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

[AA[i:i+SS] for i in range(len(AA))[::SS]]

其中AA是数组，SS是块大小。例如：

>>> AA=range(10,21);SS=3
>>> [AA[i:i+SS] for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
# or [range(10, 13), range(13, 16), range(16, 19), range(19, 21)] in py3

要扩展py3中的范围，请执行以下操作

(py3) >>> [list(AA[i:i+SS]) for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]

python-pydash包可能是一个不错的选择。

from pydash.arrays import chunk
ids = ['22', '89', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '1']
chunk_ids = chunk(ids,5)
print(chunk_ids)
# output: [['22', '89', '2', '3', '4'], ['5', '6', '7', '8', '9'], ['10', '11', '1']]

有关更多签出pydash块列表的信息

根据这个答案，得票最多的答案在最后留下一个“矮子”。这是我的解决方案，可以在没有矮子的情况下，尽可能地获得大小均匀的块。它基本上试图准确选择应该拆分列表的小数点，但只需将其舍入到最接近的整数：

from __future__ import division  # not needed in Python 3
def n_even_chunks(l, n):
    """Yield n as even chunks as possible from l."""
    last = 0
    for i in range(1, n+1):
        cur = int(round(i * (len(l) / n)))
        yield l[last:cur]
        last = cur

演示：

>>> pprint.pprint(list(n_even_chunks(list(range(100)), 9)))
[[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, 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]]
>>> pprint.pprint(list(n_even_chunks(list(range(100)), 11)))
[[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, 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]]

与排名前几的答案进行比较：

>>> pprint.pprint(list(chunks(list(range(100)), 100//9)))
[[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, 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]]
>>> pprint.pprint(list(chunks(list(range(100)), 100//11)))
[[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, 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]]

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]]

简单而优雅

L = range(1, 1000)
print [L[x:x+10] for x in xrange(0, len(L), 10)]

或者如果您愿意：

def chunks(L, n): return [L[x: x+n] for x in xrange(0, len(L), n)]
chunks(L, 10)