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

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] ]

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]

更新

Kelly发现这是一个重复的答案。但我在这里留下这个作为反例,因为我包含了一个毫无意义的最小值。

所以如果你想用min来避免IndexError,没有必要,range会帮你处理这种情况。


旧的答案

奇怪的是,当n > len(l)返回[]时,下面的句柄在语义上是正确的。

>>> l = [0, 1, 2, 3, 4]

>>> n = 2
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1], [1, 2], [2, 3], [3, 4]]
>>>
>>> n = 3
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2], [1, 2, 3], [2, 3, 4]]
>>>
>>> n = 4
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2, 3], [1, 2, 3, 4]]
>>>
>>> n = 5
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> [[0, 1, 2, 3, 4]]
>>>
>>> n = 10 # n > len(l)
>>> [l[i: i + min(n, len(l)-i)] for i in range(len(l)-n+1)]
>>> []
>>> 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]

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)]