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Accepted
Андрей Котов
Asked:2022-09-24 00:00:23 +0000 UTC2022-09-24 00:00:23 +0000 UTC

多线程列表处理

  • 772

我有一个需要在 n 个线程中处理的列表。我写了这个脚本，但它不正确。这里的线程数将等于列表中的条目数。但我不知道如何制作n个线程

import codecs
import os
import threading


def pr(str):
    print(str + threading.currentThread().getName())


def soap(id_number):
    for number in id_number:
        t = threading.Thread(target=pr, args=(number,))
        t.start()


if __name__ == '__main__':
    id_file = codecs.open(os.path.join(os.path.dirname(os.path.realpath(__file__)), "card.txt"), "r", encoding='utf-8')
    lines = id_file.readlines()
    id_file.close()
    soap(lines)
    exit(0)
python

1 个回答

  1. Best Answer

    将列表分成几部分并将其传递给流：

    import threading

def pr(batch):
    for el in batch:
        print('{}: {}'.format(threading.currentThread().getName(), el))

def soap(id_number, n):
    batch_size = (len(id_number) + n - 1) / n
    for i in range(0, len(id_number), batch_size):
        batch = id_number[i:i+batch_size]
        t = threading.Thread(target=pr, args=(batch,))
        t.start()

soap([0,1,2,3,4,5,6,7,8,9], 3)
    • 1

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