考虑到@iman对@JakeBiesinger回答的评论，我重新组合了它，使其具有不同数量的线程:

from multiprocessing.pool import ThreadPool

def foo(bar, baz):
    print 'hello {0}'.format(bar)
    return 'foo' + baz

numOfThreads = 3 
results = []

pool = ThreadPool(numOfThreads)

for i in range(0, numOfThreads):
    results.append(pool.apply_async(foo, ('world', 'foo'))) # tuple of args for foo)

# do some other stuff in the main process
# ...
# ...

results = [r.get() for r in results]
print results

pool.close()
pool.join()

根据上面提到的，下面是适用于Python3的更通用的解决方案。

import threading

class ThreadWithReturnValue(threading.Thread):
    def __init__(self, *init_args, **init_kwargs):
        threading.Thread.__init__(self, *init_args, **init_kwargs)
        self._return = None
    def run(self):
        self._return = self._target(*self._args, **self._kwargs)
    def join(self):
        threading.Thread.join(self)
        return self._return

使用

        th = ThreadWithReturnValue(target=requests.get, args=('http://www.google.com',))
        th.start()
        response = th.join()
        response.status_code  # => 200

GuySoft的想法很棒，但我认为对象不一定要从Thread继承，start()可以从接口中删除:

from threading import Thread
import queue
class ThreadWithReturnValue(object):
    def __init__(self, target=None, args=(), **kwargs):
        self._que = queue.Queue()
        self._t = Thread(target=lambda q,arg1,kwargs1: q.put(target(*arg1, **kwargs1)) ,
                args=(self._que, args, kwargs), )
        self._t.start()

    def join(self):
        self._t.join()
        return self._que.get()


def foo(bar):
    print('hello {0}'.format(bar))
    return "foo"

twrv = ThreadWithReturnValue(target=foo, args=('world!',))

print(twrv.join())   # prints foo

我知道这个线程是旧的....但我也遇到了同样的问题…如果你愿意使用thread.join()

import threading

class test:

    def __init__(self):
        self.msg=""

    def hello(self,bar):
        print('hello {}'.format(bar))
        self.msg="foo"


    def main(self):
        thread = threading.Thread(target=self.hello, args=('world!',))
        thread.start()
        thread.join()
        print(self.msg)

g=test()
g.main()

join总是返回None，我认为你应该子类化Thread来处理返回代码等。

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)