考虑使用matplotlib.cbook片段

例如：

import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
     print s

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)]

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))

def main():
  print(chunkify([1,2,3,4,5,6],2))

def chunkify(list, n):
  chunks = []
  for i in range(0, len(list), n):
    chunks.append(list[i:i+n])
  return chunks

main()

我认为这很简单，可以为您提供数组的一部分。

[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]]

不要重新发明轮子。

更新：即将发布的Python 3.12引入了itertools.batch，最终解决了这个问题。见下文。

鉴于

import itertools as it
import collections as ct

import more_itertools as mit


iterable = range(11)
n = 3

Code

itertools.batch++

list(it.batched(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

更多工具+

list(mit.chunked(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

list(mit.sliced(iterable, n))
# [range(0, 3), range(3, 6), range(6, 9), range(9, 11)]

list(mit.grouper(n, iterable))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

list(mit.windowed(iterable, len(iterable)//n, step=n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

list(mit.chunked_even(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

（或DIY，如果你愿意）

标准库

list(it.zip_longest(*[iter(iterable)] * n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]

d = {}
for i, x in enumerate(iterable):
    d.setdefault(i//n, []).append(x)
    

list(d.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

dd = ct.defaultdict(list)
for i, x in enumerate(iterable):
    dd[i//n].append(x)
    

list(dd.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

工具书类

more_itertools.chunked（相关已发布）更多intertools.slicedmore_itertools.grouper（相关文章）more_itertools.windowd（另请参见错开、zip_offset）更多intertools.chunked_evenzip_langest（相关帖子，相关帖子）setdefault（排序结果需要Python 3.6+）collections.defaultdict（排序结果需要Python 3.6+）

+第三方库，实现itertools配方等。>pip安装更多工具

++Python标准库3.12+中包含的.batched类似于more_itertools.chunked。