滚动或滑动窗口迭代器？

我需要一个可在序列/迭代器/生成器上迭代的滚动窗口（又名滑动窗口）。 （默认的 Python 迭代可以被视为一种特殊情况，其中窗口长度为 1。）我当前正在使用以下代码。我怎样才能更优雅和/或更有效地做到这一点？

def rolling_window(seq, window_size):
    it = iter(seq)
    win = [it.next() for cnt in xrange(window_size)] # First window
    yield win
    for e in it: # Subsequent windows
        win[:-1] = win[1:]
        win[-1] = e
        yield win

if __name__=="__main__":
    for w in rolling_window(xrange(6), 3):
        print w

"""Example output:   
   [0, 1, 2]
   [1, 2, 3]
   [2, 3, 4]
   [3, 4, 5]
"""

_{For the specific case of window_size == 2 (i.e., iterating over adjacent, overlapping pairs in a sequence), see also How can I iterate over overlapping (current, next) pairs of values from a list? https://stackoverflow.com/questions/5434891/.}

旧版本的 Python 文档中有一个itertools例子 http://docs.python.org/release/2.3.5/lib/itertools-example.html:

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

文档中的内容更加简洁，并且使用itertools我想效果会更好。

如果你的迭代器是一个简单的列表/元组使用指定窗口大小滑动它的简单方法是：

seq = [0, 1, 2, 3, 4, 5]
window_size = 3

for i in range(len(seq) - window_size + 1):
    print(seq[i: i + window_size])

Output:

[0, 1, 2]
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]