foo下面的函数返回一个字符串'foo'。如何获取'foo'从线程目标返回的值？

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()

上面显示的“一种显而易见的方法”不起作用：thread.join()return None。

在Python 3.2+中，stdlib concurrent.futures模块向提供了更高级别的API threading，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)

FWIW，该multiprocessing模块为此类提供了一个不错的接口Pool。而且，如果您要坚持使用线程而不是进程，则可以只使用multiprocessing.pool.ThreadPool该类作为替代品。

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

from multiprocessing.pool import ThreadPool
pool = ThreadPool(processes=1)

async_result = pool.apply_async(foo, ('world', 'foo')) # tuple of args for foo

# do some other stuff in the main process

return_val = async_result.get()  # get the return value from your function.

我见过的一种方法是将可变对象（例如列表或字典）与索引或某种其他标识符一起传递给线程的构造函数。然后，线程可以将其结果存储在该对象的专用插槽中。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?

如果您确实想join()返回被调用函数的返回值，则可以使用如下所示的Thread子类来实现：

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

由于名称修改，这有点麻烦，并且它访问特定于Thread实现的“私有”数据结构...但是它可以工作。

对于python3

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):
        print(type(self._target))
        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

Jake的答案很好，但是如果您不想使用线程池（您不知道需要多少线程，而是根据需要创建它们），那么内置的一种在线程之间传输信息的好方法队列类，因为它提供线程安全性。

我创建了以下装饰器，以使其与线程池类似：

def threaded(f, daemon=False):
    import Queue

    def wrapped_f(q, *args, **kwargs):
        '''this function calls the decorated function and puts the 
        result in a queue'''
        ret = f(*args, **kwargs)
        q.put(ret)

    def wrap(*args, **kwargs):
        '''this is the function returned from the decorator. It fires off
        wrapped_f in a new thread and returns the thread object with
        the result queue attached'''

        q = Queue.Queue()

        t = threading.Thread(target=wrapped_f, args=(q,)+args, kwargs=kwargs)
        t.daemon = daemon
        t.start()
        t.result_queue = q        
        return t

    return wrap

然后，将其用作：

@threaded
def long_task(x):
    import time
    x = x + 5
    time.sleep(5)
    return x

# does not block, returns Thread object
y = long_task(10)
print y

# this blocks, waiting for the result
result = y.result_queue.get()
print result

装饰函数每次调用时都会创建一个新线程，并返回一个Thread对象，该对象包含将接收结果的队列。

更新

自从我发布这个答案已经有一段时间了，但是它仍然得到视图，所以我想我将对其进行更新以反映我在较新版本的Python中执行此操作的方式：

concurrent.futures模块中添加了Python 3.2，该模块为并行任务提供了高级接口。它提供ThreadPoolExecutor和ProcessPoolExecutor，因此您可以使用具有相同api的线程或进程池。

此API的一个好处是将任务提交给Executor返回值Future return会对象，该对象将以您提交的可调用对象的返回值完成。

这使得queue不需要附加对象，从而大大简化了装饰器：

_DEFAULT_POOL = ThreadPoolExecutor()

def threadpool(f, executor=None):
    @wraps(f)
    def wrap(*args, **kwargs):
        return (executor or _DEFAULT_POOL).submit(f, *args, **kwargs)

    return wrap

这将使用默认模块如果未传入，线程池执行程序。

用法与之前非常相似：

@threadpool
def long_task(x):
    import time
    x = x + 5
    time.sleep(5)
    return x

# does not block, returns Future object
y = long_task(10)
print y

# this blocks, waiting for the result
result = y.result()
print result

如果您使用的是Python 3.4+，则使用此方法（通常是Future对象）的一个非常不错的功能是可以包装返回的future并将其转换为asyncio.Futurewith asyncio.wrap_future。这使得它很容易与协程一起工作：

result = await asyncio.wrap_future(long_task(10))

如果不需要访问基础concurrent.Future对象，则可以在包装器中包含自动换行：

_DEFAULT_POOL = ThreadPoolExecutor()

def threadpool(f, executor=None):
    @wraps(f)
    def wrap(*args, **kwargs):
        return asyncio.wrap_future((executor or _DEFAULT_POOL).submit(f, *args, **kwargs))

    return wrap

然后，每当需要将cpu密集型代码或阻塞代码从事件循环线程中推出时，都可以将其放入经过修饰的函数中：

@threadpool
def some_long_calculation():
    ...

