我喜欢这个解决方案:

s = '1234567890'
o = []
while s:
    o.append(s[:2])
    s = s[2:]

我喜欢这个解决方案:

s = '1234567890'
o = []
while s:
    o.append(s[:2])
    s = s[2:]

为了完整，你可以用regex来完成:

>>> import re
>>> re.findall('..','1234567890')
['12', '34', '56', '78', '90']

对于奇数个字符，你可以这样做:

>>> import re
>>> re.findall('..?', '123456789')
['12', '34', '56', '78', '9']

你也可以这样做，简化正则表达式为更长的块:

>>> import re
>>> re.findall('.{1,2}', '123456789')
['12', '34', '56', '78', '9']

如果字符串很长，可以使用re.finditer逐块生成。

我陷入了同样的困境。

这招对我很管用:

x = "1234567890"
n = 2
my_list = []
for i in range(0, len(x), n):
    my_list.append(x[i:i+n])
print(my_list)

输出:

['12', '34', '56', '78', '90']

python中已经有一个内置的函数。

>>> from textwrap import wrap
>>> s = '1234567890'
>>> wrap(s, 2)
['12', '34', '56', '78', '90']

这是wrap的文档字符串说的:

>>> help(wrap)
'''
Help on function wrap in module textwrap:

wrap(text, width=70, **kwargs)
    Wrap a single paragraph of text, returning a list of wrapped lines.

    Reformat the single paragraph in 'text' so it fits in lines of no
    more than 'width' columns, and return a list of wrapped lines.  By
    default, tabs in 'text' are expanded with string.expandtabs(), and
    all other whitespace characters (including newline) are converted to
    space.  See TextWrapper class for available keyword args to customize
    wrapping behaviour.
'''

试试下面的代码:

from itertools import islice

def split_every(n, iterable):
    i = iter(iterable)
    piece = list(islice(i, n))
    while piece:
        yield piece
        piece = list(islice(i, n))

s = '1234567890'
print list(split_every(2, list(s)))