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']

我陷入了同样的困境。

这招对我很管用:

x = "1234567890"
n = 2
my_list = []
for i in range(0, len(x), n):
    my_list.append(x[i:i+n])
print(my_list)

输出:

['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']

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)))

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])]

我喜欢这个解决方案:

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