另一种方法是使用双参数形式的iter:

from itertools import islice

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

这可以很容易地适应使用填充(这类似于Markus Jarderot的答案):

from itertools import islice, chain, repeat

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

这些甚至可以组合为可选的填充:

_no_pad = object()
def group(it, size, pad=_no_pad):
    if pad == _no_pad:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(pad))
        sentinel = (pad,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

如果列表很大，执行效率最高的方法是使用生成器:

def get_chunk(iterable, chunk_size):
    result = []
    for item in iterable:
        result.append(item)
        if len(result) == chunk_size:
            yield tuple(result)
            result = []
    if len(result) > 0:
        yield tuple(result)

for x in get_chunk([1,2,3,4,5,6,7,8,9,10], 3):
    print x

(1, 2, 3)
(4, 5, 6)
(7, 8, 9)
(10,)

使用map()而不是zip()修复填充问题在J.F.塞巴斯蒂安的回答:

>>> def chunker(iterable, chunksize):
...   return map(None,*[iter(iterable)]*chunksize)

例子:

>>> s = '1234567890'
>>> chunker(s, 3)
[('1', '2', '3'), ('4', '5', '6'), ('7', '8', '9'), ('0', None, None)]
>>> chunker(s, 4)
[('1', '2', '3', '4'), ('5', '6', '7', '8'), ('9', '0', None, None)]
>>> chunker(s, 5)
[('1', '2', '3', '4', '5'), ('6', '7', '8', '9', '0')]

首先，我将它设计为将字符串拆分为子字符串以解析包含十六进制的字符串。 今天我把它变成复杂的，但仍然简单的生成器。

def chunker(iterable, size, reductor, condition):
    it = iter(iterable)
    def chunk_generator():
        return (next(it) for _ in range(size))
    chunk = reductor(chunk_generator())
    while condition(chunk):
        yield chunk
        chunk = reductor(chunk_generator())

参数:

明显的

Iterable是任何包含/生成/迭代输入数据的Iterable /迭代器/生成器， 当然，大小是你想要得到的块的大小，

更有趣的

reductor is a callable, which receives generator iterating over content of chunk. I'd expect it to return sequence or string, but I don't demand that. You can pass as this argument for example list, tuple, set, frozenset, or anything fancier. I'd pass this function, returning string (provided that iterable contains / generates / iterates over strings): def concatenate(iterable): return ''.join(iterable) Note that reductor can cause closing generator by raising exception. condition is a callable which receives anything what reductor returned. It decides to approve & yield it (by returning anything evaluating to True), or to decline it & finish generator's work (by returning anything other or raising exception). When number of elements in iterable is not divisible by size, when it gets exhausted, reductor will receive generator generating less elements than size. Let's call these elements lasts elements. I invited two functions to pass as this argument: lambda x:x - the lasts elements will be yielded. lambda x: len(x)==<size> - the lasts elements will be rejected. replace <size> using number equal to size

def chunker(seq, size):
    return (seq[pos:pos + size] for pos in range(0, len(seq), size))

适用于任何序列:

text = "I am a very, very helpful text"

for group in chunker(text, 7):
   print(repr(group),)
# 'I am a ' 'very, v' 'ery hel' 'pful te' 'xt'

print('|'.join(chunker(text, 10)))
# I am a ver|y, very he|lpful text

animals = ['cat', 'dog', 'rabbit', 'duck', 'bird', 'cow', 'gnu', 'fish']

for group in chunker(animals, 3):
    print(group)
# ['cat', 'dog', 'rabbit']
# ['duck', 'bird', 'cow']
# ['gnu', 'fish']

from itertools import izip_longest

def chunker(iterable, chunksize, filler):
    return izip_longest(*[iter(iterable)]*chunksize, fillvalue=filler)