inverse of izip

[Steven Bethard]

Satchidanand Haridas <sharidas <at> zeomega.com> writes:
> How about using iter() to get another solution like the following:
> 
>  >>> starzip2 = lambda it: tuple([iter(x) for x in itertools.izip(*it)])
> 
>  >>> l,m = starzip2(itertools.izip(range(10),range(10)))
> 
>  >>> l
> <tupleiterator object at 0x4016802c>
>  >>> m
> <tupleiterator object at 0x4016896c>
> 
>  >>> list(l)
> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>  >>> list(m)
> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]


Unfortunately, I think this exhausts the iterators too early because it 
applies * to the iterator:

>>> def range10():
...     for i in range(10):
...             yield i
...     print "exhausted"
...
>>> l,m = starzip2(itertools.izip(range10(),range10()))
exhausted

I believe we only get one "exhausted" because as soon as one iterator is used 
up with izip, the next iterator is discarded.  But we are hitting "exhausted" 
before we ever ask for an element from the starzip2 iterators, so it looks to 
me like all the pairs from the first iterator are read into memory before the 
second iterators are ever accessed...

Steve