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]
"""
def GetShiftingWindows(thelist, size):
return [ thelist[x:x+size] for x in range( len(thelist) - size + 1 ) ]
>> a = [1, 2, 3, 4, 5]
>> GetShiftingWindows(a, 3)
[ [1, 2, 3], [2, 3, 4], [3, 4, 5] ]
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
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.
