我需要一个滚动窗口(又名滑动窗口)可迭代的序列/迭代器/生成器。(默认的Python迭代可以被认为是一种特殊情况，其中窗口长度为1。)我目前正在使用以下代码。我怎样才能做得更优雅和/或更有效?
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]
"""
对于window_size == 2的特定情况(即，在序列中迭代相邻的重叠对)，请参见如何从列表中迭代重叠(当前，下一个)值对?
#Importing the numpy library
import numpy as np
arr = np.arange(6) #Sequence
window_size = 3
np.lib.stride_tricks.as_strided(arr, shape= (len(arr) - window_size +1, window_size),
strides = arr.strides*2)
"""Example output:
[0, 1, 2]
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
"""
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:)
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。
>>> 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]
