如何将列表分成大小均匀的块？

python Python

Me无敌

2020-05-28

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in itertools but I couldn't find anything obviously useful. Might've missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

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共5个回答
DavaidTony宝儿 2020.05.28

简单而优雅

l = range(1, 1000)
print [l[x:x+10] for x in xrange(0, len(l), 10)]

或者，如果您喜欢：

chunks = lambda l, n: [l[x: x+n] for x in xrange(0, len(l), n)]
chunks(l, 10)
小宇宙 2020.05.28

这是处理任意可迭代对象的生成器：

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：

>>> 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]]
十三 2020.05.28

我知道这有点旧，但是没有人提到numpy.array_split：

import numpy as np

lst = range(50)
np.array_split(lst, 5)
# [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])]
Tony凯 2020.05.28

如果您想要超级简单的东西：

def chunks(l, n):
    n = max(1, n)
    return (l[i:i+n] for i in range(0, len(l), n))

在Python 2.x中使用xrange()代替range()

JinJin 2020.05.28

这是一个生成所需块的生成器：

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]]

如果您使用的是Python 2，则应使用xrange()而不是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]

您也可以简单地使用列表理解而不是编写函数，尽管将这样的操作封装在命名函数中是个好主意，这样您的代码更易于理解。Python 3：

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2版本：

[lst[i:i + n] for i in xrange(0, len(lst), n)]

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