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.  New in Python 2.5, it's also guaranteed that if
        (i, j, n) and (i', j', n') are adjacent triples in the list, and
        the second is not the last triple in the list, then i+n != i' or
        j+n != j'.  IOW, adjacent triples never describe adjacent equal
        blocks.

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

        >>> s = SequenceMatcher(None, "abxcd", "abcd")
        >>> s.get_matching_blocks()
        [Match(a=0, b=0, size=2), Match(a=3, b=2, size=2), Match(a=5, b=4, size=0)]
        """

        if self.matching_blocks is not None:
            return self.matching_blocks
        la, lb = len(self.a), len(self.b)

        # This is most naturally expressed as a recursive algorithm, but
        # at least one user bumped into extreme use cases that exceeded
        # the recursion limit on their box.  So, now we maintain a list
        # ('queue`) of blocks we still need to look at, and append partial
        # results to `matching_blocks` in a loop; the matches are sorted
        # at the end.
        queue = [(0, la, 0, lb)]
        matching_blocks = []
        while queue:
            alo, ahi, blo, bhi = queue.pop()
            i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
            # a[alo:i] vs b[blo:j] unknown
            # a[i:i+k] same as b[j:j+k]
            # a[i+k:ahi] vs b[j+k:bhi] unknown
            if k:   # if k is 0, there was no matching block
                matching_blocks.append(x)
                if alo < i and blo < j:
                    queue.append((alo, i, blo, j))
                if i+k < ahi and j+k < bhi:
                    queue.append((i+k, ahi, j+k, bhi))
        matching_blocks.sort()

        # It's possible that we have adjacent equal blocks in the
        # matching_blocks list now.  Starting with 2.5, this code was added
        # to collapse them.
        i1 = j1 = k1 = 0
        non_adjacent = []
        for i2, j2, k2 in matching_blocks:
            # Is this block adjacent to i1, j1, k1?
            if i1 + k1 == i2 and j1 + k1 == j2:
                # Yes, so collapse them -- this just increases the length of
                # the first block by the length of the second, and the first
                # block so lengthened remains the block to compare against.
                k1 += k2
            else:
                # Not adjacent.  Remember the first block (k1==0 means it's
                # the dummy we started with), and make the second block the
                # new block to compare against.
                if k1:
                    non_adjacent.append((i1, j1, k1))
                i1, j1, k1 = i2, j2, k2
        if k1:
            non_adjacent.append((i1, j1, k1))

        non_adjacent.append( (la, lb, 0) )
        self.matching_blocks = map(Match._make, non_adjacent)
        return self.matching_blocks
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.  New in Python 2.5, it's also guaranteed that if
        (i, j, n) and (i', j', n') are adjacent triples in the list, and
        the second is not the last triple in the list, then i+n != i' or
        j+n != j'.  IOW, adjacent triples never describe adjacent equal
        blocks.

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

        >>> s = SequenceMatcher(None, "abxcd", "abcd")
        >>> s.get_matching_blocks()
        [Match(a=0, b=0, size=2), Match(a=3, b=2, size=2), Match(a=5, b=4, size=0)]
        """

        if self.matching_blocks is not None:
            return self.matching_blocks
        la, lb = len(self.a), len(self.b)

        # This is most naturally expressed as a recursive algorithm, but
        # at least one user bumped into extreme use cases that exceeded
        # the recursion limit on their box.  So, now we maintain a list
        # ('queue`) of blocks we still need to look at, and append partial
        # results to `matching_blocks` in a loop; the matches are sorted
        # at the end.
        queue = [(0, la, 0, lb)]
        matching_blocks = []
        while queue:
            alo, ahi, blo, bhi = queue.pop()
            i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
            # a[alo:i] vs b[blo:j] unknown
            # a[i:i+k] same as b[j:j+k]
            # a[i+k:ahi] vs b[j+k:bhi] unknown
            if k:   # if k is 0, there was no matching block
                matching_blocks.append(x)
                if alo < i and blo < j:
                    queue.append((alo, i, blo, j))
                if i+k < ahi and j+k < bhi:
                    queue.append((i+k, ahi, j+k, bhi))
        matching_blocks.sort()

