example. I need this behaviour to fold an N-Dimensional Array down to a 1D array recursively — WobbaFetttttt
– WobbaFetttttt, Commented Feb 10, 2017 at 18:43
"because the end goal is to apply this unfolding until it is a 1D array" - then why not just flatten your array to 1D directly? A single flatten or ravel call should be all you need. — user2357112
– user2357112, Commented Feb 10, 2017 at 18:44
Have you run the sample code that you have included in the question? It attempts to create a huge array. — Warren Weckesser
– Warren Weckesser, Commented Feb 10, 2017 at 18:58
yea its a randomly sized array, just change the 100 if you cannot run it on your device — WobbaFetttttt
– WobbaFetttttt, Commented Feb 10, 2017 at 19:04

hpaulj · Accepted Answer · 2017-02-10 22:12:43Z

0

Am I right in deducing that you want flatten the last 2 dimensions of the array?

In [96]: shape = numpy.random.randint(10, size=numpy.random.randint(10))+1
In [97]: shape
Out[97]: array([2, 7, 2])
In [98]: newshape=tuple(shape[:-2])+(-1,)
In [99]: arr = np.arange(np.prod(shape)).reshape(shape)
In [100]: arr.shape
Out[100]: (2, 7, 2)
In [101]: arr.reshape(newshape).shape
Out[101]: (2, 14)
In [102]: arr.reshape(newshape)
Out[102]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13],
       [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]])

If you don't like the order of terms in the last dimension, you may need to transpose

In [109]: np.swapaxes(arr, -1,-2).reshape(newshape)
Out[109]: 
array([[ 0,  2,  4,  6,  8, 10, 12,  1,  3,  5,  7,  9, 11, 13],
       [14, 16, 18, 20, 22, 24, 26, 15, 17, 19, 21, 23, 25, 27]])

I can't test it against your code because range(0,len(shape)-1, -1) is an empty range.

I don't think you want

In [112]: np.concatenate(arr,axis=-1).shape
Out[112]: (7, 4)
In [113]: np.concatenate((arr[0,...],arr[1,...]), axis=-1)
Out[113]: 
array([[ 0,  1, 14, 15],
       [ 2,  3, 16, 17],
       [ 4,  5, 18, 19],
       [ 6,  7, 20, 21],
       [ 8,  9, 22, 23],
       [10, 11, 24, 25],
       [12, 13, 26, 27]])

That splits the arr on the 1st axis, and then joins it on the last.

answered Feb 10, 2017 at 22:12

hpaulj

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1 Comment

WobbaFetttttt Over a year ago

thanks, not quite what i wanted, but it gave me some hints on my quest. my solutions does achieve what i want. I guess there is no 1 numpy function that achieves what I want

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