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)))

我喜欢这个解决方案:

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

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)))

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']

more_itertools。切片之前提到过。下面是more_itertools库中的另外四个选项:

s = "1234567890"

["".join(c) for c in mit.grouper(2, s)]

["".join(c) for c in mit.chunked(s, 2)]

["".join(c) for c in mit.windowed(s, 2, step=2)]

["".join(c) for c in  mit.split_after(s, lambda x: int(x) % 2 == 0)]

后面的每个选项都会产生以下输出:

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

所讨论选项的文档:grouper, chunked, windosed, split_after

使用PyPI中的更多itertools:

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