如何使用线程并行压缩迭代器？

如何解决如何使用线程并行压缩迭代器？

您可以通过创建一个生成器来实现您所要求的生成器，该生成器从apply_asyncThreadPool上的-calls中产生结果。

仅供参考，我pandas.read_csv通过指定chunksize参数得到的- iterators对该方法进行了基准测试。我创建了一个大小为csv文件的1M行的八个副本，并指定了chunksize = 100_000。

使用您提供的顺序方法读取了四个文件mt_gen，使用四个线程的池读取了以下功能的四个文件：

单螺纹〜3.68 s

多线程〜1.21 s

但这并不意味着它将改善每种硬件和数据设置的结果。

import time
import threading
from multiprocessing.dummy import Pool  # dummy uses threads


def _load_sim(x = 10e6):
    for _ in range(int(x)):
        x -= 1
    time.sleep(1)


def gen(start, stop):
    for i in range(start, stop):
        _load_sim()
        print(f'{threading.current_thread().name} yielding {i}')
        yield i


def multi_threaded(gens):
    combi_g = mt_gen(gens)
    for item in combi_g:
        print(item)


def mt_gen(gens):
    with Pool(N_WORKERS) as pool:
        while True:
            async_results = [pool.apply_async(next, args=(g,)) for g in gens]
            try:
                results = [r.get() for r in async_results]
            except stopiteration:  # needed for Python 3.7+, PEP 479, bpo-32670
                return
            yield results


if __name__ == '__main__':

    N_GENS = 10
    N_WORKERS = 4
    GEN_LENGTH = 3

    gens = [gen(x * GEN_LENGTH, (x + 1) * GEN_LENGTH) for x in range(N_GENS)]
    multi_threaded(gens)

输出：

Thread-1 yielding 0
Thread-2 yielding 3
Thread-4 yielding 6
Thread-3 yielding 9
Thread-1 yielding 12
Thread-2 yielding 15
Thread-4 yielding 18
Thread-3 yielding 21
Thread-1 yielding 24
Thread-2 yielding 27
[0, 3, 6, 9, 12, 15, 18, 21, 24, 27]
Thread-3 yielding 7
Thread-1 yielding 10
Thread-2 yielding 4
Thread-4 yielding 1
Thread-3 yielding 13
Thread-1 yielding 16
Thread-4 yielding 22
Thread-2 yielding 19
Thread-3 yielding 25
Thread-1 yielding 28
[1, 4, 7, 10, 13, 16, 19, 22, 25, 28]
Thread-1 yielding 8
Thread-4 yielding 2
Thread-3 yielding 11
Thread-2 yielding 5
Thread-1 yielding 14
Thread-4 yielding 17
Thread-3 yielding 20
Thread-2 yielding 23
Thread-1 yielding 26
Thread-4 yielding 29
[2, 5, 8, 11, 14, 17, 20, 23, 26, 29]

Process finished with exit code 0

解决方法

假设我有 N个生成器，它们生成一个项目流gs = [..] # list of generators。

我可以轻松地zip在一起，以在每个相应的发电机获得元组的发电机gs：tuple_gen = zip(*gs)。

这将依次调用next(g)每个并将结果收集到一个元组中。但是，如果每个产品的生产成本很高，我们可能希望并行化多个线程上的工作。g``gs``next(g)

我该如何实现pzip(..)呢？