14
votes

The question is the inverse of this question. I'm looking for a generic method to from the original big array from small arrays:

array([[[ 0,  1,  2],
        [ 6,  7,  8]],    
       [[ 3,  4,  5],
        [ 9, 10, 11]], 
       [[12, 13, 14],
        [18, 19, 20]],    
       [[15, 16, 17],
        [21, 22, 23]]])

->

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]])

I am currently developing a solution, will post it when it's done, would however like to see other (better) ways.

5

5 Answers

36
votes
import numpy as np
def blockshaped(arr, nrows, ncols):
    """
    Return an array of shape (n, nrows, ncols) where
    n * nrows * ncols = arr.size

    If arr is a 2D array, the returned array looks like n subblocks with
    each subblock preserving the "physical" layout of arr.
    """
    h, w = arr.shape
    return (arr.reshape(h//nrows, nrows, -1, ncols)
               .swapaxes(1,2)
               .reshape(-1, nrows, ncols))


def unblockshaped(arr, h, w):
    """
    Return an array of shape (h, w) where
    h * w = arr.size

    If arr is of shape (n, nrows, ncols), n sublocks of shape (nrows, ncols),
    then the returned array preserves the "physical" layout of the sublocks.
    """
    n, nrows, ncols = arr.shape
    return (arr.reshape(h//nrows, -1, nrows, ncols)
               .swapaxes(1,2)
               .reshape(h, w))

For example,

c = np.arange(24).reshape((4,6))
print(c)
# [[ 0  1  2  3  4  5]
#  [ 6  7  8  9 10 11]
#  [12 13 14 15 16 17]
#  [18 19 20 21 22 23]]

print(blockshaped(c, 2, 3))
# [[[ 0  1  2]
#   [ 6  7  8]]

#  [[ 3  4  5]
#   [ 9 10 11]]

#  [[12 13 14]
#   [18 19 20]]

#  [[15 16 17]
#   [21 22 23]]]

print(unblockshaped(blockshaped(c, 2, 3), 4, 6))
# [[ 0  1  2  3  4  5]
#  [ 6  7  8  9 10 11]
#  [12 13 14 15 16 17]
#  [18 19 20 21 22 23]]

Note that there is also superbatfish's blockwise_view. It arranges the blocks in a different format (using more axes) but it has the advantage of (1) always returning a view and (2) being capable of handing arrays of any dimension.

4
votes

Yet another (simple) approach:

threedarray = ...
twodarray = np.array(map(lambda x: x.flatten(), threedarray))
print(twodarray.shape)
1
votes

I hope I get you right, let's say we have a,b :

>>> a = np.array([[1,2] ,[3,4]])
>>> b = np.array([[5,6] ,[7,8]])
    >>> a
    array([[1, 2],
           [3, 4]])
    >>> b
    array([[5, 6],
           [7, 8]])

in order to make it one big 2d array use numpy.concatenate:

>>> c = np.concatenate((a,b), axis=1 )
>>> c
array([[1, 2, 5, 6],
       [3, 4, 7, 8]])
0
votes

It works for the images I tested for now. Will if further tests are made. It is however a solution which takes no account about speed and memory usage.

def unblockshaped(blocks, h, w):
    n, nrows, ncols = blocks.shape
    bpc = w/ncols
    bpr = h/nrows

    reconstructed = zeros((h,w))
    t = 0
    for i in arange(bpr):
        for j in arange(bpc):
            reconstructed[i*nrows:i*nrows+nrows,j*ncols:j*ncols+ncols] = blocks[t]
            t = t+1
    return reconstructed
0
votes

Here is a solution that one can use if someone is wishing to create tiles of a matrix:

from itertools import product
import numpy as np
def tiles(arr, nrows, ncols):
    """
    If arr is a 2D array, the returned list contains nrowsXncols numpy arrays
    with each array preserving the "physical" layout of arr.

    When the array shape (rows, cols) are not divisible by (nrows, ncols) then
    some of the array dimensions can change according to numpy.array_split.

    """
    rows, cols = arr.shape
    col_arr = np.array_split(range(cols), ncols)
    row_arr = np.array_split(range(rows), nrows)
    return [arr[r[0]: r[-1]+1, c[0]: c[-1]+1]
                     for r, c in product(row_arr, col_arr)]