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#!/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
import difflib
from bzrlib.trace import mutter
__all__ = ['PatienceSequenceMatcher', 'unified_diff', 'unified_diff_files']
def unique_lcs_py(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
# 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_py(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_py(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_py(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_py(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_py(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_py(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:
if a < next_a:
raise ValueError('Non increasing matches for a')
if b < next_b:
raise ValueError('Non increasing matches for b')
next_a = a + match_len
next_b = b + match_len
class PatienceSequenceMatcher_py(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_py(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_py._do_check_consistency:
if __debug__:
_check_consistency(self.matching_blocks)
return self.matching_blocks
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