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?
如何使用以下方法:
mylist = [1, 2, 3, 4, 5, 6, 7]
def sliding_window(l, window_size=2):
if window_size > len(l):
raise ValueError("Window size must be smaller or equal to the number of elements in the list.")
t = []
for i in xrange(0, window_size):
t.append(l[i:])
return zip(*t)
print sliding_window(mylist, 3)
输出:
[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7)]
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]
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]
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
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