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, islicedef 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
并且，有关一些双端队列/元组的性能信息：
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 loopIn [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 loopIn [36]: %timeit -n 10000 [sum(x) for x in tmpf5.ia(**kwa)]10000 loops, best of 3: 340 us per loopIn [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。
