下面的函数foo返回一个字符串'foo'。我如何才能获得从线程的目标返回的值'foo' ?
from threading import Thread
def foo(bar):
print('hello {}'.format(bar))
return 'foo'
thread = Thread(target=foo, args=('world!',))
thread.start()
return_value = thread.join()
上面所示的“一种明显的方法”不起作用:thread.join()返回None。
当前回答
最好的方法…定义一个全局变量，然后在线程函数中更改该变量。没有可以传递或取回的东西
from threading import Thread
# global var
radom_global_var = 5
def function():
global random_global_var
random_global_var += 1
domath = Thread(target=function)
domath.start()
domath.join()
print(random_global_var)
# result: 6
其他回答
另一个不需要更改现有代码的解决方案:
import Queue # Python 2.x
#from queue import Queue # Python 3.x
from threading import Thread
def foo(bar):
print 'hello {0}'.format(bar) # Python 2.x
#print('hello {0}'.format(bar)) # Python 3.x
return 'foo'
que = Queue.Queue() # Python 2.x
#que = Queue() # Python 3.x
t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
t.join()
result = que.get()
print result # Python 2.x
#print(result) # Python 3.x
它也可以很容易地调整到多线程环境:
import Queue # Python 2.x
#from queue import Queue # Python 3.x
from threading import Thread
def foo(bar):
print 'hello {0}'.format(bar) # Python 2.x
#print('hello {0}'.format(bar)) # Python 3.x
return 'foo'
que = Queue.Queue() # Python 2.x
#que = Queue() # Python 3.x
threads_list = list()
t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
threads_list.append(t)
# Add more threads here
...
threads_list.append(t2)
...
threads_list.append(t3)
...
# Join all the threads
for t in threads_list:
t.join()
# Check thread's return value
while not que.empty():
result = que.get()
print result # Python 2.x
#print(result) # Python 3.x
join总是返回None，我认为你应该子类化Thread来处理返回代码等。
在Python 3.2+中，stdlib concurrent。futures模块为线程提供了一个更高级别的API，包括将返回值或异常从工作线程传递回主线程:
import concurrent.futures
def foo(bar):
print('hello {}'.format(bar))
return 'foo'
with concurrent.futures.ThreadPoolExecutor() as executor:
future = executor.submit(foo, 'world!')
return_value = future.result()
print(return_value)
最好的方法…定义一个全局变量，然后在线程函数中更改该变量。没有可以传递或取回的东西
from threading import Thread
# global var
radom_global_var = 5
def function():
global random_global_var
random_global_var += 1
domath = Thread(target=function)
domath.start()
domath.join()
print(random_global_var)
# result: 6
这是我根据@Kindall的回答创建的版本。
这个版本使得您所要做的就是输入带有参数的命令来创建新线程。
这是用Python 3.8做的:
from threading import Thread
from typing import Any
def test(plug, plug2, plug3):
print(f"hello {plug}")
print(f'I am the second plug : {plug2}')
print(plug3)
return 'I am the return Value!'
def test2(msg):
return f'I am from the second test: {msg}'
def test3():
print('hello world')
def NewThread(com, Returning: bool, *arguments) -> Any:
"""
Will create a new thread for a function/command.
:param com: Command to be Executed
:param arguments: Arguments to be sent to Command
:param Returning: True/False Will this command need to return anything
"""
class NewThreadWorker(Thread):
def __init__(self, group = None, target = None, name = None, args = (), kwargs = None, *,
daemon = None):
Thread.__init__(self, group, target, name, args, kwargs, daemon = daemon)
self._return = None
def run(self):
if self._target is not None:
self._return = self._target(*self._args, **self._kwargs)
def join(self):
Thread.join(self)
return self._return
ntw = NewThreadWorker(target = com, args = (*arguments,))
ntw.start()
if Returning:
return ntw.join()
if __name__ == "__main__":
print(NewThread(test, True, 'hi', 'test', test2('hi')))
NewThread(test3, True)
