third_party/stdlib/difflib.py
def get_grouped_opcodes(self, n=3):
        """ Isolate change clusters by eliminating ranges with no changes.

        Return a generator of groups with up to n lines of context.
        Each group is in the same format as returned by get_opcodes().

        >>> from pprint import pprint
        >>> a = map(str, range(1,40))
        >>> b = a[:]
        >>> b[8:8] = ['i']     # Make an insertion
        >>> b[20] += 'x'       # Make a replacement
        >>> b[23:28] = []      # Make a deletion
        >>> b[30] += 'y'       # Make another replacement
        >>> pprint(list(SequenceMatcher(None,a,b).get_grouped_opcodes()))
        [[('equal', 5, 8, 5, 8), ('insert', 8, 8, 8, 9), ('equal', 8, 11, 9, 12)],
         [('equal', 16, 19, 17, 20),
          ('replace', 19, 20, 20, 21),
          ('equal', 20, 22, 21, 23),
          ('delete', 22, 27, 23, 23),
          ('equal', 27, 30, 23, 26)],
         [('equal', 31, 34, 27, 30),
          ('replace', 34, 35, 30, 31),
          ('equal', 35, 38, 31, 34)]]
        """

        codes = self.get_opcodes()
        if not codes:
            codes = [("equal", 0, 1, 0, 1)]
        # Fixup leading and trailing groups if they show no changes.
        if codes[0][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[0]
            codes[0] = tag, max(i1, i2-n), i2, max(j1, j2-n), j2
        if codes[-1][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[-1]
            codes[-1] = tag, i1, min(i2, i1+n), j1, min(j2, j1+n)

        nn = n + n
        group = []
        for tag, i1, i2, j1, j2 in codes:
            # End the current group and start a new one whenever
            # there is a large range with no changes.
            if tag == 'equal' and i2-i1 > nn:
                group.append((tag, i1, min(i2, i1+n), j1, min(j2, j1+n)))
                yield group
                group = []
                i1, j1 = max(i1, i2-n), max(j1, j2-n)
            group.append((tag, i1, i2, j1 ,j2))
        if group and not (len(group)==1 and group[0][0] == 'equal'):
            yield group
def get_grouped_opcodes(self, n=3):
        """ Isolate change clusters by eliminating ranges with no changes.

        Return a generator of groups with up to n lines of context.
        Each group is in the same format as returned by get_opcodes().

        >>> from pprint import pprint
        >>> a = map(str, range(1,40))
        >>> b = a[:]
        >>> b[8:8] = ['i']     # Make an insertion
        >>> b[20] += 'x'       # Make a replacement
        >>> b[23:28] = []      # Make a deletion
        >>> b[30] += 'y'       # Make another replacement
        >>> pprint(list(SequenceMatcher(None,a,b).get_grouped_opcodes()))
        [[('equal', 5, 8, 5, 8), ('insert', 8, 8, 8, 9), ('equal', 8, 11, 9, 12)],
         [('equal', 16, 19, 17, 20),
          ('replace', 19, 20, 20, 21),
          ('equal', 20, 22, 21, 23),
          ('delete', 22, 27, 23, 23),
          ('equal', 27, 30, 23, 26)],
         [('equal', 31, 34, 27, 30),
          ('replace', 34, 35, 30, 31),
          ('equal', 35, 38, 31, 34)]]
        """

        codes = self.get_opcodes()
        if not codes:
            codes = [("equal", 0, 1, 0, 1)]
        # Fixup leading and trailing groups if they show no changes.
        if codes[0][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[0]
            codes[0] = tag, max(i1, i2-n), i2, max(j1, j2-n), j2
        if codes[-1][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[-1]
            codes[-1] = tag, i1, min(i2, i1+n), j1, min(j2, j1+n)

        nn = n + n
        group = []
        for tag, i1, i2, j1, j2 in codes:
            # End the current group and start a new one whenever
            # there is a large range with no changes.
            if tag == 'equal' and i2-i1 > nn:
                group.append((tag, i1, min(i2, i1+n), j1, min(j2, j1+n)))
                yield group
                group = []
                i1, j1 = max(i1, i2-n), max(j1, j2-n)
            group.append((tag, i1, i2, j1 ,j2))
        if group and not (len(group)==1 and group[0][0] == 'equal'):
            yield group
def get_grouped_opcodes ( self , n = 3 ) : codes = self . get_opcodes ( ) if not codes : codes = [ ( "equal" , 0 , 1 , 0 , 1 ) ] # Fixup leading and trailing groups if they show no changes. if codes [ 0 ] [ 0 ] == 'equal' : tag , i1 , i2 , j1 , j2 = codes [ 0 ] codes [ 0 ] = tag , max ( i1 , i2 - n ) , i2 , max ( j1 , j2 - n ) , j2 if codes [ - 1 ] [ 0 ] == 'equal' : tag , i1 , i2 , j1 , j2 = codes [ - 1 ] codes [ - 1 ] = tag , i1 , min ( i2 , i1 + n ) , j1 , min ( j2 , j1 + n ) nn = n + n group = [ ] for tag , i1 , i2 , j1 , j2 in codes : # End the current group and start a new one whenever # there is a large range with no changes. if tag == 'equal' and i2 - i1 > nn : group . append ( ( tag , i1 , min ( i2 , i1 + n ) , j1 , min ( j2 , j1 + n ) ) ) yield group group = [ ] i1 , j1 = max ( i1 , i2 - n ) , max ( j1 , j2 - n ) group . append ( ( tag , i1 , i2 , j1 , j2 ) ) if group and not ( len ( group ) == 1 and group [ 0 ] [ 0 ] == 'equal' ) : yield group
Isolate change clusters by eliminating ranges with no changes.

        Return a generator of groups with up to n lines of context.
        Each group is in the same format as returned by get_opcodes().

        >>> from pprint import pprint
        >>> a = map(str, range(1,40))
        >>> b = a[:]
        >>> b[8:8] = ['i']     # Make an insertion
        >>> b[20] += 'x'       # Make a replacement
        >>> b[23:28] = []      # Make a deletion
        >>> b[30] += 'y'       # Make another replacement
        >>> pprint(list(SequenceMatcher(None,a,b).get_grouped_opcodes()))
        [[('equal', 5, 8, 5, 8), ('insert', 8, 8, 8, 9), ('equal', 8, 11, 9, 12)],
         [('equal', 16, 19, 17, 20),
          ('replace', 19, 20, 20, 21),
          ('equal', 20, 22, 21, 23),
          ('delete', 22, 27, 23, 23),
          ('equal', 27, 30, 23, 26)],
         [('equal', 31, 34, 27, 30),
          ('replace', 34, 35, 30, 31),
          ('equal', 35, 38, 31, 34)]]
Isolate change clusters by eliminating ranges with no changes .