这是一个老问题，但是对于那些仍然感兴趣的人来说，在这个页面中有一个使用生成器的窗口滑块的伟大实现(Adrian Rosebrock)。

它是OpenCV的一个实现，但是你可以很容易地将它用于任何其他目的。对于渴望的人，我将粘贴代码在这里，但为了更好地理解它，我建议访问原始页面。

def sliding_window(image, stepSize, windowSize):
    # slide a window across the image
    for y in xrange(0, image.shape[0], stepSize):
        for x in xrange(0, image.shape[1], stepSize):
            # yield the current window
            yield (x, y, image[y:y + windowSize[1], x:x + windowSize[0]])

提示:您可以在迭代生成器时检查窗口的.shape，以丢弃那些不符合您需求的窗口

干杯

如何使用以下方法:

mylist = [1, 2, 3, 4, 5, 6, 7]

def sliding_window(l, window_size=2):
    if window_size > len(l):
        raise ValueError("Window size must be smaller or equal to the number of elements in the list.")

    t = []
    for i in xrange(0, window_size):
        t.append(l[i:])

    return zip(*t)

print sliding_window(mylist, 3)

输出:

[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7)]

为什么不

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

它被记录在Python文档中。 您可以轻松地将其扩展到更宽的窗口。

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]

更新

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)]
>>> []