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The code I wrote performs a mean blur on an image (I hardcoded it as zebra.jpg for testing purposes). My problem is that for an image of 39KB image it take minutes to perform, is there any way of making this code more efficient? Preferably using built in python modules. Also is there any way of improving my code in other ways?

from PIL import Image
img = Image.open('zebra.jpg')

img_w = img.size[0]
img_h = img.size[1]

# The bigger the kernel size, the more intense the blur
kernel = [[1]*10]*10

outputIm = Image.new("RGB", (img_w, img_h))
d = []
for y in range(0, int(img_h)):
    for x in range(0, int(img_w)):
        r, g, b, count = 0, 0, 0, 0
        index_y = int((len(kernel[0]) - 1) / 2.0) * -1

        for kernel_offset_y in kernel:
            index_x = int((len(kernel_offset_y) - 1) / 2.0) * -1

            for kernel_val in kernel_offset_y:

                if img_w > x + index_x+1 > 0 and img_h > y + index_y+1 > 0:
                    temp_r, temp_g, temp_b = img.getpixel((int(x + index_x), int(y + index_y)))
                    r += temp_r * kernel_val
                    g += temp_g * kernel_val
                    b += temp_b * kernel_val
                    count += 1
                    index_x += 1
            index_y += 1

        if (r > 0):
            r = r / count
        if (g > 0):
            g = g / count
        if (b > 0):
            b = b / count

        d.append((r,g,b))

outputIm.putdata([tuple(pixel) for pixel in d])
outputIm.save('blurred.jpg')
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    \$\begingroup\$ Try using pypy as well, as it is very good at speeding up this sort of code. \$\endgroup\$
    – matsjoyce
    Commented May 28, 2017 at 21:44
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    \$\begingroup\$ Usually FFT is an answer (definitely better than a brute-forsish approach you took). I am not sure it is an ultimate answer for your particular filter. \$\endgroup\$
    – vnp
    Commented May 29, 2017 at 4:47
  • \$\begingroup\$ How would I go about implementing FFT here? @vnp \$\endgroup\$
    – mark
    Commented May 29, 2017 at 12:24
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    \$\begingroup\$ As usual, FFT the signal, FFT the kernel, (matrix-)multiply the results, and FFT back. \$\endgroup\$
    – vnp
    Commented May 30, 2017 at 6:55
  • \$\begingroup\$ Do you have a link to where I can read more on FFT on images? Because I don't really understand what you mean @vnp \$\endgroup\$
    – mark
    Commented May 30, 2017 at 7:03

1 Answer 1

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Unnecessary type conversions

There are a couple of unnecessary type conversions, for example:

index_y = int((len(kernel[0]) - 1) / 2.0) * -1

Casting to int truncates, but so does integer division. So you can simplify the above with:

index_y = -(len(kernel_offset_y) - 1) // 2

In Python 2 integer division truncates by default, so ... / 2 would work too. But not in Python 3, where the behavior was changed. Using // works in Python 2.7 and Python 3 consistently.

I also replaced the * -1 with a simpler - prefix.


On this line, x, index_x, y and index_y are all integers, so no need to convert them to int:

temp_r, temp_g, temp_b = img.getpixel((int(x + index_x), int(y + index_y)))

This is also unnecessary:

outputIm.putdata([tuple(pixel) for pixel in d])

Since d is already a list of tuples, you could write simply:

outputIm.putdata(d)

Simplify

The default start value of range is 0, so instead of range(0, k) you could write range(k).

Instead of x = x / y you could write x /= y.

The parentheses are unnecessary in if (expr):.

Formatting

Instead of this:

kernel = [[1]*10]*10

PEP8 recommends this writing style:

kernel = [[1] * 10] * 10
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    \$\begingroup\$ Awesome input, but is there any way to make the 4 nested for loops any more efficient? \$\endgroup\$
    – mark
    Commented May 28, 2017 at 19:04
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    \$\begingroup\$ @mark There is. Use linear algebra libraries like numpy. Unless you're doing this for learning purposes - use opencv or numpy opencv-python-tutroals.readthedocs.io/en/latest/py_tutorials/… \$\endgroup\$ Commented May 28, 2017 at 22:23
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    \$\begingroup\$ I did this for learning purpose so using opencv isn't really my goal, but do you have any input on how I would implement a version with numpy?@user1685095 \$\endgroup\$
    – mark
    Commented May 29, 2017 at 10:21

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