在图像上找到3x3滑动窗口

Question

I have an image.

I want to obtain a 3x3 window (neighbouring pixels) for every pixel in the image.

I have this Python code:

for x in range(2,r-1,1):
    for y in range(2,c-1,1):
        mask5=numpy.array([cv.Get2D(copy_img,x-1,y-1),cv.Get2D(copy_img,x-1,y),cv.Get2D(copy_img,x-1,y+1),cv.Get2D(copy_img,x,y-1),cv.Get2D(copy_img,x,y),cv.Get2D(copy_img,x,y+1),cv.Get2D(copy_img,x+1,y-1),cv.Get2D(copy_img,x+1,y),cv.Get2D(copy_img,x+1,y+1)])
        cent=[cv.Get2D(copy_img,x,y)]

mask5 is the 3x3 window. cent is the center pixel.

Is there a more efficient way to do this - ie using maps, iterators - anything but the two nested loops I've used?

Answer 1

This can be done faster, by reshaping and swapping axes, and then repeating over all kernel elements, like this:

im = np.arange(81).reshape(9,9)
print np.swapaxes(im.reshape(3,3,3,-1),1,2)

This gives you an array of 3*3 tiles which tessalates across the surface:

[[[[ 0  1  2]   [[ 3  4  5]   [[ 6  7  8]
   [ 9 10 11]    [12 13 14]    [15 16 17]
   [18 19 20]]   [21 22 23]]   [24 25 26]]]

 [[[27 28 29]   [[30 31 32]   [[33 34 35]
   [36 37 38]    [39 40 41]    [42 43 44]
   [45 46 47]]   [48 49 50]]   [51 52 53]]]

 [[[54 55 56]   [[57 58 59]   [[60 61 62]
   [63 64 65]    [66 67 68]    [69 70 71]
   [72 73 74]]   [75 76 77]]   [78 79 80]]]]

To get the overlapping tiles we need to repeat this 8 further times, but 'wrapping' the array, by using a combination of vstack and column_stack . Note that the right and bottom tile arrays wrap around (which may or may not be what you want, depending on how you are treating edge conditions):

im =  np.vstack((im[1:],im[0]))
im =  np.column_stack((im[:,1:],im[:,0]))
print np.swapaxes(im.reshape(3,3,3,-1),1,2)

#Output:
[[[[10 11 12]   [[13 14 15]   [[16 17  9]
   [19 20 21]    [22 23 24]    [25 26 18]
   [28 29 30]]   [31 32 33]]   [34 35 27]]]

 [[[37 38 39]   [[40 41 42]   [[43 44 36]
   [46 47 48]    [49 50 51]    [52 53 45]
   [55 56 57]]   [58 59 60]]   [61 62 54]]]

 [[[64 65 66]   [[67 68 69]   [[70 71 63]
   [73 74 75]    [76 77 78]    [79 80 72]
   [ 1  2  3]]   [ 4  5  6]]   [ 7  8  0]]]]

Doing it this way you wind up with 9 sets of arrays, so you then need to zip them back together. This, and all the reshaping generalises to this (for arrays where the dimensions are divisible by 3):

def new(im):
    rows,cols = im.shape
    final = np.zeros((rows, cols, 3, 3))
    for x in (0,1,2):
        for y in (0,1,2):
            im1 = np.vstack((im[x:],im[:x]))
            im1 = np.column_stack((im1[:,y:],im1[:,:y]))
            final[x::3,y::3] = np.swapaxes(im1.reshape(rows/3,3,cols/3,-1),1,2)
    return final

Comparing this new function to looping through all the slices (below), using timeit , its about 4 times faster, for a 300*300 array.

def old(im):
    rows,cols = im.shape
    s = []
    for x in xrange(1,rows):
        for y in xrange(1,cols):
            s.append(im[x-1:x+2,y-1:y+2])
    return s

Answer 2

I think the following does what you are after. The loop is only over the 9 elements. I'm sure there is a way of vectorizing it, but it's probably not worth the effort.

import numpy

im = numpy.random.randint(0,50,(5,7))

# idx_2d contains the indices of each position in the array
idx_2d = numpy.mgrid[0:im.shape[0],0:im.shape[1]]

# We break that into 2 sub arrays
x_idx = idx_2d[1]
y_idx = idx_2d[0]

# The mask is used to ignore the edge values (or indeed any values).
mask = numpy.ones(im.shape, dtype='bool')
mask[0, :] = False
mask[:, 0] = False
mask[im.shape[0] - 1, :] = False
mask[:, im.shape[1] - 1] = False

# We create and fill an array that contains the lookup for every
# possible 3x3 array.
idx_array = numpy.zeros((im[mask].size, 3, 3), dtype='int64')

# Compute the flattened indices for each position in the 3x3 grid
for n in range(0, 3):
    for m in range(0, 3):
        # Compute the flattened indices for each position in the 
        # 3x3 grid
        idx = (x_idx + (n-1)) + (y_idx  + (m-1)) * im.shape[1]

        # mask it, and write it to the big array
        idx_array[:, m, n] = idx[mask]


# sub_images contains every valid 3x3 sub image
sub_images = im.ravel()[idx_array]

# Finally, we can flatten and sort each sub array quickly
sorted_sub_images = numpy.sort(sub_images.reshape((idx[mask].size, 9)))

Answer 3

Try the following code as matlab function im2col(...)

import numpy as np

def im2col(Im, block, style='sliding'):
    """block = (patchsize, patchsize)
        first do sliding
    """
    bx, by = block
    Imx, Imy = Im.shape
    Imcol = []
    for j in range(0, Imy):
        for i in range(0, Imx):
            if (i+bx <= Imx) and (j+by <= Imy):
                Imcol.append(Im[i:i+bx, j:j+by].T.reshape(bx*by))
            else:
                break
    return np.asarray(Imcol).T

if __name__ == '__main__':
    Im = np.reshape(range(6*6), (6,6))
    patchsize = 3
    print Im
    out =  im2col(Im, (patchsize, patchsize))
    print out
    print out.shape
    print len(out)