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

这里是多线程,有一个简单的例子会很有帮助。您可以运行它并轻松了解多线程在Python中的工作方式。我使用了一个锁来防止访问其他线程,直到前面的线程完成它们的工作。通过使用这行代码,

t锁定=线程。有界信号量(值=4)

您可以一次允许多个进程,并保留将在稍后或完成之前的进程后运行的其余线程。

import threading
import time

#tLock = threading.Lock()
tLock = threading.BoundedSemaphore(value=4)
def timer(name, delay, repeat):
    print  "\r\nTimer: ", name, " Started"
    tLock.acquire()
    print "\r\n", name, " has the acquired the lock"
    while repeat > 0:
        time.sleep(delay)
        print "\r\n", name, ": ", str(time.ctime(time.time()))
        repeat -= 1

    print "\r\n", name, " is releaseing the lock"
    tLock.release()
    print "\r\nTimer: ", name, " Completed"

def Main():
    t1 = threading.Thread(target=timer, args=("Timer1", 2, 5))
    t2 = threading.Thread(target=timer, args=("Timer2", 3, 5))
    t3 = threading.Thread(target=timer, args=("Timer3", 4, 5))
    t4 = threading.Thread(target=timer, args=("Timer4", 5, 5))
    t5 = threading.Thread(target=timer, args=("Timer5", 0.1, 5))

    t1.start()
    t2.start()
    t3.start()
    t4.start()
    t5.start()

    print "\r\nMain Complete"

if __name__ == "__main__":
    Main()

其他回答

下面的代码可以运行10个线程同时打印0到99之间的数字:

from threading import Thread

def test():
    for i in range(0, 100):
        print(i)

thread_list = []

for _ in range(0, 10):
    thread = Thread(target=test)
    thread_list.append(thread)

for thread in thread_list:
    thread.start()

for thread in thread_list:
    thread.join()

下面的代码是上述代码循环版本的简写,运行10个线程,同时打印0到99之间的数字:

from threading import Thread

def test():
    [print(i) for i in range(0, 100)]

thread_list = [Thread(target=test) for _ in range(0, 10)]

[thread.start() for thread in thread_list]

[thread.join() for thread in thread_list]

结果如下:

...
99
83
97
84
98
99
85
86
87
88
...

以前的解决方案都没有在我的GNU/Linux服务器上使用多个内核(我没有管理员权限)。他们只是在一个核心上跑步。

我使用较低级别的os.fork接口来派生多个进程。这是对我有用的代码:

from os import fork

values = ['different', 'values', 'for', 'threads']

for i in range(len(values)):
    p = fork()
    if p == 0:
        my_function(values[i])
        break

这很容易理解。这里有两种简单的线程处理方法。

import time
from concurrent.futures import ThreadPoolExecutor, as_completed
import threading

def a(a=1, b=2):
    print(a)
    time.sleep(5)
    print(b)
    return a+b

def b(**kwargs):
    if "a" in kwargs:
        print("am b")
    else:
        print("nothing")
        
to_do=[]
executor = ThreadPoolExecutor(max_workers=4)
ex1=executor.submit(a)
to_do.append(ex1)
ex2=executor.submit(b, **{"a":1})
to_do.append(ex2)

for future in as_completed(to_do):
    print("Future {} and Future Return is {}\n".format(future, future.result()))

print("threading")

to_do=[]
to_do.append(threading.Thread(target=a))
to_do.append(threading.Thread(target=b, kwargs={"a":1}))

for threads in to_do:
    threads.start()
    
for threads in to_do:
    threads.join()

使用线程/多处理的最简单方法是使用更多高级库,如autothread。

import autothread
from time import sleep as heavyworkload

@autothread.multithreaded() # <-- This is all you need to add
def example(x: int, y: int):
    heavyworkload(1)
    return x*y

现在,您可以为函数提供int列表。Autothread将为您处理所有事务,并只提供并行计算的结果。

result = example([1, 2, 3, 4, 5], 10)

作为第二个anwser的python3版本:

import queue as Queue
import threading
import urllib.request

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

theurls = ["http://google.com", "http://yahoo.com", "http://www.python.org","https://wiki.python.org/moin/"]

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

    s = q.get()
    
def non_thread_func():
    for u in theurls:
        get_url(q,u)
        

    s = q.get()
   

您可以测试它:

start = time.time()
thread_func()
end = time.time()
print(end - start)

start = time.time()
non_thread_func()
end = time.time()
print(end - start)

non_thread_func()花费的时间应该是thread_func()的4倍