Chris Angelico wrote:
> On Thu, Feb 9, 2012 at 2:55 PM, Steven D'Aprano
> <steve+comp.lang.pyt...@pearwood.info> wrote:
>> If your data is humongous but only available lazily, buy more memory :)
>
> Or if you have a huge iterable and only need a small index into it,
> snag those first few entries into a list, then yield everything else,
> then yield the saved ones:
> def cycle(seq,n):
> seq=iter(seq)
> lst=[next(seq) for i in range(n)]
> try:
> while True: yield next(seq)
> except StopIteration:
> for i in lst: yield i
I think that should be spelt
def cycle2(seq, n):
seq = iter(seq)
head = [next(seq) for i in range(n)]
for item in seq:
yield item
for item in head:
yield item
or, if you are into itertools,
def cycle3(seq, n):
seq = iter(seq)
return chain(seq, list(islice(seq, n)))
$ python -m timeit -s'from tmp import cycle; data = range(1000); start=10'
'for item in cycle(data, 10): pass'
1000 loops, best of 3: 358 usec per loop
$ python -m timeit -s'from tmp import cycle2; data = range(1000); start=10'
'for item in cycle2(data, 10): pass'
1000 loops, best of 3: 172 usec per loop
$ python -m timeit -s'from tmp import cycle3; data = range(1000); start=10'
'for item in cycle3(data, 10): pass'
10000 loops, best of 3: 56.5 usec per loop
For reference:
$ python -m timeit -s'data = range(1000); start=10' 'for item in
data[start:] + data[:start]: pass'
10000 loops, best of 3: 56.4 usec per loop
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