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Kelly发现这是一个重复的答案。但我在这里留下这个作为反例，因为我包含了一个毫无意义的最小值。

所以如果你想用min来避免IndexError，没有必要，range会帮你处理这种情况。

旧的答案

奇怪的是，当n > len(l)返回[]时，下面的句柄在语义上是正确的。

>>> l = [0, 1, 2, 3, 4]

>>> n = 2
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1], [1, 2], [2, 3], [3, 4]]
>>>
>>> n = 3
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2], [1, 2, 3], [2, 3, 4]]
>>>
>>> n = 4
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2, 3], [1, 2, 3, 4]]
>>>
>>> n = 5
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2, 3, 4]]
>>>
>>> n = 10 # n > len(l)
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> []

这里是一个泛化，增加了对step, fillvalue参数的支持:

from collections import deque
from itertools import islice

def sliding_window(iterable, size=2, step=1, fillvalue=None):
    if size < 0 or step < 1:
        raise ValueError
    it = iter(iterable)
    q = deque(islice(it, size), maxlen=size)
    if not q:
        return  # empty iterable or size == 0
    q.extend(fillvalue for _ in range(size - len(q)))  # pad to size
    while True:
        yield iter(q)  # iter() to avoid accidental outside modifications
        try:
            q.append(next(it))
        except StopIteration: # Python 3.5 pep 479 support
            return
        q.extend(next(it, fillvalue) for _ in range(step - 1))

它每次产生块大小的项目，每次迭代滚动步骤位置，在必要时用fillvalue填充每个块。示例:size=4, step=3, fillvalue='*':

 [a b c d]e f g h i j k l m n o p q r s t u v w x y z
  a b c[d e f g]h i j k l m n o p q r s t u v w x y z
  a b c d e f[g h i j]k l m n o p q r s t u v w x y z
  a b c d e f g h i[j k l m]n o p q r s t u v w x y z
  a b c d e f g h i j k l[m n o p]q r s t u v w x y z
  a b c d e f g h i j k l m n o[p q r s]t u v w x y z
  a b c d e f g h i j k l m n o p q r[s t u v]w x y z
  a b c d e f g h i j k l m n o p q r s t u[v w x y]z
  a b c d e f g h i j k l m n o p q r s t u v w x[y z * *]

有关step参数的用例示例，请参见用python有效地处理大型.txt文件。

这似乎是为collections.deque定制的，因为您实际上有一个FIFO(添加到一端，从另一端删除)。然而，即使你使用列表，你也不应该切片两次;相反，您应该只从列表中弹出(0)并追加()新项。

下面是一个基于deque的优化实现:

from collections import deque

def window(seq, n=2):
    it = iter(seq)
    win = deque((next(it, None) for _ in xrange(n)), maxlen=n)
    yield win
    append = win.append
    for e in it:
        append(e)
        yield win

在我的测试中，它在大多数时候都轻松击败了这里发布的其他所有东西，尽管pillmuncher的tee版本在大可迭代对象和小窗口方面击败了它。在较大的窗口上，deque再次以原始速度领先。

Access to individual items in the deque may be faster or slower than with lists or tuples. (Items near the beginning are faster, or items near the end if you use a negative index.) I put a sum(w) in the body of my loop; this plays to the deque's strength (iterating from one item to the next is fast, so this loop ran a a full 20% faster than the next fastest method, pillmuncher's). When I changed it to individually look up and add items in a window of ten, the tables turned and the tee method was 20% faster. I was able to recover some speed by using negative indexes for the last five terms in the addition, but tee was still a little faster. Overall I would estimate that either one is plenty fast for most uses and if you need a little more performance, profile and pick the one that works best.

在旧版本的Python文档中有一个itertools示例:

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

文档中的那个更简洁一点，我想它使用了itertools来达到更好的效果。

如果你的迭代器是一个简单的列表/元组，用指定的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])

输出:

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

deque窗口的一个轻微修改版本，使其成为一个真正的滚动窗口。因此，它开始只填充一个元素，然后增长到它的最大窗口大小，然后缩小，因为它的左边缘接近结束:

from collections import deque
def window(seq, n=2):
    it = iter(seq)
    win = deque((next(it, None) for _ in xrange(1)), maxlen=n)
    yield win
    append = win.append
    for e in it:
        append(e)
        yield win
    for _ in xrange(len(win)-1):
        win.popleft()
        yield win

for wnd in window(range(5), n=3):
    print(list(wnd))

这给了

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

修改了DiPaolo的答案，允许任意填充和可变步长

import itertools
def window(seq, n=2,step=1,fill=None,keep=0):
    "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(itertools.islice(it, n))    
    if len(result) == n:
        yield result
    while True:        
#         for elem in it:        
        elem = tuple( next(it, fill) for _ in range(step))
        result = result[step:] + elem        
        if elem[-1] is fill:
            if keep:
                yield result
            break
        yield result