# this will suspend while the function is executed on a threadpool
result = await some_long_calculation()

另一个不需要更改现有代码的解决方案：

import Queue
from threading import Thread

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

que = Queue.Queue()

t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
t.join()
result = que.get()
print result

还可以轻松地将其调整为多线程环境：

import Queue
from threading import Thread

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

que = Queue.Queue()
threads_list = list()

t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
threads_list.append(t)

# Add more threads here
...
threads_list.append(t2)
...
threads_list.append(t3)
...

# Join all the threads
for t in threads_list:
    t.join()

# Check thread's return value
while not que.empty():
    result = que.get()
    print result

Parris / kindall的答案 join / return移植到Python 3 的答案：

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

请注意，Thread该类在Python 3中的实现方式有所不同。

我偷了kindall的答案并整理了一下。

关键部分是将* args和** kwargs添加到join（）中以处理超时

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

下面的更新的答案

这是我最受欢迎的答案，因此我决定使用将同时在py2和py3上运行的代码进行更新。

另外，我看到这个问题的许多答案表明对Thread.join（）缺乏理解。有些人完全无法处理timeoutarg。但是，当您拥有（1）可以返回的目标函数None并且（2）您还传递了（timeout arg给join（）。请参阅“测试4”以了解这种极端情况。

与py2和py3一起使用的ThreadWithReturn类：

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

if sys.version_info >= (3, 0):
    _thread_target_key = '_target'
    _thread_args_key = '_args'
    _thread_kwargs_key = '_kwargs'
else:
    _thread_target_key = '_Thread__target'
    _thread_args_key = '_Thread__args'
    _thread_kwargs_key = '_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 not target is 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))

您能否确定在测试4中可能遇到的特殊情况？

问题在于，我们希望GiveMe（）返回None（请参见测试2），但是我们也希望join（）如果超时则返回None。

returned is None 意味着：

（1）这就是GiveMe（）返回的结果，或者

（2）join（）超时

这个例子很简单，因为我们知道GiveMe（）将始终返回None。但是在实际情况下（目标可能合法返回None或其他），我们希望显式检查发生了什么。

以下是解决这种情况的方法：

# 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()

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())

我对这个问题的解决方案是将函数和线程包装在一个类中。不需要使用池，队列或c类型变量传递。这也是非阻塞的。您改为查看状态。请参阅代码末尾有关如何使用它的示例。

import threading

class ThreadWorker():
    '''
    The basic idea is given a function create an object.
    The object can then run the function in a thread.
    It provides a wrapper to start it,check its status,and get data out the function.
    '''
    def __init__(self,func):
        self.thread = None
        self.data = None
        self.func = self.save_data(func)

    def save_data(self,func):
        '''modify function to save its returned data'''
        def new_func(*args, **kwargs):
            self.data=func(*args, **kwargs)

        return new_func

    def start(self,params):
        self.data = None
        if self.thread is not None:
            if self.thread.isAlive():
                return 'running' #could raise exception here

        #unless thread exists and is alive start or restart it
        self.thread = threading.Thread(target=self.func,args=params)
        self.thread.start()
        return 'started'

    def status(self):
        if self.thread is None:
            return 'not_started'
        else:
            if self.thread.isAlive():
                return 'running'
            else:
                return 'finished'

    def get_results(self):
        if self.thread is None:
            return 'not_started' #could return exception
        else:
            if self.thread.isAlive():
                return 'running'
            else:
                return self.data

def add(x,y):
    return x +y

add_worker = ThreadWorker(add)
print add_worker.start((1,2,))
print add_worker.status()
print add_worker.get_results()

join总是返回None，我想你应该子类Thread来处理返回码等等。

考虑到@iman对@JakeBiesinger答案的评论，我将其重新组成为具有多个线程：

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()

