Detailed introduction to multithreading in python (code example)

This article brings you a detailed introduction (code example) about multi-threading in Python. It has certain reference value. Friends in need can refer to it. I hope it will be helpful to you.

This article records the problems encountered in learning Python and some common usages. Note that the Python version of this development environment is 2.7.

1. Python file naming

When naming python files, you must be careful not to conflict with the system's default module name, otherwise an error will be reported.
As shown in the following example, when learning threads, the file name is threading.py. The Python script is completely normal and there is no problem. As a result, the following error is reported: AttributeError: 'module' object has no attribute 'xxx'.

threading.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_test.py
@time: 18/8/25 09:14
"""

import threading

# 获取已激活的线程数
print(threading.active_count())

Execution:

➜  baseLearn python threading/threading.py
Traceback (most recent call last):
  File "threading/threading.py", line 9, in <module>
    import threading
  File "/Users/kaiyiwang/Code/python/baseLearn/threading/threading.py", line 12, in <module>
    print(threading.active_count())
AttributeError: 'module' object has no attribute 'active_count'
➜  baseLearn

Problem location:

View import library Source file, found that the source file exists and has no errors, and the .pyc file of the source file also exists

Problem solution:

• 1.When naming the py script, do not use the same words as python reserved words, module names, etc.

• 2. Delete the .pyc file of the library (because py The script will generate a .pyc file every time it is run; if the .pyc file has been generated, if the code is not updated, pyc will still be used when running, so delete the .pyc file) and rerun the code; or find a program that can In the environment where the code is running, copy and replace the .pyc file of the current machine.

Rename the script file name to threading_test.py, and then execute it, it will not An error was reported.

➜  baseLearn python threading/threading_test.py
1
➜  baseLearn

2. Multi-threading threading

Multi-threading is an effective way to speed up program calculations. Python’s multi-threading modulethreading is quick and easy to get started. Let’s start with this section. How everyone uses it.

1. Add thread

threading_test.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_test.py
@time: 18/8/25 09:14
"""

import threading

# 获取已激活的线程数
# print(threading.active_count())

# 查看所有线程信息
# print(threading.enumerate())

# 查看现在正在运行的线程
# print(threading.current_thread())

def thread_job():
    print('This is a thread of %s' % threading.current_thread())


def main():
    thread = threading.Thread(target=thread_job,)  # 定义线程
    thread.start() # 让线程开始工作

if __name__ == '__main__':
    main()

2. Join function

The result of not adding join()

We letT1 increase the time consuming of thread work

threading_join.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_join.py
@time: 18/8/25 09:14
"""

import threading
import time

def thread_job():
    print('T1 start\n')
    for i in range(10):
        time.sleep(0.1) # 任务时间0.1s
    print("T1 finish\n")


def main():
    added_thread = threading.Thread(target=thread_job, name='T1')  # 定义线程
    added_thread.start() # 让线程开始工作
    print("all done\n")

if __name__ == '__main__':
    main()

The expected output results are in order Execute in sequence:

T1 start
T1 finish
all done

But the actual running result is:

➜  baseLearn python threading/threading_join.py
T1 start
all done


T1 finish

➜  baseLearn

The result of joining()

The thread task is output before it is completedall done. If you want to follow the order, you can call join on the thread after starting it:

added_thread.start()
added_thread.join()
print("all done\n")

Print results:

➜  baseLearn python threading/threading_join.py
T1 start

T1 finish

all done

Full script file:

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_join.py
@time: 18/8/25 09:14
"""

import threading
import time

def thread_job():
    print('T1 start\n')
    for i in range(10):
        time.sleep(0.1) # 任务时间0.1s
    print("T1 finish\n")


def main():
    added_thread = threading.Thread(target=thread_job, name='T1')  # 定义线程
    added_thread.start() # 让线程开始工作
    added_thread.join()
    print("all done\n")

if __name__ == '__main__':
    main()

small Test your skills

If you add two threads, what is the printed output?

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_join.py
@time: 18/8/25 09:14
"""

import threading
import time

def T1_job():
    print('T1 start\n')
    for i in range(10):
        time.sleep(0.1) # 任务时间0.1s
    print("T1 finish\n")

def T2_job():
    print("T2 start\n")
    print("T2 finish\n")

def main():
    thread_1 = threading.Thread(target=T1_job, name='T1')  # 定义线程
    thread_2 = threading.Thread(target=T2_job, name='T2')  # 定义线程
    thread_1.start()  # 开启T1
    thread_2.start()  # 开启T2
    print("all done\n")

if __name__ == '__main__':
    main()

The "one" result output is:

T1 start

T2 start

T2 finish

all done

T1 finish

Now neither T1 nor T2 has join. Note that "one" is said here because of the appearance of all done It completely depends on the execution speed of the two threads. It is entirely possible that T2 finish appears after all done. We cannot tolerate this messy execution method, so we need to use join to control it.

