还有一个解决方案

def make_chunks(data, chunk_size): 
    while data:
        chunk, data = data[:chunk_size], data[chunk_size:]
        yield chunk

>>> for chunk in make_chunks([1, 2, 3, 4, 5, 6, 7], 2):
...     print chunk
... 
[1, 2]
[3, 4]
[5, 6]
[7]
>>> 

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,)]

另一个更明确的版本。

def chunkList(initialList, chunkSize):
    """
    This function chunks a list into sub lists 
    that have a length equals to chunkSize.

    Example:
    lst = [3, 4, 9, 7, 1, 1, 2, 3]
    print(chunkList(lst, 3)) 
    returns
    [[3, 4, 9], [7, 1, 1], [2, 3]]
    """
    finalList = []
    for i in range(0, len(initialList), chunkSize):
        finalList.append(initialList[i:i+chunkSize])
    return finalList

这适用于v2/v3，可内联，基于生成器，仅使用标准库：

import itertools
def split_groups(iter_in, group_size):
    return ((x for _, x in item) for _, item in itertools.groupby(enumerate(iter_in), key=lambda x: x[0] // group_size))

一个简单的解决方案

OP已请求“相等大小的块”。我将“等尺寸”理解为“平衡”尺寸：如果尺寸不可能相等（例如，23/5），我们正在寻找尺寸大致相同的物品组。

这里的输入是：

项目列表：input_list（例如，23个数字的列表）要拆分这些项目的组数：n个组（例如5个）

输入：

input_list = list(range(23))
n_groups = 5

连续元素组：

approx_sizes = len(input_list)/n_groups 

groups_cont = [input_list[int(i*approx_sizes):int((i+1)*approx_sizes)] 
               for i in range(n_groups)]

“每N个”元素组：

groups_leap = [input_list[i::n_groups] 
               for i in range(n_groups)]

后果

print(len(input_list))

print('Contiguous elements lists:')
print(groups_cont)

print('Leap every "N" items lists:')
print(groups_leap)

将输出：23连续元素列表：[[0, 1, 2, 3], [4, 5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16, 17], [18, 19, 20, 21, 22]]跳过每“N”个项目列表：[[0, 5, 10, 15, 20], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18], [4, 9, 14, 19]]

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]]