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

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)

其他回答

定义你的目标 1)采取一个论点q 2)用q.put(foo)替换return foo的任何语句;返回

一个函数

def func(a):
    ans = a * a
    return ans

将成为

def func(a, q):
    ans = a * a
    q.put(ans)
    return

然后你就可以这样做了

from Queue import Queue
from threading import Thread

ans_q = Queue()
arg_tups = [(i, ans_q) for i in xrange(10)]

threads = [Thread(target=func, args=arg_tup) for arg_tup in arg_tups]
_ = [t.start() for t in threads]
_ = [t.join() for t in threads]
results = [q.get() for _ in xrange(len(threads))]

你可以使用函数装饰器/包装器来实现它,这样你就可以使用现有的函数作为目标,而不需要修改它们,但要遵循这个基本方案。

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 (!)

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

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

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'