我正在努力理解Python中的线程。我看过文档和示例,但坦率地说,许多示例过于复杂,我很难理解它们。

如何清楚地显示为多线程划分的任务?


当前回答

import threading
import requests

def send():

  r = requests.get('https://www.stackoverlow.com')

thread = []
t = threading.Thread(target=send())
thread.append(t)
t.start()

其他回答

Python 3具有启动并行任务的功能。这使我们的工作更容易。

它有线程池和进程池。

以下内容提供了一个见解:

ThreadPoolExecutor示例(源代码)

import concurrent.futures
import urllib.request

URLS = ['http://www.foxnews.com/',
        'http://www.cnn.com/',
        'http://europe.wsj.com/',
        'http://www.bbc.co.uk/',
        'http://some-made-up-domain.com/']

# Retrieve a single page and report the URL and contents
def load_url(url, timeout):
    with urllib.request.urlopen(url, timeout=timeout) as conn:
        return conn.read()

# We can use a with statement to ensure threads are cleaned up promptly
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
    # Start the load operations and mark each future with its URL
    future_to_url = {executor.submit(load_url, url, 60): url for url in URLS}
    for future in concurrent.futures.as_completed(future_to_url):
        url = future_to_url[future]
        try:
            data = future.result()
        except Exception as exc:
            print('%r generated an exception: %s' % (url, exc))
        else:
            print('%r page is %d bytes' % (url, len(data)))

ProcessPoolExecutor(源)

import concurrent.futures
import math

PRIMES = [
    112272535095293,
    112582705942171,
    112272535095293,
    115280095190773,
    115797848077099,
    1099726899285419]

def is_prime(n):
    if n % 2 == 0:
        return False

    sqrt_n = int(math.floor(math.sqrt(n)))
    for i in range(3, sqrt_n + 1, 2):
        if n % i == 0:
            return False
    return True

def main():
    with concurrent.futures.ProcessPoolExecutor() as executor:
        for number, prime in zip(PRIMES, executor.map(is_prime, PRIMES)):
            print('%d is prime: %s' % (number, prime))

if __name__ == '__main__':
    main()

注意:对于Python中的实际并行化,您应该使用多处理模块来分叉并行执行的多个进程(由于全局解释器锁,Python线程提供了交织,但实际上它们是串行执行的,而不是并行执行的,并且仅在交织I/O操作时有用)。

然而,如果您只是在寻找交错(或者正在执行可以并行化的I/O操作,尽管存在全局解释器锁),那么线程模块就是开始的地方。作为一个非常简单的例子,让我们考虑通过并行对子范围求和来对大范围求和的问题:

import threading

class SummingThread(threading.Thread):
     def __init__(self,low,high):
         super(SummingThread, self).__init__()
         self.low=low
         self.high=high
         self.total=0

     def run(self):
         for i in range(self.low,self.high):
             self.total+=i


thread1 = SummingThread(0,500000)
thread2 = SummingThread(500000,1000000)
thread1.start() # This actually causes the thread to run
thread2.start()
thread1.join()  # This waits until the thread has completed
thread2.join()
# At this point, both threads have completed
result = thread1.total + thread2.total
print result

请注意,以上是一个非常愚蠢的示例,因为它绝对没有I/O,并且由于全局解释器锁,虽然在CPython中交错执行(增加了上下文切换的开销),但仍将串行执行。

这里有一个简单的示例:您需要尝试一些替代URL,并返回第一个URL的内容以进行响应。

import Queue
import threading
import urllib2

# Called by each thread
def get_url(q, url):
    q.put(urllib2.urlopen(url).read())

theurls = ["http://google.com", "http://yahoo.com"]

q = Queue.Queue()

for u in theurls:
    t = threading.Thread(target=get_url, args = (q,u))
    t.daemon = True
    t.start()

s = q.get()
print s

在这种情况下,线程被用作一种简单的优化:每个子线程都在等待URL解析和响应,以将其内容放入队列;每个线程都是一个守护进程(如果主线程结束,则不会保持进程运行——这比不结束更常见);主线程启动所有子线程,在队列中执行get以等待其中一个线程完成put,然后发出结果并终止(这将删除所有可能仍在运行的子线程,因为它们是守护进程线程)。

Python中线程的正确使用总是与I/O操作相关(因为CPython无论如何都不使用多个内核来运行CPU绑定的任务,线程的唯一原因是在等待一些I/O时不会阻塞进程)。顺便说一句,队列几乎总是将工作分配给线程和/或收集工作结果的最佳方式,而且它们本质上是线程安全的,因此它们使您不用担心锁、条件、事件、信号量和其他线程间协调/通信概念。

import threading
import requests

def send():

  r = requests.get('https://www.stackoverlow.com')

thread = []
t = threading.Thread(target=send())
thread.append(t)
t.start()

给定函数f,如下所示:

import threading
threading.Thread(target=f).start()

向f传递参数

threading.Thread(target=f, args=(a,b,c)).start()