在滑动对中迭代列表的最 Pythonic 最有效的方法是什么？这是一个相关的例子：
>>> l
['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> for x, y in itertools.izip(l, l[1::2]): print x, y
...
a b
b d
c f
这是成对的迭代，但是我们如何才能在滑动对上进行迭代呢？含义迭代对：
a b
b c
c d
d e
etc.
这是对对的迭代，除了每次将对滑动 1 个元素而不是 2 个元素。谢谢。
你可以更简单。只需压缩列表并将列表偏移一。
In [4]: zip(l, l[1:])
Out[4]: [('a', 'b'), ('b', 'c'), ('c', 'd'), ('d', 'e'), ('e', 'f'), ('f', 'g')]
怎么样：
for x, y in itertools.izip(l, l[1:]): print x, y
这是我不久前为类似场景编写的一个小生成器：
def pairs(items):
items_iter = iter(items)
prev = next(items_iter)
for item in items_iter:
yield prev, item
prev = item
这是一个适用于迭代器/生成器以及列表的任意大小滑动窗口的函数
def sliding(seq, n):
return izip(*starmap(islice, izip(tee(seq, n), count(0), repeat(None))))
不过，内森的解决方案可能更有效。
通过在列表中添加两个后续条目定义的时间显示在下方，并按从最快到最慢的顺序排列。
吉尔
In [69]: timeit.repeat("for x,y in itertools.izip(l, l[1::1]): x + y", setup=setup, number=1000)
Out[69]: [1.029047966003418, 0.996290922164917, 0.998831033706665]
杰夫·里迪
In [70]: timeit.repeat("for x,y in sliding(l,2): x+y", setup=setup, number=1000)
Out[70]: [1.2408790588378906, 1.2099130153656006, 1.207326889038086]
阿莱斯塔尼斯
In [66]: timeit.repeat("for i in range(0, len(l)-1): l[i] + l[i+1]", setup=setup, number=1000)
Out[66]: [1.3387370109558105, 1.3243639469146729, 1.3245630264282227]
奇穆利格
In [68]: timeit.repeat("for x,y in zip(l, l[1:]): x+y", setup=setup, number=1000)
Out[68]: [1.4756009578704834, 1.4369518756866455, 1.5067830085754395]
内森·维拉斯库萨
In [63]: timeit.repeat("for x,y in pairs(l): x+y", setup=setup, number=1000)
Out[63]: [2.254757881164551, 2.3750967979431152, 2.302199125289917]
SR2222
注意减少的重复次数......
In [60]: timeit.repeat("for x,y in SubsequenceIter(l,2): x+y", setup=setup, number=100)
Out[60]: [1.599524974822998, 1.5634570121765137, 1.608154058456421]
设置代码：
setup="""
from itertools import izip, starmap, islice, tee, count, repeat
l = range(10000)
def sliding(seq, n):
return izip(*starmap(islice, izip(tee(seq, n), count(0), repeat(None))))
class SubsequenceIter(object):
def __init__(self, iterable, subsequence_length):
self.iterator = iter(iterable)
self.subsequence_length = subsequence_length
self.subsequence = [0]
def __iter__(self):
return self
def next(self):
self.subsequence.pop(0)
while len(self.subsequence) < self.subsequence_length:
self.subsequence.append(self.iterator.next())
return self.subsequence
def pairs(items):
items_iter = iter(items)
prev = items_iter.next()
for item in items_iter:
yield (prev, item)
prev = item
"""
不完全是最有效的，但非常灵活：
class SubsequenceIter(object):
def __init__(self, iterable, subsequence_length):
self.iterator = iter(iterable)
self.subsequence_length = subsequence_length
self.subsequence = [0]
def __iter__(self):
return self
def next(self):
self.subsequence.pop(0)
while len(self.subsequence) < self.subsequence_length:
self.subsequence.append(self.iterator.next())
return self.subsequence
用法：
for x, y in SubsequenceIter(l, 2):
print x, y
无需导入，只要提供对象列表或字符串即可；任何带有var[indexing]. 经测试python 3.6
# This will create windows with all but 1 overlap
def ngrams_list(a_list, window_size=5, skip_step=1):
return list(zip(*[a_list[i:] for i in range(0, window_size, skip_step)]))
for 循环本身就创建了这个a_list字母表（如图所示window = 5，OP 想要window=2：
['ABCDEFGHIJKLMNOPQRSTUVWXYZ',
'BCDEFGHIJKLMNOPQRSTUVWXYZ',
'CDEFGHIJKLMNOPQRSTUVWXYZ',
'DEFGHIJKLMNOPQRSTUVWXYZ',
'EFGHIJKLMNOPQRSTUVWXYZ']
zip(*result_of_for_loop)将收集所有完整的垂直列作为结果。如果你想要少于一个重叠：
# You can sample that output to get less overlap:
def sliding_windows_with_overlap(a_list, window_size=5, overlap=2):
zip_output_as_list = ngrams_list(a_list, window_size)])
return zip_output_as_list[::overlap+1]
使用它跳过以&overlap=2开头的列，并选择BCD
[('A', 'B', 'C', 'D', 'E'),
('D', 'E', 'F', 'G', 'H'),
('G', 'H', 'I', 'J', 'K'),
('J', 'K', 'L', 'M', 'N'),
('M', 'N', 'O', 'P', 'Q'),
('P', 'Q', 'R', 'S', 'T'),
('S', 'T', 'U', 'V', 'W'),
('V', 'W', 'X', 'Y', 'Z')]
编辑：看起来这与@chmullig 提供的类似，带有选项