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']

一如既往，为那些喜欢一句俏皮话的人

n = 2  
line = "this is a line split into n characters"  
line = [line[i * n:i * n+n] for i,blah in enumerate(line[::n])]

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

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']

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']

使用groupby的解决方案:

from itertools import groupby, chain, repeat, cycle

text = "wwworldggggreattecchemggpwwwzaz"
n = 3
c = cycle(chain(repeat(0, n), repeat(1, n)))
res = ["".join(g) for _, g in groupby(text, lambda x: next(c))]
print(res)

输出:

['www', 'orl', 'dgg', 'ggr', 'eat', 'tec', 'che', 'mgg', 'pww', 'wza', 'z']