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.
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Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia
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