Question

列表为，

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

我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:]):
    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]

预期输出

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

Answer 1

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]

您还可以使用itertools中的grouper recipe：

>>> 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

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 3

一个解决方案是

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

Answer 4

In [180]: lst = range(1,11)

In [181]: for i in zip(*[iter(lst)]*2):
   .....:     print i
   .....:
(1, 2)
(3, 4)
(5, 6)
(7, 8)
(9, 10)

Answer 5

错误：

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]

假设列表具有偶数个元素

Answer 6

>>> l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 
>>> map(None,*[iter(l)]*2)
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
>>>

Answer 7

对于那些来这里要走很长一步并且不想在前面使用大量内存的人，您可以这样做。

Python 2.7：

import itertools

def step_indices(length, step):
    from_indices = xrange(0, length, step)
    to_indices = itertools.chain(xrange(step, length, step), [None])
    for i, j in itertools.izip(from_indices, to_indices):
        yield i, j

the_list = range(1234)
for i, j in step_indices(len(the_list), 100):
    up_to_100_values = the_list[i:j]

Python 3：

import itertools

def step_indices(length, step):
    from_indices = range(0, length, step)
    to_indices = itertools.chain(range(step, length, step), [None])
    for i, j in zip(from_indices, to_indices):
        yield i, j

the_list = list(range(1234))
for i, j in step_indices(len(the_list), 100):
    up_to_100_values = the_list[i:j]

Answer 8

使用map的另一种解决方案（语法与其他答案略有不同）。在元素数量为奇数的列表中，最后一个元素与None配对：

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>> map(None, a[::2], a[1::2])
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10), (11, None)]

您可以修改函数以返回例如列表而不是元组：

>>> map(lambda x,y:[x,y], a[::2], a[1::2])
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11, None]]

>>> for e in map(lambda x,y:[x,y], a[::2], a[1::2]):
...     print e
... 
[1, 2]
[3, 4]
[5, 6]
[7, 8]
[9, 10]
[11, None]

Answer 9

我知道这是一个古老的问题，只是一种不同的方法。

l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 
chunks = [l[x:x+2] for x in range(0,len(l),2)]
print(chunks)

一次迭代列表中的每2个元素

9 个答案: