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I've got two binary files. They look something like this, but the data is more random:

File A:

FF FF FF FF 00 00 00 00 FF FF 44 43 42 41 FF FF ...

File B:

41 42 43 44 00 00 00 00 44 43 42 41 40 39 38 37 ...

What I'd like is to call something like:

>>> someDiffLib.diff(file_a_data, file_b_data)

And receive something like:

[Match(pos=4, length=4)]

Indicating that in both files the bytes at position 4 are the same for 4 bytes. The sequence 44 43 42 41 would not match because they're not in the same positions in each file.

Is there a library that will do the diff for me? Or should I just write the loops to do the comparison?

share|improve this question
    
docs.python.org/2/library/difflib.html - first result in google for "diff in python" –  Andrey Apr 3 '13 at 21:26
    
possible duplicate of difference between two strings in python/php –  Andrey Apr 3 '13 at 21:27
    
@Andrey thanks, I tried that, but it appears that get_matching_blocks() doesn't check if the bytes are in the same spot in each files, just that the sequence exists in each file. Otherwise, yeah, that's pretty much what I want. –  omghai2u Apr 3 '13 at 21:28
    
So you want to get a list of every position where a match starts and the length of that match, and you don't care about sections of the file that would match if they were lined up properly? –  Kyle Strand Apr 3 '13 at 21:30
2  
@KyleStrand yes, I think so. Although I'm not sure what "lined up properly" would mean in this case. In my example above, I do not want the 44 43 42 41 to match because they're in different positions; if that's what you mean. –  omghai2u Apr 3 '13 at 21:31

1 Answer 1

up vote 7 down vote accepted

You can use itertools.groupby() for this, here is an example:

from itertools import groupby

# this just sets up some byte strings to use, Python 2.x version is below
# instead of this you would use f1 = open('some_file', 'rb').read()
f1 = bytes(int(b, 16) for b in 'FF FF FF FF 00 00 00 00 FF FF 44 43 42 41 FF FF'.split())
f2 = bytes(int(b, 16) for b in '41 42 43 44 00 00 00 00 44 43 42 41 40 39 38 37'.split())

matches = []
for k, g in groupby(range(min(len(f1), len(f2))), key=lambda i: f1[i] == f2[i]):
    if k:
        pos = next(g)
        length = len(list(g)) + 1
        matches.append((pos, length))

Or the same thing as above using a list comprehension:

matches = [(next(g), len(list(g))+1)
           for k, g in groupby(range(min(len(f1), len(f2))), key=lambda i: f1[i] == f2[i])
               if k]

Here is the setup for the example if you are using Python 2.x:

f1 = ''.join(chr(int(b, 16)) for b in 'FF FF FF FF 00 00 00 00 FF FF 44 43 42 41 FF FF'.split())
f2 = ''.join(chr(int(b, 16)) for b in '41 42 43 44 00 00 00 00 44 43 42 41 40 39 38 37'.split())
share|improve this answer
    
Hot. I'm loving what you're doing there. I was hoping for a beautiful answer like this. –  omghai2u Apr 3 '13 at 21:47

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