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Home/ Questions/Q 263911
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T22:38:30+00:00 2026-05-11T22:38:30+00:00

i have an issue i could use some help with, i have python list

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i have an issue i could use some help with, i have python list that looks like this:

fail = [
['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\b\\include', 'Test.java']
['b5cc17d3a35877ca8b76f0b2e07497039c250696', 'ron\\c', 'apa1.txt']
['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', 'knark.txt']
['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\d', 'Sourcecheck.py']
['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\a\\include', 'svin.txt']
['b5cc17d3a35877ca8b76f0b2e07497039c250696', 'ron\\a', 'apa2.txt']
['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', 'apa.txt']

sha1 value, directory, filename

What i want is to separate this content in two different lists based on the sha1 value and directory. For example.

['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', 'apa.txt']
['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', 'knark.txt']

i want to add to the list duplicate = [], because it’s in the same directory with the same sha1 value (and only that directory). The rest of the entries i want to add to another list, say diff = [], because the sha1 value is the same but the directories differ.

I’m kinda lost with the logic here so all that help i could get would be thankful!

EDIT: Fixed a typo, last value (filename) was in some cases a 1-list element, which was 100% incorrect, thansk to SilentGhost to become aware of this issue.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T22:38:30+00:00Added an answer on May 11, 2026 at 10:38 pm
    duplicate = []
    # Sort the list so we can compare adjacent values
    fail.sort()
    #if you didn't want to modify the list in place you can use:
    #sortedFail = sorted(fail)
    #      and then use sortedFail in the rest of the code instead of fail
    for i, x in enumerate(fail):
        if i+1 == len(fail):
            #end of the list
            break
        if x[:2] == fail[i+1][:2]:
            if x not in duplicate:
                duplicate.add(x)
            if fail[i+1] not in duplicate:
                duplicate.add(fail[i+1])
    # diff is just anything not in duplicate as far as I can tell from the explanation
    diff = [d for d in fail if d not in duplicate]
    

    With your example input

    duplicate: [
                  ['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', ['apa.txt']], 
                  ['95d1543adea47e88923c3d4ad56e9f65c2b40c76', 'ron\\c', 'knark.txt']
               ]
    
    diff: [
              ['b5cc17d3a35877ca8b76f0b2e07497039c250696', 'ron\\a', ['apa2.txt']], 
              ['b5cc17d3a35877ca8b76f0b2e07497039c250696', 'ron\\c', 'apa1.txt'], 
              ['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\a\\include', ['svin.txt']],
              ['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\b\\include', 'Test.java'],
              ['da39a3ee5e6b4b0d3255bfef95601890afd80709', 'ron\\d', 'Sourcecheck.py']
          ]
    

    So perhaps I missed something, but I think this is what you were asking for.

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