Consume an iterable

[Muhammad Alkarouri]

Thanks everyone, but not on my machine (Python 2.6.1, OS X 10.6) it's
not:


In [1]: from itertools import count, islice

In [2]: from collections import deque

In [3]: i1=count()

In [4]: def consume1(iterator, n):
   ...:     deque(islice(iterator, n), maxlen=0)
   ...:
   ...:

In [5]: i2=count()

In [6]: def consume2(iterator, n):
   ...:     for _ in islice(iterator, n): pass
   ...:
   ...:

In [7]: timeit consume1(i1, 10)
1000000 loops, best of 3: 1.63 us per loop

In [8]: timeit consume2(i2, 10)
1000000 loops, best of 3: 846 ns per loop


Can somebody please test whether it is only my machine or is this
reproducible?

(Thanks Jan for making me actually carry the test)