def map_reduce(iterable, keyfunc, valuefunc=None, reducefunc=None):
    """Return a dictionary that maps the items in *iterable* to categories
    defined by *keyfunc*, transforms them with *valuefunc*, and
    then summarizes them by category with *reducefunc*.

    *valuefunc* defaults to the identity function if it is unspecified.
    If *reducefunc* is unspecified, no summarization takes place:

        >>> keyfunc = lambda x: x.upper()
        >>> result = map_reduce('abbccc', keyfunc)
        >>> sorted(result.items())
        [('A', ['a']), ('B', ['b', 'b']), ('C', ['c', 'c', 'c'])]

    Specifying *valuefunc* transforms the categorized items:

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

    Specifying *reducefunc* summarizes the categorized items:

        >>> keyfunc = 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.

    """
    valuefunc = (lambda x: x) if (valuefunc is None) else valuefunc

    ret = defaultdict(list)
    for item in iterable:
        key = keyfunc(item)
        value = valuefunc(item)
        ret[key].append(value)

    if reducefunc is not None:
        for key, value_list in ret.items():
            ret[key] = reducefunc(value_list)

    ret.default_factory = None
    return ret
Return a dictionary that maps the items in *iterable* to categories
    defined by *keyfunc*, transforms them with *valuefunc*, and
    then summarizes them by category with *reducefunc*.

    *valuefunc* defaults to the identity function if it is unspecified.
    If *reducefunc* is unspecified, no summarization takes place:

        >>> keyfunc = lambda x: x.upper()
        >>> result = map_reduce('abbccc', keyfunc)
        >>> sorted(result.items())
        [('A', ['a']), ('B', ['b', 'b']), ('C', ['c', 'c', 'c'])]

    Specifying *valuefunc* transforms the categorized items:

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

    Specifying *reducefunc* summarizes the categorized items:

        >>> keyfunc = 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.

    
product/runtime/src/main/python/pkg_resources/_vendor/more_itertools/more.py