Using numpy to construct an array with rows extracted from another 2D array as 2×2 blocks

Suppose I have the following 2D array:

x = np.array([[10,20,30,40], [50,60,70,80],[90,100,110,120]])  
print(x)

array([[ 10,  20,  30,  40],
       [ 50,  60,  70,  80],
       [ 90, 100, 110, 120]])

I would like to construct a new array, y, that contains the following:

print(y)
array([[ 10,  20,  60,  50],
       [ 20,  30,  70,  60],
       [ 30,  40,  80,  70],
       [ 50,  60,  100, 90],
       [ 60,  70,  110, 100],
       [ 70,  80,  120, 110]])

I could achieve that using Python for loops as follows:

n_rows, n_cols = x.shape
y = []
for i in range(n_rows-1): 
     for j in range(n_cols-1): 
         row = [x[i,j],x[i,j+1],x[i+1, j+1],x[i+1,j]] 
         y.append(row) 
y = np.array(y)

I wonder if there is a faster way that takes advantage of Numpy functions and avoid using Python loops.

Thank you

>Solution :

First, create a sliding_window_view into x with the 2×2 boxes you want to see:

b = np.lib.stride_tricks.sliding_window_view(x, (2, 2))

Each of the innermost 2×2 arrays contains an unraveled version of what you want, but with the second part of the array reversed. So far we didn’t copy any data. Now make a copy by raveling the last dimension. The reshape will always make a copy here because b is highly non-contiguous:

c = b.reshape(*b.shape[:2], 4)

Swap the last two columns:

c[..., 2:] = c[..., -1:1:-1]

Now ravel the leading dimensions:

y = c.reshape(-1, c.shape[-1])

If you have a version of numpy that is older whan 1.20, you can replace the definition of b with

b = np.lib.stride_tricks.as_strided(x, shape=(x.shape[0] - 1, x.shape[1] - 1, 2, 2), strides=x.strides * 2)