另一种从列表生成固定长度窗口的简单方法

from collections import deque

def window(ls,window_size=3):
    window = deque(maxlen=window_size)

    for element in ls:
        
        if len(window)==window_size:
            yield list(window)
        window.append(element)

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

for w in window(ls):
    print(w)

>>> n, m = 6, 3
>>> k = n - m+1
>>> print ('{}\n'*(k)).format(*[range(i, i+m) for i in xrange(k)])
[0, 1, 2]
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]

另一种从列表生成固定长度窗口的简单方法

from collections import deque

def window(ls,window_size=3):
    window = deque(maxlen=window_size)

    for element in ls:
        
        if len(window)==window_size:
            yield list(window)
        window.append(element)

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

for w in window(ls):
    print(w)

多个迭代器!

def window(seq, size, step=1):
    # initialize iterators
    iters = [iter(seq) for i in range(size)]
    # stagger iterators (without yielding)
    [next(iters[i]) for j in range(size) for i in range(-1, -j-1, -1)]
    while(True):
        yield [next(i) for i in iters]
        # next line does nothing for step = 1 (skips iterations for step > 1)
        [next(i) for i in iters for j in range(step-1)]

next(it)在序列结束时引发StopIteration，出于一些我无法理解的很酷的原因，yield语句在这里除外它，函数返回，忽略没有形成完整窗口的剩余值。

无论如何，这是目前为止行数最少的解决方案，它唯一的要求是seq实现__iter__或__getitem__，并且除了@dansalmo的解决方案之外，不依赖于itertools或集合:)

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)]

这里有一行。我对它进行了计时，它与顶部答案的性能相当，并且随着更大的seq逐渐变得更好，len(seq) = 20时慢20%，len(seq) = 10000时慢7%

zip(*[seq[i:(len(seq) - n - 1 + i)] for i in range(n)])