itertools.izip brokeness

[rurpy at yahoo.com]

Bengt Richter wrote:
> On 5 Jan 2006 15:48:26 GMT, Antoon Pardon <apardon at forel.vub.ac.be> wrote:
>
> >On 2006-01-04, rurpy at yahoo.com <rurpy at yahoo.com> wrote:
> >><rurpy at yahoo.com> wrote:
> >>> But here is my real question...
> >>> Why isn't something like this in itertools, or why shouldn't
> >>> it go into itertools?
> >>
> >>
> >>   4) If a need does arise, it can be met by __builtins__.map() or by
> >>      writing:  chain(iterable, repeat(None)).
> >>
> >> Yes, if youre a python guru.  I don't even understand the
> >> code presented in this thread that uses chain/repeat,
> >
> >And it wouldn't work in this case. chain(iterable, repeat(None))
> >changes your iterable into an iterator that first gives you
> >all elements in the iterator and when these are exhausted
> >will continue giving the repeat parameter. e.g.
> >
> >  chain([3,5,8],repeat("Bye")
> >
> >Will produce  3, 5 and 8 followed by an endless stream
> >of "Bye".
> >
> >But if you do this with all iterables, and you have to
> >because you don't know which one is the smaller, all
> >iterators will be infinite and izip will never stop.
>
> But you can fix that (only test is what you see ;-) :
>
>  >>> from itertools import repeat, chain, izip
>  >>> it = iter(lambda z=izip(chain([3,5,8],repeat("Bye")), chain([11,22],repeat("Bye"))):z.next(), ("Bye","Bye"))
>  >>> for t in it: print t
>  ...
>  (3, 11)
>  (5, 22)
>  (8, 'Bye')
>
> (Feel free to generalize ;-)

Which just reinforces my original point: if leaving
out a feature is justified by the existence of some
alternate method, then that method must be equally
obvious as the missing feature, or must be documented
as an idiom.  Otherwise, the justification fails.

Is the above code as obvious as
  izip([3,5,8],[11,22],sentinal='Bye')?
(where the sentinal keyword causes izip to iterate
to the longest argument.)