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,)]

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]]

我专门为此写了一个小图书馆，这里有。库的分块函数特别有效，因为它是作为生成器实现的，因此在某些情况下可以节省大量内存。它也不依赖切片表示法，因此可以使用任意迭代器。

import iterlib

print list(iterlib.chunked(xrange(1, 1000), 10))
# prints [(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), (11, 12, 13, 14, 15, 16, 17, 18, 19, 20), ...]

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 ...

不调用len（），这对大型列表很有用：

def splitter(l, n):
    i = 0
    chunk = l[:n]
    while chunk:
        yield chunk
        i += n
        chunk = l[i:i+n]

这是可迭代的：

def isplitter(l, n):
    l = iter(l)
    chunk = list(islice(l, n))
    while chunk:
        yield chunk
        chunk = list(islice(l, n))

上述产品的功能风味：

def isplitter2(l, n):
    return takewhile(bool,
                     (tuple(islice(start, n))
                            for start in repeat(iter(l))))

OR:

def chunks_gen_sentinel(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return iter(imap(tuple, continuous_slices).next,())

OR:

def chunks_gen_filter(n, seq):
    continuous_slices = imap(islice, repeat(iter(seq)), repeat(0), repeat(n))
    return takewhile(bool,imap(tuple, continuous_slices))

代码：

def split_list(the_list, chunk_size):
    result_list = []
    while the_list:
        result_list.append(the_list[:chunk_size])
        the_list = the_list[chunk_size:]
    return result_list

a_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

print split_list(a_list, 3)

结果：

[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]