c = lambda x: x.upper()
        >>> valuefunc = lambda x: 1
        >>> reducefunc = sum
        >>> result = map_reduce('abbccc', keyfunc, valuefunc, reducefunc)
        >>> sorted(result.items())
        [('A', 1), ('B', 2), ('C', 3)]

    You may want to filter the input iterable before applying the map/reduce
    procedure:

        >>> all_items = range(30)
        >>> items = [x for x in all_items if 10 <= x <= 20]  # Filter
        >>> keyfunc = lambda x: x % 2  # Evens map to 0; odds to 1
        >>> categories = map_reduce(items, keyfunc=keyfunc)
        >>> sorted(categories.items())
        [(0, [10, 12, 14, 16, 18, 20]), (1, [11, 13, 15, 17, 19])]
        >>> summaries = map_reduce(items, keyfunc=keyfunc, reducefunc=sum)
        >>> sorted(summaries.items())
        [(0, 90), (1, 75)]

    Note that all items in the iterable are gathered into a list before the
    summarization step, which may require significant storage.

    The returned object is a :obj:`collections.defaultdict` with the
    ``default_factory`` set to ``None``, such that it behaves like a normal
    dictionary.

    Nc