python GIL全局解释器锁,多线程多进程效率比较,进程池,协程,TCP服务端实现协程

GIL全局解释器锁

'''
python解释器:
    - Cpython C语言
    - Jpython  java
    ...

1、GIL: 全局解释器锁
    - 翻译: 在同一个进程下开启的多线程,同一时刻只能有一个线程执行,因为Cpython的内存管理不是线程安全。

    - GIL全局解释器锁,本质上就是一把互斥锁,保证数据安全。

定义:
In CPython, the global interpreter lock, or GIL, is a mutex that prevents multiple
native threads from executing Python bytecodes at once. This lock is necessary mainly
because CPython’s memory management is not thread-safe. (However, since the GIL
exists, other features have grown to depend on the guarantees that it enforces.)

结论:在Cpython解释器中,同一个进程下开启的多线程,同一时刻只能有一个线程执行,无法利用多核优势。

# 为什么要有全局解释器锁:
    - 没有锁:

2、GIL全局解释器锁的优缺点:
    1.优点:
        保证数据的安全
    2.缺点:
        单个进程下,开启多个线程,牺牲执行效率,无法实现并行,只能实现并发。

        - IO密集型, 多线程
        - 计算密集型,多进程
'''

import time
from threading import Thread

n = 100

def task():
    global n
    m = n
    time.sleep(3)   #遇到IO,保存状态+切换,其他线程继续争抢GIL
    n = m-1

if __name__ == '__main__':
    list1 = []
    for line in range(10):
        t = Thread(target=task)
        t.start()
        list1.append(t)

    for t in list1:
        t.join()

    print(n)    # 99

 多线程多进程效率比较

单核情况下都用多线程效率高

'''
IO密集型下使用多线程.
计算密集型下使用多进程.

IO密集型任务,每个任务4s
    - 单核:
        - 开启线程比进程节省资源。
        
    - 多核:
        - 多线程: 
            - 开启4个子线程: 16s 
            
        - 多进程:
            - 开启4个进程: 16s + 申请开启资源消耗的时间  
    
计算密集型任务,每个任务4s
    - 单核:
        - 开启线程比进程节省资源。
        
    - 多核:
        - 多线程:
            - 开启4个子线程: 16s
            
        - 多进程:
            - 开启多个进程: 4s
        
计算密集型: 多进程
    假设100份原材料同时到达工厂,聘请100个人同时制造,效率最高

IO密集型: 多线程
    假设100份原材料,只有40份了,其他还在路上,聘请40个人同时制造。

'''
from threading import Thread
from multiprocessing import Process
import time

# 计算密集型任务
def task1():
    # 计算1000000次 += 1
    i = 10
    for line in range(10000000):
        i += 1


# IO密集型任务
def task2():
    time.sleep(3)


if __name__ == '__main__':
    # 1、测试多进程:
    # 测试计算密集型
    # start_time = time.time()
    # list1 = []
    # for line in range(6):
    #     p = Process(target=task1)
    #     p.start()
    #     list1.append(p)
    #
    # for p in list1:
    #     p.join()
    # end_time = time.time()
    # # 消耗时间: 4.204240560531616
    # print(f'计算密集型消耗时间: {end_time - start_time}')
    #
    # # 测试IO密集型
    # start_time = time.time()
    # list1 = []
    # for line in range(6):
    #     p = Process(target=task2)
    #     p.start()
    #     list1.append(p)
    #
    # for p in list1:
    #     p.join()
    # end_time = time.time()
    # # 消耗时间: 4.382250785827637
    # print(f'IO密集型消耗时间: {end_time - start_time}')


    # 2、测试多线程:
    # 测试计算密集型
    start_time = time.time()
    list1 = []
    for line in range(6):
        p = Thread(target=task1)
        p.start()
        list1.append(p)

    for p in list1:
        p.join()
    end_time = time.time()
    # 消耗时间: 5.737328052520752
    print(f'计算密集型消耗时间: {end_time - start_time}')

    # 测试IO密集型
    start_time = time.time()
    list1 = []
    for line in range(6):
        p = Thread(target=task2)
        p.start()
        list1.append(p)

    for p in list1:
        p.join()
    end_time = time.time()
    # 消耗时间: 3.004171848297119
    print(f'IO密集型消耗时间: {end_time - start_time}')

进程池

# 线程池
from concurrent.futures import ProcessPoolExecutor
import time
# 池子对象: 内部可以帮你提交50个启动进程的任务
p_pool = ProcessPoolExecutor(50)


def task1(n):
    print(f'from task1...{n}')
    time.sleep(10)


if __name__ == '__main__':
    n = 1
    while True:
        # 参数1: 函数名
        # 参数2: 函数的参数1
        # 参数3: 函数的参数2
        # submit(参数1, 参数2, 参数3)
        p_pool.submit(task1, n)
        n += 1

wait方法可以让主线程阻塞,直到满足设定的要求。

from concurrent.futures import ThreadPoolExecutor, wait, ALL_COMPLETED, FIRST_COMPLETED
import time

