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?
我的两个版本的窗口实现
from typing import Sized, Iterable
def window(seq: Sized, n: int, strid: int = 1, drop_last: bool = False):
for i in range(0, len(seq), strid):
res = seq[i:i + n]
if drop_last and len(res) < n:
break
yield res
def window2(seq: Iterable, n: int, strid: int = 1, drop_last: bool = False):
it = iter(seq)
result = []
step = 0
for i, ele in enumerate(it):
result.append(ele)
result = result[-n:]
if len(result) == n:
if step % strid == 0:
yield result
step += 1
if not drop_last:
yield result
深度学习中滑动窗口数据的优化函数
def SlidingWindow(X, window_length, stride):
indexer = np.arange(window_length)[None, :] + stride*np.arange(int(len(X)/stride)-window_length+4)[:, None]
return X.take(indexer)
应用于多维数组
import numpy as np
def SlidingWindow(X, window_length, stride1):
stride= X.shape[1]*stride1
window_length = window_length*X.shape[1]
indexer = np.arange(window_length)[None, :] + stride1*np.arange(int(len(X)/stride1)-window_length-1)[:, None]
return X.take(indexer)
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]
