~bzr-pqm/bzr/bzr.dev

#!/usr/bin/env python
# Copyright (C) 2005 Bram Cohen, Copyright (C) 2005, 2006 Canonical Ltd
#
# This program is free software; you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation; either version 2 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program; if not, write to the Free Software
# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA  02111-1307  USA


from bisect import bisect
from copy import copy
import difflib
import os
import sys
import time

from bzrlib.trace import mutter


__all__ = ['PatienceSequenceMatcher', 'unified_diff', 'unified_diff_files']


def unique_lcs(a, b):
    """Find the longest common subset for unique lines.

    :param a: An indexable object (such as string or list of strings)
    :param b: Another indexable object (such as string or list of strings)
    :return: A list of tuples, one for each line which is matched.
            [(line_in_a, line_in_b), ...]

    This only matches lines which are unique on both sides.
    This helps prevent common lines from over influencing match
    results.
    The longest common subset uses the Patience Sorting algorithm:
    http://en.wikipedia.org/wiki/Patience_sorting
    """
    # set index[line in a] = position of line in a unless
    # unless a is a duplicate, in which case it's set to None
    index = {}
    for i in xrange(len(a)):
        line = a[i]
        if line in index:
            index[line] = None
        else:
            index[line]= i
    # make btoa[i] = position of line i in a, unless
    # that line doesn't occur exactly once in both, 
    # in which case it's set to None
    btoa = [None] * len(b)
    index2 = {}
    for pos, line in enumerate(b):
        next = index.get(line)
        if next is not None:
            if line in index2:
                # unset the previous mapping, which we now know to
                # be invalid because the line isn't unique
                btoa[index2[line]] = None
                del index[line]
            else:
                index2[line] = pos
                btoa[pos] = next
    # this is the Patience sorting algorithm
    # see http://en.wikipedia.org/wiki/Patience_sorting
    backpointers = [None] * len(b)
    stacks = []
    lasts = []
    k = 0
    for bpos, apos in enumerate(btoa):
        if apos is None:
            continue
        # as an optimization, check if the next line comes at the end,
        # because it usually does
        if stacks and stacks[-1] < apos:
            k = len(stacks)
        # as an optimization, check if the next line comes right after
        # the previous line, because usually it does
        elif stacks and stacks[k] < apos and (k == len(stacks) - 1 or 
                                              stacks[k+1] > apos):
            k += 1
        else:
            k = bisect(stacks, apos)
        if k > 0:
            backpointers[bpos] = lasts[k-1]
        if k < len(stacks):
            stacks[k] = apos
            lasts[k] = bpos
        else:
            stacks.append(apos)
            lasts.append(bpos)
    if len(lasts) == 0:
        return []
    result = []
    k = lasts[-1]
    while k is not None:
        result.append((btoa[k], k))
        k = backpointers[k]
    result.reverse()
    return result


def recurse_matches(a, b, alo, blo, ahi, bhi, answer, maxrecursion):
    """Find all of the matching text in the lines of a and b.

    :param a: A sequence
    :param b: Another sequence
    :param alo: The start location of a to check, typically 0
    :param ahi: The start location of b to check, typically 0
    :param ahi: The maximum length of a to check, typically len(a)
    :param bhi: The maximum length of b to check, typically len(b)
    :param answer: The return array. Will be filled with tuples
                   indicating [(line_in_a, line_in_b)]
    :param maxrecursion: The maximum depth to recurse.
                         Must be a positive integer.
    :return: None, the return value is in the parameter answer, which
             should be a list

