Skip to content
Advertisement

Implement MATLAB’s im2col ‘sliding’ in Python

Q: How to speed this up?

Below is my implementation of Matlab’s im2col ‘sliding’ with the additional feature of returning every n’th column. The function takes an image (or any 2 dim array) and slides from left to right, top to bottom, picking off every overlapping sub-image of a given size, and returning an array whose columns are the sub-images.

import numpy as np

def im2col_sliding(image, block_size, skip=1):

    rows, cols = image.shape
    horz_blocks = cols - block_size[1] + 1
    vert_blocks = rows - block_size[0] + 1

    output_vectors = np.zeros((block_size[0] * block_size[1], horz_blocks * vert_blocks))
    itr = 0
    for v_b in xrange(vert_blocks):
        for h_b in xrange(horz_blocks):
            output_vectors[:, itr] = image[v_b: v_b + block_size[0], h_b: h_b + block_size[1]].ravel()
            itr += 1

    return output_vectors[:, ::skip]

example:

a = np.arange(16).reshape(4, 4)
print a
print im2col_sliding(a, (2, 2))  # return every overlapping 2x2 patch
print im2col_sliding(a, (2, 2), 4)  # return every 4th vector

returns:

[[ 0  1  2  3]
 [ 4  5  6  7]
 [ 8  9 10 11]
 [12 13 14 15]]
[[  0.   1.   2.   4.   5.   6.   8.   9.  10.]
 [  1.   2.   3.   5.   6.   7.   9.  10.  11.]
 [  4.   5.   6.   8.   9.  10.  12.  13.  14.]
 [  5.   6.   7.   9.  10.  11.  13.  14.  15.]]
[[  0.   5.  10.]
 [  1.   6.  11.]
 [  4.   9.  14.]
 [  5.  10.  15.]]

The performance isn’t great, especially considering whether I call im2col_sliding(big_matrix, (8, 8)) (62001 columns) or im2col_sliding(big_matrix, (8, 8), 10) (6201 columns; keeping only every 10th vector) it will take the same amount of time [where big_matrix is of size 256 x 256].

I’m looking for any ideas to speed this up.

Advertisement

Answer

Approach #1

We could use some broadcasting here to get all the indices of all those sliding windows in one go and thus with indexing achieve a vectorized solution. This is inspired by Efficient Implementation of im2col and col2im.

Here’s the implementation –

def im2col_sliding_broadcasting(A, BSZ, stepsize=1):
    # Parameters
    M,N = A.shape
    col_extent = N - BSZ[1] + 1
    row_extent = M - BSZ[0] + 1
    
    # Get Starting block indices
    start_idx = np.arange(BSZ[0])[:,None]*N + np.arange(BSZ[1])
    
    # Get offsetted indices across the height and width of input array
    offset_idx = np.arange(row_extent)[:,None]*N + np.arange(col_extent)
    
    # Get all actual indices & index into input array for final output
    return np.take (A,start_idx.ravel()[:,None] + offset_idx.ravel()[::stepsize])

Approach #2

Using newly gained knowledge of NumPy array strides that lets us create such sliding windows, we would have another efficient solution –

def im2col_sliding_strided(A, BSZ, stepsize=1):
    # Parameters
    m,n = A.shape
    s0, s1 = A.strides    
    nrows = m-BSZ[0]+1
    ncols = n-BSZ[1]+1
    shp = BSZ[0],BSZ[1],nrows,ncols
    strd = s0,s1,s0,s1
    
    out_view = np.lib.stride_tricks.as_strided(A, shape=shp, strides=strd)
    return out_view.reshape(BSZ[0]*BSZ[1],-1)[:,::stepsize]

Approach #3

The strided method listed in the previous approach has been incorporated into scikit-image module for a less messier, like so –

from skimage.util import view_as_windows as viewW

def im2col_sliding_strided_v2(A, BSZ, stepsize=1):
    return viewW(A, (BSZ[0],BSZ[1])).reshape(-1,BSZ[0]*BSZ[1]).T[:,::stepsize]

Sample runs –

In [106]: a      # Input array
Out[106]: 
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19]])

In [107]: im2col_sliding_broadcasting(a, (2,3))
Out[107]: 
array([[ 0,  1,  2,  5,  6,  7, 10, 11, 12],
       [ 1,  2,  3,  6,  7,  8, 11, 12, 13],
       [ 2,  3,  4,  7,  8,  9, 12, 13, 14],
       [ 5,  6,  7, 10, 11, 12, 15, 16, 17],
       [ 6,  7,  8, 11, 12, 13, 16, 17, 18],
       [ 7,  8,  9, 12, 13, 14, 17, 18, 19]])

In [108]: im2col_sliding_broadcasting(a, (2,3), stepsize=2)
Out[108]: 
array([[ 0,  2,  6, 10, 12],
       [ 1,  3,  7, 11, 13],
       [ 2,  4,  8, 12, 14],
       [ 5,  7, 11, 15, 17],
       [ 6,  8, 12, 16, 18],
       [ 7,  9, 13, 17, 19]])

Runtime test

In [183]: a = np.random.randint(0,255,(1024,1024))

In [184]: %timeit im2col_sliding(img, (8,8), skip=1)
     ...: %timeit im2col_sliding_broadcasting(img, (8,8), stepsize=1)
     ...: %timeit im2col_sliding_strided(img, (8,8), stepsize=1)
     ...: %timeit im2col_sliding_strided_v2(img, (8,8), stepsize=1)
     ...: 
1 loops, best of 3: 1.29 s per loop
1 loops, best of 3: 226 ms per loop
10 loops, best of 3: 84.5 ms per loop
10 loops, best of 3: 111 ms per loop

In [185]: %timeit im2col_sliding(img, (8,8), skip=4)
     ...: %timeit im2col_sliding_broadcasting(img, (8,8), stepsize=4)
     ...: %timeit im2col_sliding_strided(img, (8,8), stepsize=4)
     ...: %timeit im2col_sliding_strided_v2(img, (8,8), stepsize=4)
     ...: 
1 loops, best of 3: 1.31 s per loop
10 loops, best of 3: 104 ms per loop
10 loops, best of 3: 84.4 ms per loop
10 loops, best of 3: 109 ms per loop

Around 16x speedup there with the strided method over the original loopy version!

User contributions licensed under: CC BY-SA
10 People found this is helpful
Advertisement