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

使用列表综合:

l = [1,2,3,4,5,6,7,8,9,10,11,12]
k = 5 #chunk size
print [tuple(l[x:y]) for (x, y) in [(x, x+k) for x in range(0, len(l), k)]]

其他回答

我非常喜欢tzot和J.F.Sebastian提出的Python文档版本,但它有两个缺点:

它不是很明确我通常不希望在最后一个块中有填充值

我在代码中经常使用这个:

from itertools import islice

def chunks(n, iterable):
    iterable = iter(iterable)
    while True:
        yield tuple(islice(iterable, n)) or iterable.next()

更新:一个懒块版本:

from itertools import chain, islice

def chunks(n, iterable):
   iterable = iter(iterable)
   while True:
       yield chain([next(iterable)], islice(iterable, n-1))

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]

def chunks(iterable,n):
    """assumes n is an integer>0
    """
    iterable=iter(iterable)
    while True:
        result=[]
        for i in range(n):
            try:
                a=next(iterable)
            except StopIteration:
                break
            else:
                result.append(a)
        if result:
            yield result
        else:
            break

g1=(i*i for i in range(10))
g2=chunks(g1,3)
print g2
'<generator object chunks at 0x0337B9B8>'
print list(g2)
'[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81]]'

一个简单的解决方案

OP已请求“相等大小的块”。我将“等尺寸”理解为“平衡”尺寸:如果尺寸不可能相等(例如,23/5),我们正在寻找尺寸大致相同的物品组。

这里的输入是:

项目列表:input_list(例如,23个数字的列表)要拆分这些项目的组数:n个组(例如5个)

输入:

input_list = list(range(23))
n_groups = 5

连续元素组:

approx_sizes = len(input_list)/n_groups 

groups_cont = [input_list[int(i*approx_sizes):int((i+1)*approx_sizes)] 
               for i in range(n_groups)]

“每N个”元素组:

groups_leap = [input_list[i::n_groups] 
               for i in range(n_groups)]

后果

print(len(input_list))

print('Contiguous elements lists:')
print(groups_cont)

print('Leap every "N" items lists:')
print(groups_leap)

将输出:23连续元素列表:[[0, 1, 2, 3], [4, 5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16, 17], [18, 19, 20, 21, 22]]跳过每“N”个项目列表:[[0, 5, 10, 15, 20], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18], [4, 9, 14, 19]]

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]]