a sequence question

[Nick Coghlan]

Steven Bethard wrote:
> I think you can write that second one so that it works for iterables 
> without a __len__:
> 
> py> def padded_partition(iterable, part_len, pad_val=None):
> ...     itr = itertools.chain(
> ...         iter(iterable), itertools.repeat(pad_val, part_len - 1))
> ...     return itertools.izip(*[itr]*part_len)
> ...
> py> list(padded_partition(itertools.islice(itertools.count(), 10), 2))
> [(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]
> py> list(padded_partition(itertools.islice(itertools.count(), 10), 3))
> [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]
> 
> I just unconditionally pad the iterable with 1 less than the partition 
> size...  I think that works right, but I haven't tested it any more than 
> what's shown.

I think you're right - I was looking at padding unconditionally, but because I 
was padding with the actual partition length, it didn't work correctly when the 
padding wasn't needed.

Padding with one less than the partition length fixes that quite neatly.

Cheers,
Nick.

-- 
Nick Coghlan   |   ncoghlan at email.com   |   Brisbane, Australia
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