清单为，
l = [1,2,3,4,5,6,7,8,9,0]
我如何一次遍历每两个元素？
我正在尝试这个
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]
预期输出为
[1,2]
[3, 4]
[5, 6]
[7, 8]
[9,10]
您可以使用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]
您也可以使用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]
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句