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
这是一个很老的问题，但我想分享一个简单的解决方案，它对我的开发过程有帮助。
这个答案背后的方法论是这样一个事实，即“新的”目标函数，内部是将原始函数的结果(通过__init__函数传递)通过所谓的闭包分配给包装器的结果实例属性。
这允许包装器类保留返回值以供调用者随时访问。
注意:这个方法不需要使用线程的任何mangded方法或私有方法。线程类，虽然没有考虑屈服函数(OP没有提到屈服函数)。
享受吧!
from threading import Thread as _Thread
class ThreadWrapper:
def __init__(self, target, *args, **kwargs):
self.result = None
self._target = self._build_threaded_fn(target)
self.thread = _Thread(
target=self._target,
*args,
**kwargs
)
def _build_threaded_fn(self, func):
def inner(*args, **kwargs):
self.result = func(*args, **kwargs)
return inner
此外，你可以用下面的代码运行pytest(假设你已经安装了它)来演示结果:
import time
from commons import ThreadWrapper
def test():
def target():
time.sleep(1)
return 'Hello'
wrapper = ThreadWrapper(target=target)
wrapper.thread.start()
r = wrapper.result
assert r is None
time.sleep(2)
r = wrapper.result
assert r == 'Hello'
其他回答
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。
FWIW，多处理模块使用Pool类提供了一个很好的接口。如果您希望坚持使用线程而不是进程，可以直接使用multiprocessing.pool.ThreadPool类作为替代。
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)
Kindall在Python3中的回答
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
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'
