如何将任意长度的列表拆分为大小相等的块?


请参阅如果数据结果将直接用于循环,并且不需要存储,则如何以块形式遍历列表。

对于字符串输入的同一问题,请参见每n个字符拆分字符串?。相同的技术通常适用,但也有一些变化。


当前回答

python-pydash包可能是一个不错的选择。

from pydash.arrays import chunk
ids = ['22', '89', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '1']
chunk_ids = chunk(ids,5)
print(chunk_ids)
# output: [['22', '89', '2', '3', '4'], ['5', '6', '7', '8', '9'], ['10', '11', '1']]

有关更多签出pydash块列表的信息

其他回答

toolz库具有如下分区函数:

from toolz.itertoolz.core import partition

list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]

我意识到这个问题已经过时了(在谷歌上被它绊倒了),但肯定像下面这样的问题比任何复杂的建议都要简单和清晰得多,而且只使用切片:

def chunker(iterable, chunksize):
    for i,c in enumerate(iterable[::chunksize]):
        yield iterable[i*chunksize:(i+1)*chunksize]

>>> for chunk in chunker(range(0,100), 10):
...     print list(chunk)
... 
[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]
... etc ...

senderle答案的一个线性版本:

from itertools import islice
from functools import partial

seq = [1,2,3,4,5,6,7]
size = 3
result = list(iter(partial(lambda it: tuple(islice(it, size)), iter(seq)), ()))
assert result == [(1, 2, 3), (4, 5, 6), (7,)]

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)]

与任何可迭代的内部数据是生成器对象(不是列表)一个衬垫

In [259]: get_in_chunks = lambda itr,n: ( (v for _,v in g) for _,g in itertools.groupby(enumerate(itr),lambda (ind,_): ind/n))

In [260]: list(list(x) for x in get_in_chunks(range(30),7))
Out[260]:
[[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]]