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']

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']

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

我喜欢这个解决方案:

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

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']

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']