干杯，

伙计

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。（我也将代码修改为与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)

这返回 {'world!': 'foo'}

如果使用函数输入作为结果字典的键，则保证每个唯一的输入都会在结果中给出一个条目

我正在使用此包装器，该包装器可以轻松地打开任何函数以在其中运行Thread-照顾其返回值或异常。它不会增加Queue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

使用范例

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.")

threading模块注意事项

线程函数的舒适的返回值和异常处理是“ Pythonic”的常见需求，并且确实应该已经由threading模块提供-可能直接在标准Thread类中提供。ThreadPool对于简单的任务来说有太多的开销-3个管理线程，很多官僚作风。不幸Thread的是，其布局最初是从Java复制的-例如，您仍然可以从仍然无效的1st（！）构造函数参数中看到该布局group。

将目标定义为
1）接受参数q
2）将任何语句替换return foo为q.put(foo); return

所以一个功能

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

会成为

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

然后您将照此进行

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))]

而且，您可以使用函数装饰器/包装器来制作它，这样就可以target不修改而使用现有功能，而是遵循此基本方案。

如上所述，多处理池比基本线程慢得多。使用一些答案中提出的队列是一种非常有效的选择。我将它与字典配合使用，以便能够运行许多小线程并通过将它们与字典结合来调理多个答案：

#!/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)

GuySoft的想法很棒，但是我认为对象不一定必须继承自Thread，并且可以从接口中删除start（）：

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

一种常见的解决方案是foo使用类似这样的装饰器包装函数

result = queue.Queue()

def task_wrapper(*args):
    result.put(target(*args))

然后整个代码可能像这样

result = queue.Queue()

def task_wrapper(*args):
    result.put(target(*args))

threads = [threading.Thread(target=task_wrapper, args=args) for args in args_list]

for t in threads:
    t.start()
    while(True):
        if(len(threading.enumerate()) < max_num):
            break
for t in threads:
    t.join()
return result

注意

一个重要的问题是返回值可能是无序的。（实际上，return value不一定将保存到queue，因为您可以选择任意线程安全的数据结构）

为什么不只使用全局变量？

import threading


class myThread(threading.Thread):
    def __init__(self, ind, lock):
        threading.Thread.__init__(self)
        self.ind = ind
        self.lock = lock

    def run(self):
        global results
        with self.lock:
            results.append(self.ind)



results = []
lock = threading.Lock()
threads = [myThread(x, lock) for x in range(1, 4)]
for t in threads:
    t.start()
for t in threads:
    t.join()
print(results)

Kindall在Python3中的答案

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

    def run(self):
        try:
            if self._target:
                self._return = self._target(*self._args, **self._kwargs)
        finally:
            del self._target, self._args, self._kwargs 

    def join(self,timeout=None):
        Thread.join(self,timeout)
        return self._return

如果仅要从函数调用中验证True或False，我发现一个更简单的解决方案是更新全局列表。

import threading

lists = {"A":"True", "B":"True"}

def myfunc(name: str, mylist):
    for i in mylist:
        if i == 31:
            lists[name] = "False"
            return False
        else:
            print("name {} : {}".format(name, i))

t1 = threading.Thread(target=myfunc, args=("A", [1, 2, 3, 4, 5, 6], ))
t2 = threading.Thread(target=myfunc, args=("B", [11, 21, 31, 41, 51, 61], ))
t1.start()
t2.start()
t1.join()
t2.join()

for value in lists.values():
    if value == False:
        # Something is suspicious 
        # Take necessary action 

如果您想查找任何一个线程是否返回了错误的状态以采取必要的操作，这将对您有所帮助。