Compare current pixel value with the previous one on numpy array

Question

Is possible to implement this image filtering process in numpy array ? I need to check if the pixel in the previous column and previous row is differente of the current pixel.

width, height = orig_bin.size
pixels = orig_bin.load()

delta = 50
begin = 10
min_w = 30
max_w = 260
min_h = 10
max_h = 40

w_range = range(begin, width - min_w - delta)
h_range = range(begin, height - min_h - delta)

is_changing = False
for x in w_range:
    for y in h_range:
        change_pixel = False
        current_pixel = pixels[x,y]
        if current_pixel != pixels[x, y+1]:
            change_pixel = True

        if current_pixel != pixels[x+1, y]:
            change_pixel = True

        if change_pixel:
            pixels[x,y] = (0,0,0)
        else:
            pixels[x,y] = (255,255,255)

Best regards, Emilio

Is there a problem with this implementation or are you looking to optimize? What's the question here specifically? — mgilson
– mgilson, Commented Feb 27, 2014 at 19:58

Community · Accepted Answer · 2017-05-23 10:26:11Z

2

Here's one approach. Take an example image:

Render of Utah teapot model by Martin Newell

You didn't say where your orig_bin came from, so I've used scipy.misc.imread:

from scipy.misc import imread, imsave
img = imread('input.png')

First, create a mask for pixels that are different from the pixel above (this uses an idea from Bi Rico's answer):

up   = (img[1:,1:] != img[:-1,1:]).any(axis=2)

Note that imread loads images in row-major order, so the first NumPy axis is the vertical axis. See "Multidimensional Array Indexing Order Issues" for an explanation.

Similarly, create a mask for pixels that are different from the pixel to the left:

left = (img[1:,1:] != img[1:,:-1]).any(axis=2)

Combine these to get a mask for pixels that are different to either the pixel above or left:

mask = numpy.zeros(img.shape[:2], dtype=bool)
mask[1:,1:] = left | up

Create a black output image of the right size; then set the mask to white:

output = numpy.zeros(img.shape)
output[mask] = (255, 255, 255)
imsave('output.png', output)

And here's the result:

Output white

or if you want the colours to be the other way round, invert the mask:

output[~mask] = (255, 255, 255)

Output black

edited May 23, 2017 at 10:26

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answered Feb 27, 2014 at 20:59

Gareth Rees

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2 Comments

Gareth Rees Over a year ago

Good spot! Fixed; and I adopted your idea of computing a smaller mask.

user3362112 Over a year ago

Gareth Rees, Is possible to invert the colors ? The mask use (0,0,0) and the color is filled with 255,255,255 ?

Bi Rico · Accepted Answer · 2014-02-27 21:07:35Z

1

You want something like:

change = (pixels[1:, 1:] != pixels[1:, :-1]) | (pixels[1:, 1:] != pixels[:-1, 1:])

This will be binary (True, False), you'll need to multiply it by 255 if you want your result to be 0/255. You also might need to run and any on the last dimension if you array is (x, y, 3).

There are other ways you could re-cast this problem, for example a convolution with [1, -1] might get you most of the way there.

edited Feb 27, 2014 at 21:07

answered Feb 27, 2014 at 20:44

Bi Rico

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2 Comments

Gareth Rees Over a year ago

I get ValueError: operands could not be broadcast together when I try this.

Bi Rico Over a year ago

@GarethRees, the left and right side of | should have the same shape. I've fixed the issue.

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