使用列表综合：

l = [1,2,3,4,5,6,7,8,9,10,11,12]
k = 5 #chunk size
print [tuple(l[x:y]) for (x, y) in [(x, x+k) for x in range(0, len(l), k)]]

def main():
  print(chunkify([1,2,3,4,5,6],2))

def chunkify(list, n):
  chunks = []
  for i in range(0, len(list), n):
    chunks.append(list[i:i+n])
  return chunks

main()

我认为这很简单，可以为您提供数组的一部分。

def split_seq(seq, num_pieces):
    start = 0
    for i in xrange(num_pieces):
        stop = start + len(seq[i::num_pieces])
        yield seq[start:stop]
        start = stop

用法：

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

for seq in split_seq(seq, 3):
    print seq

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]]

没有魔力，但简单而正确：

def chunks(iterable, n):
    """Yield successive n-sized chunks from iterable."""
    values = []
    for i, item in enumerate(iterable, 1):
        values.append(item)
        if i % n == 0:
            yield values
            values = []
    if values:
        yield values

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]]