On Jan 22, 1:19 pm, Alan Isaac <[EMAIL PROTECTED]> wrote:
> I suppose my question should have been,
> is there an obviously faster way?
> Anyway, of the four ways below, the
> first is substantially fastest. Is
> there an obvious reason why?
Can you post your results?
I get different ones (pairs1 and pairs2 rewritten slightly to avoid
unnecessary indirection).
====== pairs.py ===========
from itertools import *
def pairs1(x):
return izip(islice(x,0,None,2),islice(x,1,None,2))
def pairs2(x):
xiter = iter(x)
while True:
yield xiter.next(), xiter.next()
def pairs3(x):
for i in range( len(x)//2 ):
yield x[2*i], x[2*i+1],
def pairs4(x):
xiter = iter(x)
return izip(xiter,xiter)
def compare():
import timeit
for i in '1234':
t = timeit.Timer('list(pairs.pairs%s(l))' % i,
'import pairs; l=range(1000)')
print 'pairs%s: %s' % (i, t.timeit(10000))
if __name__ == '__main__':
compare()
=====================
marigold:python arno$ python pairs.py
pairs1: 0.789824962616
pairs2: 4.08462786674
pairs3: 2.90438890457
pairs4: 0.536775827408
pairs4 wins.
--
Arnaud
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