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Home/ Questions/Q 509625
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T07:01:48+00:00 2026-05-13T07:01:48+00:00

Trying to solve a problem of preventing duplicate images to be uploaded. I have

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Trying to solve a problem of preventing duplicate images to be uploaded.

I have two JPGs. Looking at them I can see that they are in fact identical. But for some reason they have different file size (one is pulled from a backup, the other is another upload) and so they have a different md5 checksum.

How can I efficiently and confidently compare two images in the same sense as a human would be able to see that they are clearly identical?

Example: http://static.peterbe.com/a.jpg and http://static.peterbe.com/b.jpg

Update

I wrote this script:

import math, operator
from PIL import Image
def compare(file1, file2):
    image1 = Image.open(file1)
    image2 = Image.open(file2)
    h1 = image1.histogram()
    h2 = image2.histogram()
    rms = math.sqrt(reduce(operator.add,
                           map(lambda a,b: (a-b)**2, h1, h2))/len(h1))
    return rms

if __name__=='__main__':
    import sys
    file1, file2 = sys.argv[1:]
    print compare(file1, file2)

Then I downloaded the two visually identical images and ran the script. Output:

58.9830484122

Can anybody tell me what a suitable cutoff should be?

Update II

The difference between a.jpg and b.jpg is that the second one has been saved with PIL:

b=Image.open('a.jpg')
b.save(open('b.jpg','wb'))

This apparently applies some very very light quality modifications. I’ve now solved my problem by applying the same PIL save to the file being uploaded without doing anything with it and it now works!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T07:01:48+00:00Added an answer on May 13, 2026 at 7:01 am

    There is a OSS project that uses WebDriver to take screen shots and then compares the images to see if there are any issues (http://code.google.com/p/fighting-layout-bugs/)). It does it by openning the file into a stream and then comparing every bit.

    You may be able to do something similar with PIL.

    EDIT:

    After more research I found

    h1 = Image.open("image1").histogram()
    h2 = Image.open("image2").histogram()
    
    rms = math.sqrt(reduce(operator.add,
        map(lambda a,b: (a-b)**2, h1, h2))/len(h1))
    

    on http://snipplr.com/view/757/compare-two-pil-images-in-python/ and http://effbot.org/zone/pil-comparing-images.htm

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