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]

我怎样才能一次迭代每两个元素？

我在试试这个,

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

获得输出就像,

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

预期的产出是

[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itertools中的grouper配方:

>>> 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]

Answer 2

有什么问题：

l = [1, 2, 3, 4, 5, 6, 7, 8]
for j in range(0, len(l), 2):
        print(l[j: j + 2])

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

假设列表有偶数个元素

Answer 3

您可以按照自己的方式进行操作,只需在切片中添加一个步骤部分即可使两个切片都跳过一个数字:

for v, w in zip(l[::2],l[1::2]):  # No need to end at -1 because that's the default
    print [v, w]

但我喜欢辅助生成器:

def pairwise(iterable):
    i = iter(iterable)
    while True:
       yield i.next(), i.next()

for v, w in pairwise(l):
    print v, w

Answer 4

一个解决方案是

for v, w in zip(l[::2],l[1::2]):
    print [v, w]