如果您知道列表大小：

def SplitList(mylist, chunk_size):
    return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]

如果没有（迭代器）：

def IterChunks(sequence, chunk_size):
    res = []
    for item in sequence:
        res.append(item)
        if len(res) >= chunk_size:
            yield res
            res = []
    if res:
        yield res  # yield the last, incomplete, portion

在后一种情况下，如果您可以确保序列始终包含给定大小的整数个块（即没有不完整的最后一个块），则可以用更漂亮的方式重新表述。

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

这项任务可以在公认答案中使用生成器轻松完成。我正在添加实现长度方法的类实现，这可能对某些人有用。我需要知道进度（使用tqdm），所以生成器应该返回块的数量。

class ChunksIterator(object):
    def __init__(self, data, n):
        self._data = data
        self._l = len(data)
        self._n = n

    def __iter__(self):
        for i in range(0, self._l, self._n):
            yield self._data[i:i + self._n]

    def __len__(self):
        rem = 1 if self._l % self._n != 0 else 0
        return self._l // self._n + rem

用法：

it = ChunksIterator([1,2,3,4,5,6,7,8,9], 2)
print(len(it))
for i in it:
  print(i)

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]
>>> 

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]]

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 )]