我认为这比itertools版本更短，更可读:

def split_by_n(seq, n):
    '''A generator to divide a sequence into chunks of n units.'''
    while seq:
        yield seq[:n]
        seq = seq[n:]

print(list(split_by_n('1234567890', 2)))

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']

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。

另一个使用groupby和index//n作为键来分组字母的解决方案:

from itertools import groupby

text = "abcdefghij"
n = 3

result = []
for idx, chunk in groupby(text, key=lambda x: x.index//n):
    result.append("".join(chunk))

# result = ['abc', 'def', 'ghi', 'j']

>>> from functools import reduce
>>> from operator import add
>>> from itertools import izip
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x)]
['12', '34', '56', '78', '90']
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x, x)]
['123', '456', '789']

使用PyPI中的更多itertools:

>>> from more_itertools import sliced
>>> list(sliced('1234567890', 2))
['12', '34', '56', '78', '90']