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?
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)]
有一个库可以完全满足你的需要:
import more_itertools
list(more_itertools.windowed([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],n=3, step=3))
Out: [(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)]
我喜欢t ():
from itertools import tee, izip
def window(iterable, size):
iters = tee(iterable, size)
for i in xrange(1, size):
for each in iters[i:]:
next(each, None)
return izip(*iters)
for each in window(xrange(6), 3):
print list(each)
给:
[0, 1, 2]
[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]