Let’s try adding thread_1.join() after T1 starts and before T2 starts. :

thread_1.start()
thread_1.join() # notice the difference!
thread_2.start()
print("all done\n")

Print results:

T1 start

T1 finish

T2 start
all done

T2 finish

You can see , T2 will wait for T1 to finish before starting to run.

3. Queue to store process results

Implementation function

The code implements the function, transfers the data in the data list, uses four threads to process it, and saves the result in the Queue , after the thread is executed, obtain the stored result from the Queue

Define a Queue in the multi-threaded function to save the return value, replaces return, Define a multi-threaded list and initialize a multi-dimensional data list for processing:

threading_queue.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_queue.py
@time: 18/8/25 09:14
"""

import threading
import time
from queue import Queue

def job(l, q):
    for i in range(len(l)):
        l[i] = l[i] ** 2
    q.put(l) #多线程调用的函数不能用return返回值

def multithreading():
    q = Queue()  #q中存放返回值，代替return的返回值
    threads = []
    data = [[1,2,3],[3,4,5],[4,4,4],[5,5,5]]

    for i in range(4): #定义四个线程
        t = threading.Thread(target=job, args=(data[i], q))  #Thread首字母要大写，被调用的job函数没有括号，只是一个索引，参数在后面
        t.start() #开始线程
        threads.append(t) #把每个线程append到线程列表中

    for thread in threads:
        thread.join()

    results = []
    for _ in range(4):
        results.append(q.get()) #q.get()按顺序从q中拿出一个值
    print(results)


if __name__ == '__main__':
    multithreading()

When executing the above script, an error occurred:

➜  baseLearn python threading/threading_queue.py
Traceback (most recent call last):
  File "threading/threading_queue.py", line 11, in <module>
    from queue import Queue
ImportError: No module named queue

I checked the reason and it was caused by the python version:
Solution: No module named 'Queue'

On Python 2, the module is named Queue, on Python 3, it was renamed to follow PEP8 guidelines (all lowercase for module names), making it queue. The class remains Queue on all versions (following PEP8).

Typically, the way you'd write version portable imports would be to do:

In python3, we can quote like this:

try:
    import queue
except ImportError:
    import Queue as queue

In python2, we can quote like this:

from Queue import Queue

Print:

baseLearn python ./threading/threading_queue.py
[[1, 4, 9], [9, 16, 25], [16, 16, 16], [25, 25, 25]]

Full code:
threading_queue.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_queue.py
@time: 18/8/25 09:14
"""

import threading
# import time
from Queue import Queue

def job(l, q):
    for i in range(len(l)):
        l[i] = l[i] ** 2
    q.put(l) #多线程调用的函数不能用return返回值

def multithreading():
    q = Queue()  #q中存放返回值，代替return的返回值
    threads = []
    data = [[1,2,3],[3,4,5],[4,4,4],[5,5,5]]

    for i in range(4): #定义四个线程
        t = threading.Thread(target=job, args=(data[i], q))  #Thread首字母要大写，被调用的job函数没有括号，只是一个索引，参数在后面
        t.start() #开始线程
        threads.append(t) #把每个线程append到线程列表中

    for thread in threads:
        thread.join()

    results = []
    for _ in range(4):
        results.append(q.get()) #q.get()按顺序从q中拿出一个值
    print(results)


if __name__ == '__main__':
    multithreading()

4. GIL efficiency issue

What is GIL?

This time we will take a look at why Python's multi-threading is sometimes not particularly ideal. The main reason is that there is a necessary link in the design of Python, which is Global Interpreter Lock (GIL). This thing allows Python to still process only one thing at a time.

Explanation of GIL:

尽管Python完全支持多线程编程， 但是解释器的C语言实现部分在完全并行执行时并不是线程安全的。 实际上，解释器被一个全局解释器锁保护着，它确保任何时候都只有一个Python线程执行。 GIL最大的问题就是Python的多线程程序并不能利用多核CPU的优势 （比如一个使用了多个线程的计算密集型程序只会在一个单CPU上面运行）。在讨论普通的GIL之前，有一点要强调的是GIL只会影响到那些严重依赖CPU的程序（比如计算型的）。 如果你的程序大部分只会涉及到I/O，比如网络交互，那么使用多线程就很合适， 因为它们大部分时间都在等待。实际上，你完全可以放心的创建几千个Python线程， 现代操作系统运行这么多线程没有任何压力，没啥可担心的。

测试GIL

我们创建一个 job, 分别用 threading 和 一般的方式执行这段程序. 并且创建一个 list 来存放我们要处理的数据. 在 Normal 的时候, 我们这个 list 扩展4倍, 在 threading 的时候, 我们建立4个线程, 并对运行时间进行对比.

