Extract blocks or patches from NumPy Array

问题:

I have a 2-d numpy array as follows:

a = np.array([[1,5,9,13],
              [2,6,10,14],
              [3,7,11,15],
              [4,8,12,16]]

I want to extract it into patches of 2 by 2 sizes with out repeating the elements.

The answer should exactly be the same. This can be 3-d array or list with the same order of elements as below:

[[[1,5],
 [2,6]],   

 [[3,7],
 [4,8]],

 [[9,13],
 [10,14]],

 [[11,15],
 [12,16]]]

How can do it easily?

In my real problem the size of a is (36, 72). I can not do it one by one. I want programmatic way of doing it.

回答1:

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

In [60]: x
Out[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.shape
Out[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]])

回答2:

Using scikit-image:

import numpy as np
from skimage.util import view_as_blocks

a = 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)))

回答3:

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

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

Sample cases

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

In [94]: a
Out[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 two cases with blocksizes of (2,3) and (3,3).

Case #1 :

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]]])

Case #2 :

In [96]: extract_blocks(a, (3,3)) # Blocksize : (3,3)
Out[96]: 
array([[[2, 2, 6],
        [1, 0, 1],
        [4, 0, 0]],

       [[1, 3, 6],
        [0, 0, 3],
        [4, 1, 7]],

       [[3, 2, 4],
        [8, 0, 7],
        [1, 5, 6]],

       [[7, 2, 4],
        [3, 4, 6],
        [2, 1, 8]]])