Idiom for default values when unpacking a tuple

[Steven Bethard]

Carlos Ribeiro wrote:
> How about having a iterator function that is guaranteed to always
> return a fixed number of elements, regardless of the size of the
> sequence? I've checked it, and it does not exist in itertools.
> Something like this (simple-minded, just a proof of concept, and
> highly optimizable in at least a hundred different ways :-):
> 
> def iterfixed(seq, times, defaultitem=None):
>     for i in range(times):
>         if i < len(seq):
>             yield seq[i]
>         else:
>             yield defaultitem

Well, it doesn't quite exist in itertools, but it's there with just a 
simple composition:

 >>> def iterfixed(seq, times, defaultitem=None):
... 	return it.islice(it.chain(iter(seq), it.repeat(defaultitem)), times)
...
 >>> tuple(iterfixed((1,2,3,4), 3))
(1, 2, 3)
 >>> tuple(iterfixed((1,2,3,4), 6))
(1, 2, 3, 4, None, None)
 >>> a,b,c = iterfixed((1,2,3,4), 3)
 >>> a,b,c
(1, 2, 3)
 >>> a,b,c,d,e,f = iterfixed((1,2,3,4), 6)
 >>> a,b,c,d,e,f
(1, 2, 3, 4, None, None)

> The only catch is that, if you have only one parameter, then all you
> will get is the generator itself. But that's a corner case, and not
> the intended use anyway.

Not exactly sure what you mean here.  If you only have one item in your 
unpack tuple, I believe things still work, e.g.:

 >>> a, = iterfixed((1,2,3,4), 1)
 >>> a
1

But I'm probably just misunderstanding your statement...

Steve