CODEWITHSUNDEEP
pythondiff
omghai2u
omghai2u

Reputation: 273

Diffing Binary Files In Python

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?

Upvotes: 2

Views: 6307

Answers (2)

Kevin
Kevin

Reputation: 2050

The provided itertools.groupby solution works fine, but it's pretty slow.

I wrote a pretty naive attempt using numpy and tested it versus the other solution on a particular 16MB file I happened to have, and it was about 42x faster on my machine. Someone familiar with numpy could likely improve this significantly.

import numpy as np

def compare(path1, path2):
    x,y = np.fromfile(path1, np.int8), np.fromfile(path2, np.int8)
    length = min(x.size, y.size)
    x,y = x[:length], y[:length]

    z = np.where(x == y)[0]
    if(z.size == 0) : return z

    borders = np.append(np.insert(np.where(np.diff(z) != 1)[0] + 1, 0, 0), len(z))
    lengths = borders[1:] - borders[:-1]
    starts = z[borders[:-1]]
    return np.array([starts, lengths]).T

Upvotes: 3

Andrew Clark
Andrew Clark

Reputation: 208475

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())

Upvotes: 11

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