我知道这有点过时，但还没有人提到numpy.array_split：

import numpy as np

lst = range(50)
np.array_split(lst, 5)

结果：

[array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
 array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
 array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
 array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
 array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]

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

任何可迭代的通用分块器，使用户可以选择如何在结尾处处理部分分块。

在Python 3上测试。

分块.py

from enum import Enum

class PartialChunkOptions(Enum):
    INCLUDE = 0
    EXCLUDE = 1
    PAD = 2
    ERROR = 3

class PartialChunkException(Exception):
    pass

def chunker(iterable, n, on_partial=PartialChunkOptions.INCLUDE, pad=None):
    """
    A chunker yielding n-element lists from an iterable, with various options
    about what to do about a partial chunk at the end.

    on_partial=PartialChunkOptions.INCLUDE (the default):
                     include the partial chunk as a short (<n) element list

    on_partial=PartialChunkOptions.EXCLUDE
                     do not include the partial chunk

    on_partial=PartialChunkOptions.PAD
                     pad to an n-element list 
                     (also pass pad=<pad_value>, default None)

    on_partial=PartialChunkOptions.ERROR
                     raise a RuntimeError if a partial chunk is encountered
    """

    on_partial = PartialChunkOptions(on_partial)        

    iterator = iter(iterable)
    while True:
        vals = []
        for i in range(n):
            try:
                vals.append(next(iterator))
            except StopIteration:
                if vals:
                    if on_partial == PartialChunkOptions.INCLUDE:
                        yield vals
                    elif on_partial == PartialChunkOptions.EXCLUDE:
                        pass
                    elif on_partial == PartialChunkOptions.PAD:
                        yield vals + [pad] * (n - len(vals))
                    elif on_partial == PartialChunkOptions.ERROR:
                        raise PartialChunkException
                    return
                return
        yield vals

测试.py

import chunker

chunk_size = 3

for it in (range(100, 107),
          range(100, 109)):

    print("\nITERABLE TO CHUNK: {}".format(it))
    print("CHUNK SIZE: {}".format(chunk_size))

    for option in chunker.PartialChunkOptions.__members__.values():
        print("\noption {} used".format(option))
        try:
            for chunk in chunker.chunker(it, chunk_size, on_partial=option):
                print(chunk)
        except chunker.PartialChunkException:
            print("PartialChunkException was raised")
    print("")

test.py的输出

ITERABLE TO CHUNK: range(100, 107)
CHUNK SIZE: 3

option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106]

option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]

option PartialChunkOptions.PAD used
[100, 101, 102]
[103, 104, 105]
[106, None, None]

option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
PartialChunkException was raised


ITERABLE TO CHUNK: range(100, 109)
CHUNK SIZE: 3

option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]

option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]

option PartialChunkOptions.PAD used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]

option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]

下面是一个处理任意可迭代项的生成器：

def split_seq(iterable, size):
    it = iter(iterable)
    item = list(itertools.islice(it, size))
    while item:
        yield item
        item = list(itertools.islice(it, size))

例子：

>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[0, 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]]

我很惊讶没有人想到使用iter的双参数形式：

from itertools import islice

def chunk(it, size):
    it = iter(it)
    return iter(lambda: tuple(islice(it, size)), ())

演示：

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]

这适用于任何可迭代的对象，并延迟生成输出。它返回元组而不是迭代器，但我认为它还是有一定的优雅。它也不会垫；如果您需要填充，上面的一个简单变体就足够了：

from itertools import islice, chain, repeat

def chunk_pad(it, size, padval=None):
    it = chain(iter(it), repeat(padval))
    return iter(lambda: tuple(islice(it, size)), (padval,) * size)

演示：

>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk_pad(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

与基于izip_longest的解决方案一样，上面的解决方案也始终适用。据我所知，对于可选pad的函数，没有单行或双线itertools配方。通过结合以上两种方法，这一方法非常接近：

_no_padding = object()

def chunk(it, size, padval=_no_padding):
    if padval == _no_padding:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(padval))
        sentinel = (padval,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

演示：

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]

我相信这是提议的提供可选填充的最短的分块器。

正如Tomasz Gandor所观察到的，如果两个填充块遇到一长串填充值，它们会意外停止。以下是以合理方式解决该问题的最后一个变体：

_no_padding = object()
def chunk(it, size, padval=_no_padding):
    it = iter(it)
    chunker = iter(lambda: tuple(islice(it, size)), ())
    if padval == _no_padding:
        yield from chunker
    else:
        for ch in chunker:
            yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))

演示：

>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]

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。