python - 一次迭代列表中的每 2 个元素

Question

列表为

l = [1,2,3,4,5,6,7,8,9,0]

如何一次迭代每两个元素？

我正在尝试，

for v, w in zip(l[:-1],l[1:]):
    print [v, w]

得到的输出就像，

[1, 2]
[2, 3]
[3, 4]
[4, 5]
[5, 6]
[6, 7]
[7, 8]
[8, 9]
[9, 0]

预期输出是

[1,2]
[3, 4]
[5, 6]
[7, 8]
[9,10]

Answer 1

你可以使用iter:

>>> seq = [1,2,3,4,5,6,7,8,9,10]
>>> it = iter(seq)
>>> for x in it:
...     print (x, next(it))
...     
[1, 2]
[3, 4]
[5, 6]
[7, 8]
[9, 10]

您还可以使用 grouper recipe来自 itertools:

>>> from itertools import izip_longest
>>> def grouper(iterable, n, fillvalue=None):
...         "Collect data into fixed-length chunks or blocks"
...         # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
...         args = [iter(iterable)] * n
...         return izip_longest(fillvalue=fillvalue, *args)
... 
>>> for x, y in grouper(seq, 2):
...     print (x, y)
...     
[1, 2]
[3, 4]
[5, 6]
[7, 8]
[9, 10]