def chunk(input, size):
    return map(None, *([iter(input)] * size))

senderle答案的一个线性版本：

from itertools import islice
from functools import partial

seq = [1,2,3,4,5,6,7]
size = 3
result = list(iter(partial(lambda it: tuple(islice(it, size)), iter(seq)), ()))
assert result == [(1, 2, 3), (4, 5, 6), (7,)]

上面的答案（由koffein给出）有一个小问题：列表总是被分割成相等数量的分割，而不是每个分区的项目数相等。这是我的版本。“//chs+1”考虑到项目的数量可能不能完全除以分区大小，因此最后一个分区将仅被部分填充。

# Given 'l' is your list

chs = 12 # Your chunksize
partitioned = [ l[i*chs:(i*chs)+chs] for i in range((len(l) // chs)+1) ]

呵呵，单行版本

In [48]: chunk = lambda ulist, step:  map(lambda i: ulist[i:i+step],  xrange(0, len(ulist), step))

In [49]: chunk(range(1,100), 10)
Out[49]: 
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
 [21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
 [31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
 [41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
 [51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
 [61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
 [71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
 [81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
 [91, 92, 93, 94, 95, 96, 97, 98, 99]]

与任何可迭代的内部数据是生成器对象（不是列表）一个衬垫

In [259]: get_in_chunks = lambda itr,n: ( (v for _,v in g) for _,g in itertools.groupby(enumerate(itr),lambda (ind,_): ind/n))

In [260]: list(list(x) for x in get_in_chunks(range(30),7))
Out[260]:
[[0, 1, 2, 3, 4, 5, 6],
 [7, 8, 9, 10, 11, 12, 13],
 [14, 15, 16, 17, 18, 19, 20],
 [21, 22, 23, 24, 25, 26, 27],
 [28, 29]]

我意识到这个问题已经过时了（在谷歌上被它绊倒了），但肯定像下面这样的问题比任何复杂的建议都要简单和清晰得多，而且只使用切片：

def chunker(iterable, chunksize):
    for i,c in enumerate(iterable[::chunksize]):
        yield iterable[i*chunksize:(i+1)*chunksize]

>>> for chunk in chunker(range(0,100), 10):
...     print list(chunk)
... 
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
... etc ...