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
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())
其他回答
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()
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'
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'
另一个不需要更改现有代码的解决方案:
import Queue # Python 2.x
#from queue import Queue # Python 3.x
from threading import Thread
def foo(bar):
print 'hello {0}'.format(bar) # Python 2.x
#print('hello {0}'.format(bar)) # Python 3.x
return 'foo'
que = Queue.Queue() # Python 2.x
#que = Queue() # Python 3.x
t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
t.join()
result = que.get()
print result # Python 2.x
#print(result) # Python 3.x
它也可以很容易地调整到多线程环境:
import Queue # Python 2.x
#from queue import Queue # Python 3.x
from threading import Thread
def foo(bar):
print 'hello {0}'.format(bar) # Python 2.x
#print('hello {0}'.format(bar)) # Python 3.x
return 'foo'
que = Queue.Queue() # Python 2.x
#que = Queue() # Python 3.x
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 # Python 2.x
#print(result) # Python 3.x
