如何在 Python 中发送多个线程？

Question

从输入发布数据时，我希望能够接收内容列表。输入在一个文本文件中，该文件将在 python 中打开。为了加快进程，我想增加一次可以发送的线程数。我怎么能做到这一点，这是我在说什么的粗略想法：import requestsuserdata = open("data.txt", "r")usercodes = [x.strip() for x in userdata]for i in range(len(usercodes)):    thread_one = requests.post(url='https://test.com/input', params=usercodes[i])    thread_two = requests.post(url='https://test.com/input', params=usercodes[i+1])    thread_three = requests.post(url='https://test.com/input', params=usercodes[i+2])我希望所有线程同时运行，在这里程序将一个接一个地执行请求。

Answer 1

import requests

from multiprocessing import Pool

def make_request(usercode):

    requests.post(url='https://test.com/input', params=usercode)

if __name__ == '__main__':

    userdata = open("data.txt", "r")

    usercodes = [x.strip() for x in userdata]

    with Pool(multiprocessing.cpu_count()) as p:

        print(p.map(make_request, usercodes))

    p.close()

Answer 2

与concurrent.futures.ThreadPoolExecutor：

from concurrent.futures import ThreadPoolExecutor

from functools import partial

import requests

userdata = open("data.txt", "r")

usercodes = (x.strip() for x in userdata)  # keep as generator

with ThreadPoolExecutor() as pool:

    pool.map(partial(requests.post, 'https://test.com/input'), usercodes)

    userdata.close()   # closing the input file

Answer 3

Async 绝对是你的朋友。

from gevent import joinall, spawn, monkey

gevent.monkey.patch_all()

import requests

userdata = open("data.txt", "r")

usercodes = [x.strip() for x in userdata]

send_url = 'https://test.com/input'

threads = []

def send(usercode):

   requests.post(url=send_url, params=usercode)

for code in usercodes:

    threads.append(spawn(send, code))

joinall(threads)