Andrew Dalke wrote:
> Steven Bethard wrote:
>> Here's one possible solution:
>>
>> py> import itertools as it
>> py> def zipfill(*lists):
>> ... max_len = max(len(lst) for lst in lists)
>
> A limitation to this is the need to iterate over the
> lists twice, which might not be possible if one of them
> is a file iterator.
>
> Here's a clever, though not (in my opinion) elegant solution
>
> import itertools
>
> def zipfill(*seqs):
> count = [len(seqs)]
> def _forever(seq):
> for item in seq: yield item
> count[0] -= 1
> while 1: yield None
> seqs = [_forever(seq) for seq in seqs]
> while 1:
> x = [seq.next() for seq in seqs]
> if count == [0]:
> break
> yield x
> This seems a bit more elegant, though the "replace" dictionary is
> still a bit of a hack
>
> from itertools import repeat, chain, izip
>
> sentinel = object()
> end_of_stream = repeat(sentinel)
>
> def zipfill(*seqs):
> replace = {sentinel: None}.get
> seqs = [chain(seq, end_of_stream) for seq in seqs]
> for term in izip(*seqs):
> for element in term:
> if element is not sentinel:
> break
> else:
> # All sentinels
> break
>
> yield [replace(element, element) for element in term]
Combining your "clever" and your "elegant" approach to something fast
(though I'm not entirely confident it's correct):
def fillzip(*seqs):
def done_iter(done=[len(seqs)]):
done[0] -= 1
if not done[0]:
return
while 1:
yield None
seqs = [chain(seq, done_iter()) for seq in seqs]
return izip(*seqs)
Whether we ran out of active sequences is only tested once per sequence.
Fiddling with itertools is always fun, but feels a bit like reinventing the
wheel in this case. The only excuse being that you might need a lazy
map(None, ...) someday...
Peter
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