itertools模块中的配方提供了两种方法来实现这一点，具体取决于您希望如何处理最终的奇数大小的批次（保留它、用填充值填充它、忽略它或引发异常）：

from itertools import islice, izip_longest

def batched(iterable, n):
    "Batch data into lists of length n. The last batch may be shorter."
    # batched('ABCDEFG', 3) --> ABC DEF G
    it = iter(iterable)
    while True:
        batch = list(islice(it, n))
        if not batch:
            return
        yield batch

def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
    "Collect data into non-overlapping fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
    # grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
    # grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
    args = [iter(iterable)] * n
    if incomplete == 'fill':
        return zip_longest(*args, fillvalue=fillvalue)
    if incomplete == 'strict':
        return zip(*args, strict=True)
    if incomplete == 'ignore':
        return zip(*args)
    else:
        raise ValueError('Expected fill, strict, or ignore')

下面是一个处理任意可迭代项的生成器：

def split_seq(iterable, size):
    it = iter(iterable)
    item = list(itertools.islice(it, size))
    while item:
        yield item
        item = list(itertools.islice(it, size))

例子：

>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[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],
 [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]]

还可以将utilspie库的get_chunks函数用作：

>>> from utilspie import iterutils
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> list(iterutils.get_chunks(a, 5))
[[1, 2, 3, 4, 5], [6, 7, 8, 9]]

您可以通过pip安装utilspie：

sudo pip install utilspie

免责声明：我是utilspie库的创建者。

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]]

toolz库具有如下分区函数：

from toolz.itertoolz.core import partition

list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]

a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
CHUNK = 4
[a[i*CHUNK:(i+1)*CHUNK] for i in xrange((len(a) + CHUNK - 1) / CHUNK )]