# 参数times用来模拟网络请求的时间
def get_html(times):
    time.sleep(times)
    print("get page {}s finished".format(times))
    return times

executor = ThreadPoolExecutor(max_workers=2)
urls = [3, 2, 4] # 并不是真的url
all_task = [executor.submit(get_html, (url)) for url in urls]
wait(all_task, return_when=ALL_COMPLETED)
print("main")
# 执行结果 
# get page 2s finished
# get page 3s finished
# get page 4s finished
# main
wait方法接收3个参数,等待的任务序列、超时时间以及等待条件。等待条件return_when默认为ALL_COMPLETED,表明要等待所有的任务都结束。可以看到运行结果中,确实是所有任务都完成了,主线程才打印出main。等待条件还可以设置为FIRST_COMPLETED,表示第一个任务完成就停止等待。

用线程池爬取梨视频

# 线程池测试
############
# 2 爬取视频
#############
# 先查看ajax的调用路径 https://www.pearvideo.com/category_loading.jsp?reqType=5&categoryId=9&start=0
#categoryId=9 分类id
#start=0 从哪个位置开始,每次加载12个
# https://www.pearvideo.com/category_loading.jsp?reqType=5&categoryId=9&start=0

from concurrent.futures import ThreadPoolExecutor,wait,ALL_COMPLETED
import requests
import re
import time
start = time.time()

pool = ThreadPoolExecutor(10)
ret = requests.get('https://www.pearvideo.com/category_loading.jsp?reqType=5&categoryId=1&start=0')
# print(ret.text)
# 正则解析
reg = '<a href="(.*?)" class="vervideo-lilink actplay">'
video_urls=re.findall(reg,ret.text)
# print(video_urls)

def download(url):
    ret_detail = requests.get('https://www.pearvideo.com/' + url)   # 相当于文件句柄,建立连接,流的方式,不是一次性取出
    # print(ret_detail.text)
    # 正则去解析
    reg = 'srcUrl="(.*?)",vdoUrl=sr'    # 正则表达式匹配式
    mp4_url = re.findall(reg, ret_detail.text)[0]  # type:str
    # 下载视频
    video_content = requests.get(mp4_url)
    video_name = mp4_url.rsplit('/', 1)[-1]

    with open(video_name, 'wb')as f:
        for line in video_content.iter_content():
            f.write(line)

threads=[]
for url in video_urls:
    t = pool.submit(download(url),url)
    threads.append(t)

wait(threads, return_when=ALL_COMPLETED)    # 等待线程池全部完成
end = time.time()
print('time:',end-start)    # 计算时间

线程池其他方法可以参考: https://www.jianshu.com/p/b9b3d66aa0be

协程(理论 + 代码)

1.什么是协程?
  - 进程: 资源单位
  - 线程: 执行单位
  - 协程: 单线程下实现并发
    - 在IO密集型的情况下,使用协程能提高最高效率。

    注意: 协程不是任何单位,只是一个程序员YY出来的东西。

    Nignx
    500 ----> 500 ---> 250000 ---> 250000 ----> 10 ----> 2500000
    总结: 多进程 ---》 多线程 ---》 让每一个线程都实现协程.(单线程下实现并发)

    - 协程的目的:
      - 手动实现 "遇到IO切换 + 保存状态" 去欺骗操作系统,让操作系统误以为没有IO操作,将CPU的执行权限给你。

from gevent import monkey  # 猴子补丁
monkey.patch_all()  # 监听所有的任务是否有IO操作
from gevent import spawn  # spawn(任务)
from gevent import joinall
import time

def task1():
    print('start from task1...')
    time.sleep(1)
    print('end from task1...')

def task2():
    print('start from task2...')
    time.sleep(3)
    print('end from task2...')

def task3():
    print('start from task3...')
    time.sleep(5)
    print('end from task3...')


if __name__ == '__main__':

    start_time = time.time()
    sp1 = spawn(task1)
    sp2 = spawn(task2)
    sp3 = spawn(task3)
    # sp1.start()
    # sp2.start()
    # sp3.start()
    # sp1.join()
    # sp2.join()
    # sp3.join()
    joinall([sp1, sp2, sp3])

    end_time = time.time()

    print(f'消耗时间: {end_time - start_time}')

TCP服务端实现协程

# server
from gevent import monkey; monkey.patch_all()
from gevent import spawn
import socket

server = socket.socket()

server.bind(
    ('127.0.0.1', 9999)
)

server.listen(5)


# 与客户端通信
def working(conn):
    while True:
        try:
            data = conn.recv(1024)
            if len(data) == 0:
                break

            print(data.decode('utf-8'))

            conn.send(data.upper())

        except Exception as e:
            print(e)
            break

    conn.close()


# 与客户端连接
def run(server):
    while True:
        conn, addr = server.accept()
        print(addr)
        spawn(working, conn)


if __name__ == '__main__':
    print('服务端已启动...')
    g = spawn(run, server)
    g.join()
#client
from threading import Thread, current_thread
import socket


def send_get_msg():
    client = socket.socket()

    client.connect(
        ('127.0.0.1', 9999)
    )
    while True:
        client.send(f'{current_thread().name}'.encode('utf-8'))
        data = client.recv(1024)
        print(data.decode('utf-8'))


# 模拟100个用户访问服务端
if __name__ == '__main__':
    list1 = []
    for line in range(100):
        t = Thread(target=send_get_msg)
        t.start()
        list1.append(t)

    for t in list1:
        t.join()
原文地址:https://www.cnblogs.com/ludingchao/p/12013165.html