结果函数（来自问题的编辑），
来自@ agf，@ FogleBird，@ senderle的答案的想法的弗兰肯尼特，其结果看起来有些整洁：
from itertools import chain, repeat, islice
def window(seq, size=2, fill=0, fill_left=True, fill_right=False):
""" Returns a sliding window (of width n) over data from the iterable:
s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...
"""
ssize = size - 1
it = chain(
repeat(fill, ssize * fill_left),
iter(seq),
repeat(fill, ssize * fill_right))
result = tuple(islice(it, size))
if len(result) == size: # `<=` if okay to return seq if len(seq) < size
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
In [32]: kwa = dict(gen=xrange(1000), size=4, fill=-1, fill_left=True, fill_right=True)
In [33]: %timeit -n 10000 [a+b+c+d for a,b,c,d in tmpf5.ia(**kwa)]
10000 loops, best of 3: 358 us per loop
In [34]: %timeit -n 10000 [a+b+c+d for a,b,c,d in tmpf5.window(**kwa)]
10000 loops, best of 3: 368 us per loop
In [36]: %timeit -n 10000 [sum(x) for x in tmpf5.ia(**kwa)]
10000 loops, best of 3: 340 us per loop
In [37]: %timeit -n 10000 [sum(x) for x in tmpf5.window(**kwa)]
10000 loops, best of 3: 432 us per loop
但是无论如何，如果是数字，那么numpy可能更可取。
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