Extract blocks or patches from NumPy Array

Using scikit-image:

import numpy as npfrom skimage.util import view_as_blocksa = np.array([[1,5,9,13],              [2,6,10,14],              [3,7,11,15],              [4,8,12,16]])print(view_as_blocks(a, (2, 2)))

You can achieve it with a combination of np.reshape and np.swapaxes like so -

def extract_blocks(a, blocksize, keep_as_view=False):    M,N = a.shape    b0, b1 = blocksize    if keep_as_view==0:        return a.reshape(M//b0,b0,N//b1,b1).swapaxes(1,2).reshape(-1,b0,b1)    else:        return a.reshape(M//b0,b0,N//b1,b1).swapaxes(1,2)

As can be seen there are two ways to use it - With keep_as_view flag turned off (default one) or on. With keep_as_view = False, we are reshaping the swapped-axes to a final output of 3D, while with keep_as_view = True, we will keep it 4D and that will be a view into the input array and hence, virtually free on runtime. We will verify it with a sample case run later on.

Sample cases

Let's use a sample input array, like so -

In [94]: aOut[94]: array([[2, 2, 6, 1, 3, 6],       [1, 0, 1, 0, 0, 3],       [4, 0, 0, 4, 1, 7],       [3, 2, 4, 7, 2, 4],       [8, 0, 7, 3, 4, 6],       [1, 5, 6, 2, 1, 8]])

Now, let's use some block-sizes for testing. Let's use a blocksize of (2,3) with the view-flag turned off and on -

In [95]: extract_blocks(a, (2,3)) # Blocksize : (2,3)Out[95]: array([[[2, 2, 6],        [1, 0, 1]],       [[1, 3, 6],        [0, 0, 3]],       [[4, 0, 0],        [3, 2, 4]],       [[4, 1, 7],        [7, 2, 4]],       [[8, 0, 7],        [1, 5, 6]],       [[3, 4, 6],        [2, 1, 8]]])In [48]: extract_blocks(a, (2,3), keep_as_view=True)Out[48]: array([[[[2, 2, 6],         [1, 0, 1]],        [[1, 3, 6],         [0, 0, 3]]],       [[[4, 0, 0],         [3, 2, 4]],        [[4, 1, 7],         [7, 2, 4]]],       [[[8, 0, 7],         [1, 5, 6]],        [[3, 4, 6],         [2, 1, 8]]]])

Verify view with keep_as_view=True

In [20]: np.shares_memory(a, extract_blocks(a, (2,3), keep_as_view=True))Out[20]: True

Let's check out performance on a large array and verify the virtually free runtime claim as discussed earlier -

In [42]: a = np.random.rand(2000,3000)In [43]: %timeit extract_blocks(a, (2,3), keep_as_view=True)1000000 loops, best of 3: 801 ns per loopIn [44]: %timeit extract_blocks(a, (2,3), keep_as_view=False)10 loops, best of 3: 29.1 ms per loop

Here's a rather cryptic numpy one-liner to generate your 3-d array, called result1 here:

In [60]: xOut[60]: array([[2, 1, 2, 2, 0, 2, 2, 1, 3, 2],       [3, 1, 2, 1, 0, 1, 2, 3, 1, 0],       [2, 0, 3, 1, 3, 2, 1, 0, 0, 0],       [0, 1, 3, 3, 2, 0, 3, 2, 0, 3],       [0, 1, 0, 3, 1, 3, 0, 0, 0, 2],       [1, 1, 2, 2, 3, 2, 1, 0, 0, 3],       [2, 1, 0, 3, 2, 2, 2, 2, 1, 2],       [0, 3, 3, 3, 1, 0, 2, 0, 2, 1]])In [61]: result1 = x.reshape(x.shape[0]//2, 2, x.shape[1]//2, 2).swapaxes(1, 2).reshape(-1, 2, 2)

result1 is like a 1-d array of 2-d arrays:

In [68]: result1.shapeOut[68]: (20, 2, 2)In [69]: result1[0]Out[69]: array([[2, 1],       [3, 1]])In [70]: result1[1]Out[70]: array([[2, 2],       [2, 1]])In [71]: result1[5]Out[71]: array([[2, 0],       [0, 1]])In [72]: result1[-1]Out[72]: array([[1, 2],       [2, 1]])

(Sorry, I don't have time at the moment to give a detailed breakdown of how it works. Maybe later...)

Here's a less cryptic version that uses a nested list comprehension. In this case, result2 is a python list of 2-d numpy arrays:

In [73]: result2 = [x[2*j:2*j+2, 2*k:2*k+2] for j in range(x.shape[0]//2) for k in range(x.shape[1]//2)]In [74]: result2[5]Out[74]: array([[2, 0],       [0, 1]])In [75]: result2[-1]Out[75]: array([[1, 2],       [2, 1]])