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?


当前回答

我使用下面的代码作为一个简单的滑动窗口,它使用生成器来大幅提高可读性。根据我的经验,到目前为止,它的速度足以用于生物信息学序列分析。

我把它包括在这里是因为我还没有看到这种方法被使用过。同样,我对它的比较性能没有任何评价。

def slidingWindow(sequence,winSize,step=1):
"""Returns a generator that will iterate through
the defined chunks of input sequence. Input sequence
must be sliceable."""

    # Verify the inputs
    if not ((type(winSize) == type(0)) and (type(step) == type(0))):
        raise Exception("**ERROR** type(winSize) and type(step) must be int.")
    if step > winSize:
        raise Exception("**ERROR** step must not be larger than winSize.")
    if winSize > len(sequence):
        raise Exception("**ERROR** winSize must not be larger than sequence length.")

    # Pre-compute number of chunks to emit
    numOfChunks = ((len(sequence)-winSize)/step)+1

    # Do the work
    for i in range(0,numOfChunks*step,step):
        yield sequence[i:i+winSize]

其他回答

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:)

>>> 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]

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)

让我们让它变懒!

from itertools import islice, tee

def window(iterable, size): 
    iterators = tee(iterable, size) 
    iterators = [islice(iterator, i, None) for i, iterator in enumerate(iterators)]  
    yield from zip(*iterators)

list(window(range(5), 3))
# [(0, 1, 2), (1, 2, 3), (2, 3, 4)]

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)