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
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'
其他回答
考虑到@iman对@JakeBiesinger回答的评论，我重新组合了它，使其具有不同数量的线程:
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()
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'
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
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)
如上所述，多处理池比基本线程要慢得多。使用一些回答中提出的队列是一种非常有效的替代方法。我已经将它与字典一起使用，以便能够运行许多小线程，并通过将它们与字典结合来恢复多个答案:
#!/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)
