Return first element of `iterable` that evaluates true, else return None
    (or an optional default value).

    >>> first([0, False, None, [], (), 42])
    42

    >>> first([0, False, None, [], ()]) is None
    True

    >>> first([0, False, None, [], ()], default='ohai')
    'ohai'

    >>> import re
    >>> m = first(re.match(regex, 'abc') for regex in ['b.*', 'a(.*)'])
    >>> m.group(1)
    'bc'

    The optional `key` argument specifies a one-argument predicate function
    like that used for `filter()`.  The `key` argument, if supplied, must be
    in keyword form.  For example:

    >>> first([1, 1, 3, 4, 5], key=lambda x: x % 2 == 0)
    4

def first(iterable, default=None, key=None):
    """
    Return first element of `iterable` that evaluates true, else return None
    (or an optional default value).

    >>> first([0, False, None, [], (), 42])
    42

    >>> first([0, False, None, [], ()]) is None
    True

    >>> first([0, False, None, [], ()], default='ohai')
    'ohai'

    >>> import re
    >>> m = first(re.match(regex, 'abc') for regex in ['b.*', 'a(.*)'])
    >>> m.group(1)
    'bc'

    The optional `key` argument specifies a one-argument predicate function
    like that used for `filter()`.  The `key` argument, if supplied, must be
    in keyword form.  For example:

    >>> first([1, 1, 3, 4, 5], key=lambda x: x % 2 == 0)
    4

    """
    if key is None:
        for el in iterable:
            if el:
                return el
    else:
        for el in iterable:
            if key(el):
                return el

    return default