一次迭代列表中的每 2 个元素 [英] Iterate every 2 elements from list at a time

问题描述

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?

我正在尝试这个，

for v, w in zip(l[:-1],l[1:]):打印 [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]

预期输出为

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

解决方案

你可以使用iter:

<预><代码>>>>seq = [1,2,3,4,5,6,7,8,9,10]>>>it = iter(seq)>>>对于其中的 x:...打印(x，下一个(它))...[1, 2][3, 4][5, 6][7, 8][9, 10]

您还可以使用 grouper 配方 来自 itertools:

<预><代码>>>>从 itertools 导入 izip_longest>>>def grouper(iterable, n, fillvalue=None):...将数据收集到固定长度的块或块中"... # grouper('ABCDEFG', 3, 'x') -->ABC DEF Gxx... args = [iter(iterable)] * n...返回 izip_longest(fillvalue=fillvalue, *args)...>>>对于石斑鱼中的 x, y(seq, 2):... 打印 (x, y)...[1, 2][3, 4][5, 6][7, 8][9, 10]

Having list as ,

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

How can i iterate every two elements at a time ?

I am trying this ,

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

and getting output is like ,

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

Expected output is

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

解决方案

You can use 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]

You can also use the grouper recipe from 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]

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