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)

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]]

您可以使用更多的intertools.chunked_甚至与math.eil一起使用。这可能是最容易理解的吗？

from math import ceil
import more_itertools as mit
from pprint import pprint

pprint([*mit.chunked_even(range(19), ceil(19 / 5))])
# [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18]]

pprint([*mit.chunked_even(range(20), ceil(20 / 5))])
# [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]]

pprint([*mit.chunked_even(range(21), ceil(21 / 5))])
# [[0, 1, 2, 3, 4],
# [5, 6, 7, 8],
# [9, 10, 11, 12],
# [13, 14, 15, 16],
# [17, 18, 19, 20]]

pprint([*mit.chunked_even(range(3), ceil(3 / 5))])
# [[0], [1], [2]]

def chunk(lst):
    out = []
    for x in xrange(2, len(lst) + 1):
        if not len(lst) % x:
            factor = len(lst) / x
            break
    while lst:
        out.append([lst.pop(0) for x in xrange(factor)])
    return out

抽象将是

l = [1,2,3,4,5,6,7,8,9]
n = 3
outList = []
for i in range(n, len(l) + n, n):
    outList.append(l[i-n:i])

print(outList)

这将打印：

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

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)]