82

我在互联网上找不到任何有效的例子,我可以看到它们之间的区别以及为什么选择一个而不是另一个。

4

6 回答 6

93

第一个接受 0 个或多个参数,每个参数是一个可迭代对象,第二个接受一个参数,该参数预计会产生可迭代对象:

from itertools import chain

chain(list1, list2, list3)

iterables = [list1, list2, list3]
chain.from_iterable(iterables)

iterables可以是任何产生可迭代对象的迭代器:

def gen_iterables():
    for i in range(10):
        yield range(i)

itertools.chain.from_iterable(gen_iterables())

使用第二种形式通常是一种方便的情况,但因为它会懒惰地循环输入迭代器,所以它也是链接无限数量的有限迭代器的唯一方法:

def gen_iterables():
    while True:
        for i in range(5, 10):
            yield range(i)

chain.from_iterable(gen_iterables())

上面的例子将给你一个迭代,它产生一个循环的数字模式,它永远不会停止,但永远不会消耗比单个range()调用所需更多的内存。

于 2013-02-21T14:30:20.163 回答
14

我找不到任何有效的例子......在那里我可以看到它们 [chainchain.from_iterable] 之间的区别以及为什么选择一个而不是另一个

接受的答案是彻底的。对于那些寻求快速申请的人,可以考虑将几个列表展平:

list(itertools.chain(["a", "b", "c"], ["d", "e"], ["f"]))
# ['a', 'b', 'c', 'd', 'e', 'f']

您可能希望稍后重用这些列表,因此您可以创建一个可迭代的列表:

iterable = (["a", "b", "c"], ["d", "e"], ["f"])

试图

但是,传入一个可迭代的 tochain会得到一个不平坦的结果:

list(itertools.chain(iterable))
# [['a', 'b', 'c'], ['d', 'e'], ['f']]

为什么?你传入了一个项目(一个元组)。 chain分别需要每个列表。


解决方案

如果可能,您可以解压缩一个可迭代对象:

list(itertools.chain(*iterable))
# ['a', 'b', 'c', 'd', 'e', 'f']

list(itertools.chain(*iter(iterable)))
# ['a', 'b', 'c', 'd', 'e', 'f']

更一般地说,使用.from_iterable(因为它也适用于无限迭代器):

list(itertools.chain.from_iterable(iterable))
# ['a', 'b', 'c', 'd', 'e', 'f']

g = itertools.chain.from_iterable(itertools.cycle(iterable))
next(g)
# "a"
于 2018-07-17T00:55:09.550 回答
7

他们做的事情非常相似。对于少量的可迭代对象itertools.chain(*iterables)itertools.chain.from_iterable(iterables)执行类似的操作。

的关键优势from_iterables在于能够处理大量(可能无限)的可迭代对象,因为它们在调用时不需要全部可用。

于 2017-07-10T22:25:02.177 回答
4

另一种查看方式:

chain(iterable1, iterable2, iterable3, ...)当你已经知道你有什么迭代时,你可以把它们写成这些逗号分隔的参数。

chain.from_iterable(iterable)用于当您的可迭代对象(如 iterable1、iterable2、iterable3)从另一个可迭代对象获得时。

于 2018-08-20T10:34:33.900 回答
3

扩展@martijn-pieters 答案

尽管对可迭代内部项目的访问保持不变,并且在实现方面,

  • itertools_chain_from_iterable(即chain.from_iterable在 Python 中)和
  • chain_new(即chain在 Python 中)

在 CPython 实现中,都是chain_new_internal的鸭子类型


使用 是否有任何优化好处chain.from_iterable(x),其中x是可迭代的可迭代;主要目的是最终消耗扁平化的项目列表?

我们可以尝试使用以下方法对其进行基准测试:

import random
from itertools import chain
from functools import wraps
from time import time

from tqdm import tqdm

def timing(f):
    @wraps(f)
    def wrap(*args, **kw):
        ts = time()
        result = f(*args, **kw)
        te = time()
        print('func:%r args:[%r, %r] took: %2.4f sec' % (f.__name__, args, kw, te-ts))
        return result
    return wrap

def generate_nm(m, n):
    # Creates m generators of m integers between range 0 to n.
    yield iter(random.sample(range(n), n) for _ in range(m))
    

def chain_star(x):
    # Stores an iterable that will unpack and flatten the list of list.
    chain_x = chain(*x)
    # Consumes the items in the flatten iterable.
    for i in chain_x:
        pass

def chain_from_iterable(x):
    # Stores an iterable that will unpack and flatten the list of list.
    chain_x = chain.from_iterable(x)
    # Consumes the items in the flatten iterable.
    for i in chain_x:
        pass


@timing
def versus(f, n, m):
  f(generate_nm(n, m))

P/S:基准测试正在运行...等待结果。


结果

链星,m=1000,n=1000

for _ in range(10):
    versus(chain_star, 1000, 1000)

[出去]:

func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6494 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6603 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6367 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6350 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6296 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6399 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6341 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6381 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6343 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6309 sec

chain_from_iterable,m=1000,n=1000

for _ in range(10):
    versus(chain_from_iterable, 1000, 1000)

[出去]:

func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6416 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6315 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6535 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6334 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6327 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6471 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6426 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6287 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6353 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6297 sec

链星,m=10000,n=1000

func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2659 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2966 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2953 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3141 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2802 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2799 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2848 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3299 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2730 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3052 sec

chain_from_iterable, m=10000, n=1000

func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3129 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3064 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3071 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2660 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2837 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2877 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2756 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2939 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2715 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2877 sec

链星,m=100000,n=1000

func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.7874 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.3744 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.5584 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.3745 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.7982 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.4054 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.6769 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.6476 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.7397 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.8980 sec

chain_from_iterable, m=100000, n=1000

for _ in range(10):
    versus(chain_from_iterable, 100000, 1000)

[出去]:

func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7227 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7717 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7159 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7569 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7906 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.6211 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7294 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.8260 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.8356 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.9738 sec

链星,m=500000,n=1000

for _ in range(3):
    versus(chain_from_iterable, 500000, 1000)

[出去]:

func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 314.5671 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 313.9270 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 313.8992 sec

chain_from_iterable, m=500000, n=1000

for _ in range(3):
    versus(chain_from_iterable, 500000, 1000)

[出去]:

func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.8301 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.8104 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.9440 sec
于 2020-06-22T11:53:04.630 回答
0

另一种看待它的方法是使用 chain.from_iterable

当您有一个可迭代的可迭代对象时

于 2020-02-20T14:42:57.060 回答