我偷了kindall的答案，稍微整理了一下。

关键部分是为join()添加*args和**kwargs，以便处理超时

class threadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super(threadWithReturn, self).__init__(*args, **kwargs)
        
        self._return = None
    
    def run(self):
        if self._Thread__target is not None:
            self._return = self._Thread__target(*self._Thread__args, **self._Thread__kwargs)
    
    def join(self, *args, **kwargs):
        super(threadWithReturn, self).join(*args, **kwargs)
        
        return self._return

更新答案如下

这是我得到最多好评的答案，所以我决定更新可以在py2和py3上运行的代码。

此外，我看到许多对这个问题的回答都显示出对Thread.join()缺乏理解。有些完全不能处理timeout参数。但是当你有(1)一个可以返回None的目标函数并且(2)你也将timeout参数传递给join()时，还有一种极端情况你应该注意。请参阅“TEST 4”以理解这个极端情况。

ThreadWithReturn类，用于py2和py3:

import sys
from threading import Thread
from builtins import super    # https://stackoverflow.com/a/30159479

_thread_target_key, _thread_args_key, _thread_kwargs_key = (
    ('_target', '_args', '_kwargs')
    if sys.version_info >= (3, 0) else
    ('_Thread__target', '_Thread__args', '_Thread__kwargs')
)

class ThreadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self._return = None
    
    def run(self):
        target = getattr(self, _thread_target_key)
        if target is not None:
            self._return = target(
                *getattr(self, _thread_args_key),
                **getattr(self, _thread_kwargs_key)
            )
    
    def join(self, *args, **kwargs):
        super().join(*args, **kwargs)
        return self._return

一些示例测试如下所示:

import time, random

# TEST TARGET FUNCTION
def giveMe(arg, seconds=None):
    if not seconds is None:
        time.sleep(seconds)
    return arg

# TEST 1
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',))
my_thread.start()
returned = my_thread.join()
# (returned == 'stringy')

# TEST 2
my_thread = ThreadWithReturn(target=giveMe, args=(None,))
my_thread.start()
returned = my_thread.join()
# (returned is None)

# TEST 3
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=2)
# (returned is None) # because join() timed out before giveMe() finished

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

你能确定我们在测试4中可能遇到的极端情况吗?

问题是我们期望giveMe()返回None(参见TEST 2)，但我们也期望join()在超时时返回None。

None表示:

(1)这就是giveMe()返回的，或者

(2) join()超时

这个例子很简单，因为我们知道giveMe()总是返回None。但在真实的实例中(目标可能返回None或其他内容)，我们希望显式地检查发生了什么。

下面是如何解决这种极端情况:

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

if my_thread.isAlive():
    # returned is None because join() timed out
    # this also means that giveMe() is still running in the background
    pass
    # handle this based on your app's logic
else:
    # join() is finished, and so is giveMe()
    # BUT we could also be in a race condition, so we need to update returned, just in case
    returned = my_thread.join()

join总是返回None，我认为你应该子类化Thread来处理返回代码等。

Parris / kindall的answer join/return answer移植到Python 3:

from threading import Thread

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

class ThreadWithReturnValue(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


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

twrv.start()
print(twrv.join())   # prints foo

注意，Thread类在Python 3中实现的方式不同。

你可以使用ThreadPool()的pool.apply_async()来返回test()的值，如下所示:

from multiprocessing.pool import ThreadPool

def test(num1, num2):
    return num1 + num2

pool = ThreadPool(processes=1) # Here
result = pool.apply_async(test, (2, 3)) # Here
print(result.get()) # 5

并且，你也可以使用concurrent.futures.ThreadPoolExecutor()的submit()来返回test()的值，如下所示:

from concurrent.futures import ThreadPoolExecutor

def test(num1, num2):
    return num1 + num2

with ThreadPoolExecutor(max_workers=1) as executor:
    future = executor.submit(test, 2, 3) # Here
print(future.result()) # 5

并且，代替返回，你可以使用数组结果，如下所示:

from threading import Thread

def test(num1, num2, r):
    r[0] = num1 + num2 # Instead of "return"

result = [None] # Here

thread = Thread(target=test, args=(2, 3, result))
thread.start()
thread.join()
print(result[0]) # 5

而不是返回，你也可以使用队列结果，如下所示:

from threading import Thread
import queue

def test(num1, num2, q):
    q.put(num1 + num2) # Instead of "return" 

queue = queue.Queue() # Here

thread = Thread(target=test, args=(2, 3, queue))
thread.start()
thread.join()
print(queue.get()) # '5'

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()

您可以在线程函数的作用域之上定义一个可变变量，并将结果添加到该变量中。(我还修改了代码，使其与python3兼容)

returns = {}
def foo(bar):
    print('hello {0}'.format(bar))
    returns[bar] = 'foo'

from threading import Thread
t = Thread(target=foo, args=('world!',))
t.start()
t.join()
print(returns)

返回{'world!”:“foo”}

如果使用函数input作为结果字典的键，则保证每个惟一的输入都在结果中给出一个条目