        # It's possible that we have adjacent equal blocks in the
        # matching_blocks list now.  Starting with 2.5, this code was added
        # to collapse them.
        i1 = j1 = k1 = 0
        non_adjacent = []
        for i2, j2, k2 in matching_blocks:
            # Is this block adjacent to i1, j1, k1?
            if i1 + k1 == i2 and j1 + k1 == j2:
                # Yes, so collapse them -- this just increases the length of
                # the first block by the length of the second, and the first
                # block so lengthened remains the block to compare against.
                k1 += k2
            else:
                # Not adjacent.  Remember the first block (k1==0 means it's
                # the dummy we started with), and make the second block the
                # new block to compare against.
                if k1:
                    non_adjacent.append((i1, j1, k1))
                i1, j1, k1 = i2, j2, k2
        if k1:
            non_adjacent.append((i1, j1, k1))

        non_adjacent.append( (la, lb, 0) )
        self.matching_blocks = map(Match._make, non_adjacent)
        return self.matching_blocks
def get_matching_blocks ( self ) : if self . matching_blocks is not None : return self . matching_blocks la , lb = len ( self . a ) , len ( self . b ) # This is most naturally expressed as a recursive algorithm, but # at least one user bumped into extreme use cases that exceeded # the recursion limit on their box.  So, now we maintain a list # ('queue`) of blocks we still need to look at, and append partial # results to `matching_blocks` in a loop; the matches are sorted # at the end. queue = [ ( 0 , la , 0 , lb ) ] matching_blocks = [ ] while queue : alo , ahi , blo , bhi = queue . pop ( ) i , j , k = x = self . find_longest_match ( alo , ahi , blo , bhi ) # a[alo:i] vs b[blo:j] unknown # a[i:i+k] same as b[j:j+k] # a[i+k:ahi] vs b[j+k:bhi] unknown if k : # if k is 0, there was no matching block matching_blocks . append ( x ) if alo < i and blo < j : queue . append ( ( alo , i , blo , j ) ) if i + k < ahi and j + k < bhi : queue . append ( ( i + k , ahi , j + k , bhi ) ) matching_blocks . sort ( ) # It's possible that we have adjacent equal blocks in the # matching_blocks list now.  Starting with 2.5, this code was added # to collapse them. i1 = j1 = k1 = 0 non_adjacent = [ ] for i2 , j2 , k2 in matching_blocks : # Is this block adjacent to i1, j1, k1? if i1 + k1 == i2 and j1 + k1 == j2 : # Yes, so collapse them -- this just increases the length of # the first block by the length of the second, and the first # block so lengthened remains the block to compare against. k1 += k2 else : # Not adjacent.  Remember the first block (k1==0 means it's # the dummy we started with), and make the second block the # new block to compare against. if k1 : non_adjacent . append ( ( i1 , j1 , k1 ) ) i1 , j1 , k1 = i2 , j2 , k2 if k1 : non_adjacent . append ( ( i1 , j1 , k1 ) ) non_adjacent . append ( ( la , lb , 0 ) ) self . matching_blocks = map ( Match . _make , non_adjacent ) return self . matching_blocks
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.  New in Python 2.5, it's also guaranteed that if
        (i, j, n) and (i', j', n') are adjacent triples in the list, and
        the second is not the last triple in the list, then i+n != i' or
        j+n != j'.  IOW, adjacent triples never describe adjacent equal
        blocks.

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

        >>> s = SequenceMatcher(None, "abxcd", "abcd")
        >>> s.get_matching_blocks()
        [Match(a=0, b=0, size=2), Match(a=3, b=2, size=2), Match(a=5, b=4, size=0)]
Return list of triples describing matching subsequences .