threading_gil.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_gil.py
@time: 18/8/25 09:14
"""

import threading
from Queue import Queue
import copy
import time

def job(l, q):
    res = sum(l)
    q.put(l) #多线程调用的函数不能用return返回值

def multithreading(l):
    q = Queue()  #q中存放返回值，代替return的返回值
    threads = []

    for i in range(4): #定义四个线程
        t = threading.Thread(target=job, args=(copy.copy(l), q), name="T%i" % i)  #Thread首字母要大写，被调用的job函数没有括号，只是一个索引，参数在后面
        t.start() #开始线程
        threads.append(t) #把每个线程append到线程列表中

    [t.join() for t in threads]
    total = 0
    for _ in range(4):
        total = q.get() #q.get()按顺序从q中拿出一个值
    print(total)

def normal(l):
    total = sum(l)
    print(total)

if __name__ == '__main__':
    l = list(range(1000000))
    s_t = time.time()
    normal(l*4)
    print('normal:', time.time() - s_t)
    s_t = time.time()
    multithreading(l)
    print('multithreading: ', time.time() - s_t)

如果你成功运行整套程序, 你大概会有这样的输出. 我们的运算结果没错, 所以程序 threading 和 Normal 运行了一样多次的运算. 但是我们发现 threading 却没有快多少, 按理来说, 我们预期会要快3-4倍, 因为有建立4个线程, 但是并没有. 这就是其中的 GIL 在作怪.

1999998000000
normal:  0.10034608840942383
1999998000000
multithreading:  0.08421492576599121

5、线程锁Lock

不使用 Lock 的情况

threading_lock.py

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_lock.py
@time: 18/8/25 09:14
"""

import threading

# 全局变量A的值每次加1，循环10次，并打印
def job1():
    global A
    for i in range(10):
        A+=1
        print('job1',A)

# 全局变量A的值每次加10，循环10次，并打印
def job2():
    global A
    for i in range(10):
        A+=10
        print('job2',A)

# 定义两个线程，分别执行函数一和函数二
if __name__== '__main__':
   
    A=0
    t1=threading.Thread(target=job1)
    t2=threading.Thread(target=job2)
    t1.start()
    t2.start()
    t1.join()
    t2.join()

打印输出数据：

➜  baseLearn python ./threading/threading_lock.py
('job1', ('job2'1)
, (11)'job1'
('job2', 22)
('job2', 32)
('job2', 42)
('job2', 52)
('job2', 62)
('job2', 72)
('job2', 82)
('job2', 92)
('job2', 102)
, 12)
('job1', 103)
('job1', 104)
('job1', 105)
('job1', 106)
('job1', 107)
('job1', 108)
('job1', 109)
('job1', 110)

可以看出，打印的结果非常混乱

使用 Lock 的情况

lock在不同线程使用同一共享内存时，能够确保线程之间互不影响，使用lock的方法是， 在每个线程执行运算修改共享内存之前，执行lock.acquire()将共享内存上锁， 确保当前线程执行时，内存不会被其他线程访问，执行运算完毕后，使用lock.release()将锁打开， 保证其他的线程可以使用该共享内存。

函数一和函数二加锁

def job1():
    global A,lock
    lock.acquire()
    for i in range(10):
        A+=1
        print('job1',A)
    lock.release()

def job2():
    global A,lock
    lock.acquire()
    for i in range(10):
        A+=10
        print('job2',A)
    lock.release()

主函数中定义一个Lock

if __name__== '__main__':
    lock=threading.Lock()
    A=0
    t1=threading.Thread(target=job1)
    t2=threading.Thread(target=job2)
    t1.start()
    t2.start()
    t1.join()
    t2.join()

完整代码：

# -*- coding:utf-8 -*-

"""
@author: Corwien
@file: threading_lock.py
@time: 18/8/25 09:14
"""

import threading

def job1():
    global A,lock
    lock.acquire()
    for i in range(10):
        A+=1
        print('job1',A)
    lock.release()

def job2():
    global A,lock
    lock.acquire()
    for i in range(10):
        A+=10
        print('job2',A)
    lock.release()

if __name__== '__main__':
    lock = threading.Lock()
    A=0
    t1=threading.Thread(target=job1)
    t2=threading.Thread(target=job2)
    t1.start()
    t2.start()
    t1.join()
    t2.join()

打印输出：

➜  baseLearn python ./threading/threading_lock.py
('job1', 1)
('job1', 2)
('job1', 3)
('job1', 4)
('job1', 5)
('job1', 6)
('job1', 7)
('job1', 8)
('job1', 9)
('job1', 10)
('job2', 20)
('job2', 30)
('job2', 40)
('job2', 50)
('job2', 60)
('job2', 70)
('job2', 80)
('job2', 90)
('job2', 100)
('job2', 110)

从打印结果来看，使用lock后，一个一个线程执行完。使用lock和不使用lock，最后打印输出的结果是不同的。

相关推荐：

详解Python中的多线程编程

理解python多线程（python多线程简明教程）

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