def chunk(input, size):
    return map(None, *([iter(input)] * size))

这是一个生成大小均匀的块的生成器：

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[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]]

对于Python 2，使用xrange代替range：

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

下面是一行理解列表。不过，上面的方法更可取，因为使用命名函数使代码更容易理解。对于Python 3：

[lst[i:i + n] for i in range(0, len(lst), n)]

对于Python 2：

[lst[i:i + n] for i in xrange(0, len(lst), n)]

我不喜欢按块大小拆分元素的想法，例如，脚本可以将101到3个块划分为[50，50，1]。为了我的需要，我需要按比例分配，保持秩序不变。首先我写了自己的剧本，效果很好，而且很简单。但我后来看到了这个答案，剧本比我的好，我想是这样的。这是我的脚本：

def proportional_dividing(N, n):
    """
    N - length of array (bigger number)
    n - number of chunks (smaller number)
    output - arr, containing N numbers, diveded roundly to n chunks
    """
    arr = []
    if N == 0:
        return arr
    elif n == 0:
        arr.append(N)
        return arr
    r = N // n
    for i in range(n-1):
        arr.append(r)
    arr.append(N-r*(n-1))

    last_n = arr[-1]
    # last number always will be r <= last_n < 2*r
    # when last_n == r it's ok, but when last_n > r ...
    if last_n > r:
        # ... and if difference too big (bigger than 1), then
        if abs(r-last_n) > 1:
            #[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 7] # N=29, n=12
            # we need to give unnecessary numbers to first elements back
            diff = last_n - r
            for k in range(diff):
                arr[k] += 1
            arr[-1] = r
            # and we receive [3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
    return arr

def split_items(items, chunks):
    arr = proportional_dividing(len(items), chunks)
    splitted = []
    for chunk_size in arr:
        splitted.append(items[:chunk_size])
        items = items[chunk_size:]
    print(splitted)
    return splitted

items = [1,2,3,4,5,6,7,8,9,10,11]
chunks = 3
split_items(items, chunks)
split_items(['a','b','c','d','e','f','g','h','i','g','k','l', 'm'], 3)
split_items(['a','b','c','d','e','f','g','h','i','g','k','l', 'm', 'n'], 3)
split_items(range(100), 4)
split_items(range(99), 4)
split_items(range(101), 4)

和输出：

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]
[['a', 'b', 'c', 'd'], ['e', 'f', 'g', 'h'], ['i', 'g', 'k', 'l', 'm']]
[['a', 'b', 'c', 'd', 'e'], ['f', 'g', 'h', 'i', 'g'], ['k', 'l', 'm', 'n']]
[range(0, 25), range(25, 50), range(50, 75), range(75, 100)]
[range(0, 25), range(25, 50), range(50, 75), range(75, 99)]
[range(0, 25), range(25, 50), range(50, 75), range(75, 101)]

def chunk(input, size):
    return map(None, *([iter(input)] * size))

我专门为此写了一个小图书馆，这里有。库的分块函数特别有效，因为它是作为生成器实现的，因此在某些情况下可以节省大量内存。它也不依赖切片表示法，因此可以使用任意迭代器。

import iterlib

print list(iterlib.chunked(xrange(1, 1000), 10))
# prints [(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), (11, 12, 13, 14, 15, 16, 17, 18, 19, 20), ...]

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。