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 requests

userdata = 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 requestsfrom multiprocessing import Pooldef 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 ThreadPoolExecutorfrom functools import partialimport requestsuserdata = open("data.txt", "r")usercodes = (x.strip() for x in userdata)  # keep as generatorwith 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, monkeygevent.monkey.patch_all()import requestsuserdata = 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)