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']

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)))

试试这个:

s='1234567890'
print([s[idx:idx+2] for idx,val in enumerate(s) if idx%2 == 0])

输出:

['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']

这可以通过一个简单的for循环来实现。

a = '1234567890a'
result = []

for i in range(0, len(a), 2):
    result.append(a[i : i + 2])
print(result)

输出如下所示 ['12'， '34'， '56'， '78'， '90'， 'a']

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。