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'

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

我正在使用这个包装器，它可以轻松地将任何函数转换为在线程中运行-照顾它的返回值或异常。它不会增加队列开销。

def threading_func(f):
    """Decorator for running a function in a thread and handling its return
    value or exception"""
    def start(*args, **kw):
        def run():
            try:
                th.ret = f(*args, **kw)
            except:
                th.exc = sys.exc_info()
        def get(timeout=None):
            th.join(timeout)
            if th.exc:
                raise th.exc[0], th.exc[1], th.exc[2] # py2
                ##raise th.exc[1] #py3                
            return th.ret
        th = threading.Thread(None, run)
        th.exc = None
        th.get = get
        th.start()
        return th
    return start

用法示例

def f(x):
    return 2.5 * x
th = threading_func(f)(4)
print("still running?:", th.is_alive())
print("result:", th.get(timeout=1.0))

@threading_func
def th_mul(a, b):
    return a * b
th = th_mul("text", 2.5)

try:
    print(th.get())
except TypeError:
    print("exception thrown ok.")

线程模块注意事项

线程函数的舒适返回值和异常处理是“python”的常见需求，而且threading模块应该已经提供了——可能直接在标准Thread类中。对于简单的任务，ThreadPool有太多的开销——3个管理线程，很多官僚主义。不幸的是，线程的布局最初是从Java中复制的——例如，从仍然无用的构造函数参数组1 (!)

如上所述，多处理池比基本线程要慢得多。使用一些回答中提出的队列是一种非常有效的替代方法。我已经将它与字典一起使用，以便能够运行许多小线程，并通过将它们与字典结合来恢复多个答案:

#!/usr/bin/env python3

import threading
# use Queue for python2
import queue
import random

LETTERS = 'abcdefghijklmnopqrstuvwxyz'
LETTERS = [ x for x in LETTERS ]

NUMBERS = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

def randoms(k, q):
    result = dict()
    result['letter'] = random.choice(LETTERS)
    result['number'] = random.choice(NUMBERS)
    q.put({k: result})

threads = list()
q = queue.Queue()
results = dict()

for name in ('alpha', 'oscar', 'yankee',):
    threads.append( threading.Thread(target=randoms, args=(name, q)) )
    threads[-1].start()
_ = [ t.join() for t in threads ]
while not q.empty():
    results.update(q.get())

print(results)

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)

我见过的一种方法是将一个可变对象(如列表或字典)传递给线程的构造函数，同时传递一个索引或其他某种类型的标识符。然后线程可以将结果存储在该对象的专用槽中。例如:

def foo(bar, result, index):
    print 'hello {0}'.format(bar)
    result[index] = "foo"

from threading import Thread

threads = [None] * 10
results = [None] * 10

for i in range(len(threads)):
    threads[i] = Thread(target=foo, args=('world!', results, i))
    threads[i].start()

# do some other stuff

for i in range(len(threads)):
    threads[i].join()

print " ".join(results)  # what sound does a metasyntactic locomotive make?

如果你真的想要join()返回被调用函数的返回值，你可以用Thread子类来实现，如下所示:

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={}, Verbose=None):
        Thread.__init__(self, group, target, name, args, kwargs, Verbose)
        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):
        Thread.join(self)
        return self._return

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

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

这有点麻烦，因为一些名称混乱，它访问特定于线程实现的“私有”数据结构……但它确实有效。

对于Python 3:

class ThreadWithReturnValue(Thread):
    
    def __init__(self, group=None, target=None, name=None,
                 args=(), kwargs={}, Verbose=None):
        Thread.__init__(self, group, target, name, args, kwargs)
        self._return = None

    def run(self):
        if self._target is not None:
            self._return = self._target(*self._args,
                                                **self._kwargs)
    def join(self, *args):
        Thread.join(self, *args)
        return self._return