* and return an object that buckets it iterable into
    child iterables based on a *key* function.

        >>> iterable = ['a1', 'b1', 'c1', 'a2', 'b2', 'c2', 'b3']
        >>> s = bucket(iterable, key=lambda x: x[0])  # Bucket by 1st character
        >>> sorted(list(s))  # Get the keys
        ['a', 'b', 'c']
        >>> a_iterable = s['a']
        >>> next(a_iterable)
        'a1'
        >>> next(a_iterable)
        'a2'
        >>> list(s['b'])
        ['b1', 'b2', 'b3']

    The original iterable will be advanced and its items will be cached until
    they are used by the child iterables. This may require significant storage.

    By default, attempting to select a bucket to which no items belong  will
    exhaust the iterable and cache all values.
    If you specify a *validator* function, selected buckets will instead be
    checked against it.

        >>> from itertools import count
        >>> it = count(1, 2)  # Infinite sequence of odd numbers
        >>> key = lambda x: x % 10  # Bucket by last digit
        >>> validator = lambda x: x in {1, 3, 5, 7, 9}  # Odd digits only
        >>> s = bucket(it, key=key, validator=validator)
        >>> 2 in s
        False
        >>> list(s[2])
        []

    Nc