List and Dicts as default args

[Fredrik Lundh]

Stefan Migowsky <smigowsky at dspace.de> wrote:
> I was just wondering how to handle lists and dictionaries in
> a simple way as default arguments to functions. Since following
> strange behaviour occured :

when exposed to "strange" behaviour, please check the FAQ
http://www.python.org/doc/FAQ.html#6.25

or the fine manual
http://www.python.org/doc/current/ref/function.html

before posting.

> >>> def f(Index,List = [], Dict = {}):
> ...     List.append(1)
> ...     List.append(2)
> ...     print List
> ...     Dict[Index] = Index
> ...     print Dict
> >>> f(1)
> [1, 2]
> {1: 1}
> >>> f(2)
> [1, 2, 1, 2]
> {2: 2, 1: 1}
> >>> f(3)
> [1, 2, 1, 2, 1, 2]
> {3: 3, 2: 2, 1: 1}
>
> This behaviour only occurs with dictionaries and list. All other
> types are "well" behaved.

no, they're not.  default arguments are created once, when
the function object itself is created.  this means that *all*
mutable default values behave the same way -- there is a
single default value, used for all calls.  if you modify that in
place, you loose.

(to understand why, you need to dig a little deeper into what
the 'def' statement actually does.  see the language reference
for details).

> I could try the following but that doesn't look so nice:
>
> def f(Index,List = None, Dict = None):
>     if not List: List = []
>     List.append(1)
>     List.append(2)
>     print List
>     if not Dict: Dict = {}
>     Dict[Index] = Index
>     print Dict

nice or not, that's the right way to do it.

(note that using Caps for variable names isn't good python
style, but that's another story).

</F>