另一种常见的将元素分组为n长度组的方法:

>>> s = '1234567890'
>>> map(''.join, zip(*[iter(s)]*2))
['12', '34', '56', '78', '90']

这个方法直接来自zip()的文档。

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']

使用PyPI中的更多itertools:

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

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']

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