这些答案都很好，很有用，但是语法太神秘了……为什么不写一个简单的函数呢?

def SplitEvery(string, length):
    if len(string) <= length: return [string]        
    sections = len(string) / length
    lines = []
    start = 0;
    for i in range(sections):
        line = string[start:start+length]
        lines.append(line)
        start += length
    return lines

简单地叫它:

text = '1234567890'
lines = SplitEvery(text, 2)
print(lines)

# output: ['12', '34', '56', '78', '90']

你可以使用itertools中的grouper()方法:

Python 2. x:

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)

Python 3. x:

from itertools import zip_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 zip_longest(*args, fillvalue=fillvalue)

这些函数是内存高效的，并且适用于任何可迭代对象。

>>> line = '1234567890'
>>> n = 2
>>> [line[i:i+n] for i in range(0, len(line), n)]
['12', '34', '56', '78', '90']

使用PyPI中的更多itertools:

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

>>> 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']

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']