    """
    if maxrecursion < 0:
        mutter('max recursion depth reached')
        # this will never happen normally, this check is to prevent DOS attacks
        return
    oldlength = len(answer)
    if alo == ahi or blo == bhi:
        return
    last_a_pos = alo-1
    last_b_pos = blo-1
    for apos, bpos in unique_lcs(a[alo:ahi], b[blo:bhi]):
        # recurse between lines which are unique in each file and match
        apos += alo
        bpos += blo
        # Most of the time, you will have a sequence of similar entries
        if last_a_pos+1 != apos or last_b_pos+1 != bpos:
            recurse_matches(a, b, last_a_pos+1, last_b_pos+1,
                apos, bpos, answer, maxrecursion - 1)
        last_a_pos = apos
        last_b_pos = bpos
        answer.append((apos, bpos))
    if len(answer) > oldlength:
        # find matches between the last match and the end
        recurse_matches(a, b, last_a_pos+1, last_b_pos+1,
                        ahi, bhi, answer, maxrecursion - 1)
    elif a[alo] == b[blo]:
        # find matching lines at the very beginning
        while alo < ahi and blo < bhi and a[alo] == b[blo]:
            answer.append((alo, blo))
            alo += 1
            blo += 1
        recurse_matches(a, b, alo, blo,
                        ahi, bhi, answer, maxrecursion - 1)
    elif a[ahi - 1] == b[bhi - 1]:
        # find matching lines at the very end
        nahi = ahi - 1
        nbhi = bhi - 1
        while nahi > alo and nbhi > blo and a[nahi - 1] == b[nbhi - 1]:
            nahi -= 1
            nbhi -= 1
        recurse_matches(a, b, last_a_pos+1, last_b_pos+1,
                        nahi, nbhi, answer, maxrecursion - 1)
        for i in xrange(ahi - nahi):
            answer.append((nahi + i, nbhi + i))


def _collapse_sequences(matches):
    """Find sequences of lines.

    Given a sequence of [(line_in_a, line_in_b),]
    find regions where they both increment at the same time
    """
    answer = []
    start_a = start_b = None
    length = 0
    for i_a, i_b in matches:
        if (start_a is not None
            and (i_a == start_a + length) 
            and (i_b == start_b + length)):
            length += 1
        else:
            if start_a is not None:
                answer.append((start_a, start_b, length))
            start_a = i_a
            start_b = i_b
            length = 1

    if length != 0:
        answer.append((start_a, start_b, length))

    return answer


def _check_consistency(answer):
    # For consistency sake, make sure all matches are only increasing
    next_a = -1
    next_b = -1
    for a,b,match_len in answer:
        assert a >= next_a, 'Non increasing matches for a'
        assert b >= next_b, 'Not increasing matches for b'
        next_a = a + match_len
        next_b = b + match_len


class PatienceSequenceMatcher(difflib.SequenceMatcher):
    """Compare a pair of sequences using longest common subset."""

    _do_check_consistency = True

    def __init__(self, isjunk=None, a='', b=''):
        if isjunk is not None:
            raise NotImplementedError('Currently we do not support'
                                      ' isjunk for sequence matching')
        difflib.SequenceMatcher.__init__(self, isjunk, a, b)

    def get_matching_blocks(self):
        """Return list of triples describing matching subsequences.

        Each triple is of the form (i, j, n), and means that
        a[i:i+n] == b[j:j+n].  The triples are monotonically increasing in
        i and in j.

        The last triple is a dummy, (len(a), len(b), 0), and is the only
        triple with n==0.

        >>> s = PatienceSequenceMatcher(None, "abxcd", "abcd")
        >>> s.get_matching_blocks()
        [(0, 0, 2), (3, 2, 2), (5, 4, 0)]
        """
        # jam 20060525 This is the python 2.4.1 difflib get_matching_blocks 
        # implementation which uses __helper. 2.4.3 got rid of helper for
        # doing it inline with a queue.
        # We should consider doing the same for recurse_matches

        if self.matching_blocks is not None:
            return self.matching_blocks

        matches = []
        recurse_matches(self.a, self.b, 0, 0,
                        len(self.a), len(self.b), matches, 10)
        # Matches now has individual line pairs of
        # line A matches line B, at the given offsets
        self.matching_blocks = _collapse_sequences(matches)
        self.matching_blocks.append( (len(self.a), len(self.b), 0) )
        if PatienceSequenceMatcher._do_check_consistency:
            if __debug__:
                _check_consistency(self.matching_blocks)

        return self.matching_blocks


# This is a version of unified_diff which only adds a factory parameter
# so that you can override the default SequenceMatcher
# this has been submitted as a patch to python
def unified_diff(a, b, fromfile='', tofile='', fromfiledate='',
                 tofiledate='', n=3, lineterm='\n',
                 sequencematcher=None):
    r"""
    Compare two sequences of lines; generate the delta as a unified diff.

