On Nov 6, 2:52 am, Steven D'Aprano <st...@remove-this-
cybersource.com.au> wrote:
> My first attempt was this:
>
> def split(iterable, n):
> iterators = []
> for i, iterator in enumerate(itertools.tee(iterable, n)):
> iterators.append((t[i] for t in iterator))
> return tuple(iterators)
>
> But it doesn't work, as all the iterators see the same values:
Because the value of i is not evaluated until the generator is
actually run; so all the generators end up seeing only the final value
of i rather than the intended values. This is a common problem with
generator expressions that are not immediately run.
> I tried changing the t[i] to use operator.itergetter instead, but no
> luck. Finally I got this:
>
> def split(iterable, n):
> iterators = []
> for i, iterator in enumerate(itertools.tee(iterable, n)):
> f = lambda it, i=i: (t[i] for t in it)
> iterators.append(f(iterator))
> return tuple(iterators)
>
> which seems to work:
>
> >>> data = [(1,2,3), (4,5,6), (7,8,9)]
> >>> a, b, c = split(data, 3)
> >>> list(a), list(b), list(c)
>
> ([1, 4, 7], [2, 5, 8], [3, 6, 9])
>
> Is this the right approach, or have I missed something obvious?
That avoids the generator problem, but in this case you could get the
same result a bit more straight-forwardly by just using imap instead:
def split(iterable, n):
iterators = []
for i, iterator in enumerate(itertools.tee(iterable, n)):
iterators.append(itertools.imap(operator.itemgetter(i),
iterator))
return tuple(iterators)
>>> map(list, split(data, 3))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
Cheers,
Ian
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