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()

我找到的大多数答案都很长，需要熟悉其他模块或高级python特性，除非他们已经熟悉答案所谈论的一切，否则会让人感到困惑。

简化方法的工作代码:

import threading

class ThreadWithResult(threading.Thread):
    def __init__(self, group=None, target=None, name=None, args=(), kwargs={}, *, daemon=None):
        def function():
            self.result = target(*args, **kwargs)
        super().__init__(group=group, target=function, name=name, daemon=daemon)

示例代码:

import time, random


def function_to_thread(n):
    count = 0
    while count < 3:
            print(f'still running thread {n}')
            count +=1
            time.sleep(3)
    result = random.random()
    print(f'Return value of thread {n} should be: {result}')
    return result


def main():
    thread1 = ThreadWithResult(target=function_to_thread, args=(1,))
    thread2 = ThreadWithResult(target=function_to_thread, args=(2,))
    thread1.start()
    thread2.start()
    thread1.join()
    thread2.join()
    print(thread1.result)
    print(thread2.result)

main()

解释: 我想大大简化事情，所以我创建了一个ThreadWithResult类，并让它继承threading.Thread。__init__中的嵌套函数函数调用我们想要保存值的线程函数，并将该嵌套函数的结果保存为实例属性self。线程执行完成后的结果。

创建this的实例与创建threading.Thread的实例是相同的。将希望在新线程上运行的函数传递给目标参数，将函数可能需要的任何参数传递给args参数，将任何关键字参数传递给kwargs参数。

e.g.

my_thread = ThreadWithResult(target=my_function, args=(arg1, arg2, arg3))

我认为这比绝大多数答案更容易理解，而且这种方法不需要额外的导入!我加入了time和random模块来模拟线程的行为，但它们并不是实现最初问题中所要求的功能所必需的。

我知道我是在这个问题被问到很久之后才回答的，但我希望这能在未来帮助更多的人!

编辑:我创建了保存线程结果的PyPI包，允许你访问上面相同的代码，并在项目中重用它(GitHub代码在这里)。PyPI包完全扩展了线程。线程类，因此您可以设置在线程上设置的任何属性。线程在ThreadWithResult类!

上面的原始答案介绍了这个子类背后的主要思想，但要了解更多信息，请参阅这里更详细的解释(来自模块docstring)。

快速使用示例:

pip3 install -U save-thread-result     # MacOS/Linux
pip  install -U save-thread-result     # Windows

python3     # MacOS/Linux
python      # Windows

from save_thread_result import ThreadWithResult

# As of Release 0.0.3, you can also specify values for
#`group`, `name`, and `daemon` if you want to set those
# values manually.
thread = ThreadWithResult(
    target = my_function,
    args   = (my_function_arg1, my_function_arg2, ...)
    kwargs = {my_function_kwarg1: kwarg1_value, my_function_kwarg2: kwarg2_value, ...}
)

thread.start()
thread.join()
if getattr(thread, 'result', None):
    print(thread.result)
else:
    # thread.result attribute not set - something caused
    # the thread to terminate BEFORE the thread finished
    # executing the function passed in through the
    # `target` argument
    print('ERROR! Something went wrong while executing this thread, and the function you passed in did NOT complete!!')

# seeing help about the class and information about the threading.Thread super class methods and attributes available:
help(ThreadWithResult)

使用队列:

import threading, queue

def calc_square(num, out_queue1):
  l = []
  for x in num:
    l.append(x*x)
  out_queue1.put(l)


arr = [1,2,3,4,5,6,7,8,9,10]
out_queue1=queue.Queue()
t1=threading.Thread(target=calc_square, args=(arr,out_queue1))
t1.start()
t1.join()
print (out_queue1.get())

我发现做到这一点的最短和最简单的方法是利用Python类及其动态属性。您可以使用threading.current_thread()从派生线程的上下文中检索当前线程，并将返回值赋给一个属性。

import threading

def some_target_function():
    # Your code here.
    threading.current_thread().return_value = "Some return value."

your_thread = threading.Thread(target=some_target_function)
your_thread.start()
your_thread.join()

return_value = your_thread.return_value
print(return_value)

join总是返回None，我认为你应该子类化Thread来处理返回代码等。

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'