    Unified diffs are a compact way of showing line changes and a few
    lines of context.  The number of context lines is set by 'n' which
    defaults to three.

    By default, the diff control lines (those with ---, +++, or @@) are
    created with a trailing newline.  This is helpful so that inputs
    created from file.readlines() result in diffs that are suitable for
    file.writelines() since both the inputs and outputs have trailing
    newlines.

    For inputs that do not have trailing newlines, set the lineterm
    argument to "" so that the output will be uniformly newline free.

    The unidiff format normally has a header for filenames and modification
    times.  Any or all of these may be specified using strings for
    'fromfile', 'tofile', 'fromfiledate', and 'tofiledate'.  The modification
    times are normally expressed in the format returned by time.ctime().

    Example:

    >>> for line in unified_diff('one two three four'.split(),
    ...             'zero one tree four'.split(), 'Original', 'Current',
    ...             'Sat Jan 26 23:30:50 1991', 'Fri Jun 06 10:20:52 2003',
    ...             lineterm=''):
    ...     print line
    --- Original Sat Jan 26 23:30:50 1991
    +++ Current Fri Jun 06 10:20:52 2003
    @@ -1,4 +1,4 @@
    +zero
     one
    -two
    -three
    +tree
     four
    """
    if sequencematcher is None:
        sequencematcher = difflib.SequenceMatcher

    started = False
    for group in sequencematcher(None,a,b).get_grouped_opcodes(n):
        if not started:
            yield '--- %s %s%s' % (fromfile, fromfiledate, lineterm)
            yield '+++ %s %s%s' % (tofile, tofiledate, lineterm)
            started = True
        i1, i2, j1, j2 = group[0][1], group[-1][2], group[0][3], group[-1][4]
        yield "@@ -%d,%d +%d,%d @@%s" % (i1+1, i2-i1, j1+1, j2-j1, lineterm)
        for tag, i1, i2, j1, j2 in group:
            if tag == 'equal':
                for line in a[i1:i2]:
                    yield ' ' + line
                continue
            if tag == 'replace' or tag == 'delete':
                for line in a[i1:i2]:
                    yield '-' + line
            if tag == 'replace' or tag == 'insert':
                for line in b[j1:j2]:
                    yield '+' + line


def unified_diff_files(a, b, sequencematcher=None):
    """Generate the diff for two files.
    """
    # Should this actually be an error?
    if a == b:
        return []
    if a == '-':
        file_a = sys.stdin
        time_a = time.time()
    else:
        file_a = open(a, 'rb')
        time_a = os.stat(a).st_mtime

    if b == '-':
        file_b = sys.stdin
        time_b = time.time()
    else:
        file_b = open(b, 'rb')
        time_b = os.stat(b).st_mtime

    # TODO: Include fromfiledate and tofiledate
    return unified_diff(file_a.readlines(), file_b.readlines(),
                        fromfile=a, tofile=b,
                        sequencematcher=sequencematcher)


def main(args):
    import optparse
    p = optparse.OptionParser(usage='%prog [options] file_a file_b'
                                    '\nFiles can be "-" to read from stdin')
    p.add_option('--patience', dest='matcher', action='store_const', const='patience',
                 default='patience', help='Use the patience difference algorithm')
    p.add_option('--difflib', dest='matcher', action='store_const', const='difflib',
                 default='patience', help='Use python\'s difflib algorithm')

    algorithms = {'patience':PatienceSequenceMatcher, 'difflib':difflib.SequenceMatcher}

    (opts, args) = p.parse_args(args)
    matcher = algorithms[opts.matcher]

    if len(args) != 2:
        print 'You must supply 2 filenames to diff'
        return -1

    for line in unified_diff_files(args[0], args[1], sequencematcher=matcher):
        sys.stdout.write(line)
    
if __name__ == '__main__':
    sys.exit(main(sys.argv[1:]))