On Sun, 05 Nov 2006 19:35:58 +0000, Tuomas wrote:
> Thanks. My solution became:
>
> >>> def flattern(arg):
> ... result = []
> ... for item in arg:
> ... if isinstance(item, (list, tuple)):
> ... result.extend(flattern(item))
> ... else:
> ... result.append(item)
> ... return tuple(result)
> ...
> >>> def g(*arg):
> ... arg = flattern(arg)
> ... return arg
> ...
> >>> def f(*arg):
> ... return g(arg)
> ...
> >>> f('foo', 'bar')
> ('foo', 'bar')
That's the most complicated do-nothing function I've ever seen. Here is a
shorter version:
def shortf(*args):
return args
>>> f('foo', 'bar')
('foo', 'bar')
>>> shortf('foo', 'bar')
('foo', 'bar')
>>> f(1,2,3,4)
(1, 2, 3, 4)
>>> shortf(1,2,3,4)
(1, 2, 3, 4)
>>> f({}, None, 1, -1.2, "hello world")
({}, None, 1, -1.2, 'hello world')
>>> shortf({}, None, 1, -1.2, "hello world")
({}, None, 1, -1.2, 'hello world')
Actually, they aren't *quite* identical: your function rips lists apart,
which is probably not a good idea.
>>> f("foo", [1,2,3], None) # three arguments turns into five
('foo', 1, 2, 3, None)
>>> shortf("foo", [1,2,3], None) # three arguments stays three
('foo', [1, 2, 3], None)
I still don't understand why you are doing this. Can we have an example of
why you think you need to do this?
--
Steven.
--
http://mail.python.org/mailman/listinfo/python-list
