我有一个脚本,我希望一个函数与另一个函数同时运行。
我看过的示例代码:
import threading
def MyThread ( threading.thread ):
doing something........
def MyThread2 ( threading.thread ):
doing something........
MyThread().start()
MyThread2().start()
我无法正常工作。我更希望使用线程函数而不是类来实现。
感谢您的帮助。
这是工作脚本,感谢您的帮助。
from threading import Thread
class myClass():
def help(self):
os.system('./ssh.py')
def nope(self):
a = [1,2,3,4,5,6,67,78]
for i in a:
print i
sleep(1)
if __name__ == "__main__":
Yep = myClass()
thread = Thread(target = Yep.help)
thread2 = Thread(target = Yep.nope)
thread.start()
thread2.start()
thread.join()
print 'Finished'
答案 0 :(得分:271)
您不需要使用Thread
的子类来完成这项工作 - 看看我在下面发布的简单示例,看看如何:
from threading import Thread
from time import sleep
def threaded_function(arg):
for i in range(arg):
print("running")
sleep(1)
if __name__ == "__main__":
thread = Thread(target = threaded_function, args = (10, ))
thread.start()
thread.join()
print("thread finished...exiting")
这里我展示了如何使用线程模块创建一个调用普通函数作为其目标的线程。你可以看到我如何在线程构造函数中传递我需要的任何参数。
答案 1 :(得分:37)
您的代码存在一些问题:
def MyThread ( threading.thread ):
如果你真的只想用功能做这个,你有两个选择:
使用线程:
import threading
def MyThread1():
pass
def MyThread2():
pass
t1 = threading.Thread(target=MyThread1, args=[])
t2 = threading.Thread(target=MyThread2, args=[])
t1.start()
t2.start()
使用主题:
import thread
def MyThread1():
pass
def MyThread2():
pass
thread.start_new_thread(MyThread1, ())
thread.start_new_thread(MyThread2, ())
答案 2 :(得分:11)
我尝试添加另一个join(),看起来很有效。这是代码
from threading import Thread
from time import sleep
def function01(arg,name):
for i in range(arg):
print(name,'i---->',i,'\n')
print (name,"arg---->",arg,'\n')
sleep(1)
def test01():
thread1 = Thread(target = function01, args = (10,'thread1', ))
thread1.start()
thread2 = Thread(target = function01, args = (10,'thread2', ))
thread2.start()
thread1.join()
thread2.join()
print ("thread finished...exiting")
test01()
答案 3 :(得分:3)
您可以使用target
构造函数中的Thread
参数直接传入一个被调用而不是run
的函数。
答案 4 :(得分:2)
你是否覆盖了run()方法?如果您覆盖了__init__
,那么您确定要调用基地threading.Thread.__init__()
吗?
在启动两个线程之后,主线程是否继续在子线程上无限期地工作/阻塞/连接,以便主线程执行在子线程完成任务之前不会结束?
最后,你得到任何未处理的例外吗?
答案 5 :(得分:-1)
Python 3具有Launching parallel tasks的功能。这使我们的工作更加轻松。
它具有thread pooling和Process pooling的名称。
以下内容提供了一个见解:
ThreadPoolExecutor示例
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)))
另一个示例
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.ThreadPoolExecutor(max_workers=5) as executor:
for number, prime in zip(PRIMES, executor.map(is_prime, PRIMES)):
print('%d is prime: %s' % (number, prime))
if __name__ == '__main__':
main()