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
其他回答
Jake的回答很好，但如果您不想使用线程池(您不知道需要多少线程，但可以根据需要创建它们)，那么在线程之间传输信息的一个好方法是内置的Queue。队列类，因为它提供线程安全性。
我创建了以下装饰器，使其以类似于线程池的方式工作:
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
然后你就把它用作:
@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
装饰函数每次被调用时都会创建一个新线程，并返回一个thread对象，其中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，但它仍然得到了观看，所以我想我应该更新它，以反映我在新版本的Python中这样做的方式:
Python 3.2并发添加。期货模块，为并行任务提供高级接口。它提供了ThreadPoolExecutor和ProcessPoolExecutor，因此您可以使用具有相同api的线程或进程池。
该api的一个好处是将任务提交给Executor将返回一个Future对象，该对象将以您提交的可调用对象的返回值结束。
这使得附加队列对象成为不必要的，这大大简化了装饰器:
_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
如果没有传入，将使用默认的模块线程池执行器。
用法和前面的非常相似:
@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+，那么使用此方法(以及一般的Future对象)的一个非常好的特性是可以将返回的Future对象包装起来以将其转换为asyncio。使用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))
如果您不需要访问底层并发。对象，你可以在装饰器中包含wrap:
_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，当你需要将cpu密集型代码或阻塞代码从事件循环线程中推出时，你可以将它放在装饰函数中:
@threadpool
def some_long_calculation():
...
# this will suspend while the function is executed on a threadpool
result = await some_long_calculation()
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()
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()
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。
join总是返回None，我认为你应该子类化Thread来处理返回代码等。
