third_party/stdlib/difflib.py
def ratio(self):
        """Return a measure of the sequences' similarity (float in [0,1]).

        Where T is the total number of elements in both sequences, and
        M is the number of matches, this is 2.0*M / T.
        Note that this is 1 if the sequences are identical, and 0 if
        they have nothing in common.

        .ratio() is expensive to compute if you haven't already computed
        .get_matching_blocks() or .get_opcodes(), in which case you may
        want to try .quick_ratio() or .real_quick_ratio() first to get an
        upper bound.

        >>> s = SequenceMatcher(None, "abcd", "bcde")
        >>> s.ratio()
        0.75
        >>> s.quick_ratio()
        0.75
        >>> s.real_quick_ratio()
        1.0
        """

        matches = reduce(lambda sum, triple: sum + triple[-1],
                         self.get_matching_blocks(), 0)
        return _calculate_ratio(matches, len(self.a) + len(self.b))
def ratio(self):
        """Return a measure of the sequences' similarity (float in [0,1]).

        Where T is the total number of elements in both sequences, and
        M is the number of matches, this is 2.0*M / T.
        Note that this is 1 if the sequences are identical, and 0 if
        they have nothing in common.

        .ratio() is expensive to compute if you haven't already computed
        .get_matching_blocks() or .get_opcodes(), in which case you may
        want to try .quick_ratio() or .real_quick_ratio() first to get an
        upper bound.

        >>> s = SequenceMatcher(None, "abcd", "bcde")
        >>> s.ratio()
        0.75
        >>> s.quick_ratio()
        0.75
        >>> s.real_quick_ratio()
        1.0
        """

        matches = reduce(lambda sum, triple: sum + triple[-1],
                         self.get_matching_blocks(), 0)
        return _calculate_ratio(matches, len(self.a) + len(self.b))
Return a measure of the sequences' similarity (float in [0,1]).

        Where T is the total number of elements in both sequences, and
        M is the number of matches, this is 2.0*M / T.
        Note that this is 1 if the sequences are identical, and 0 if
        they have nothing in common.

        .ratio() is expensive to compute if you haven't already computed
        .get_matching_blocks() or .get_opcodes(), in which case you may
        want to try .quick_ratio() or .real_quick_ratio() first to get an
        upper bound.

        >>> s = SequenceMatcher(None, "abcd", "bcde")
        >>> s.ratio()
        0.75
        >>> s.quick_ratio()
        0.75
        >>> s.real_quick_ratio()
        1.0