下面的函数 foo
返回一个字符串 'foo'
。如何获取从线程目标返回的值 '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()
上面显示的“一种明显的方法”不起作用:thread.join()
返回了 None
。
我见过的一种方法是将可变对象(例如列表或字典)与索引或其他某种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?
如果您确实希望 join()
返回被调用函数的返回值,您可以使用 Thread
子类执行此操作,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
由于某些名称修改,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
FWIW,multiprocessing
模块使用 Pool
类为此提供了一个很好的接口。如果您想坚持使用线程而不是进程,则可以使用 multiprocessing.pool.ThreadPool
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.
multiprocess
导入的,但它们与进程无关。
processes=1
设置为多个!
在 Python 3.2+ 中,stdlib concurrent.futures
模块为 threading
提供了更高级别的 API,包括将工作线程的返回值或异常传递回主线程:
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)
futures = [executor.submit(foo, param) for param in param_list]
将保持顺序,退出 with
将允许收集结果。 [f.result() for f in futures]
Jake 的回答很好,但是如果您不想使用线程池(您不知道需要多少线程,但根据需要创建它们),那么在线程之间传输信息的好方法是内置的Queue.Queue 类,因为它提供线程安全。
我创建了以下装饰器以使其以与线程池类似的方式运行:
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,您只需将其用作:
@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,但它仍然获得了意见,所以我想我会更新它以反映我在较新版本的 Python 中执行此操作的方式:
在为并行任务提供高级接口的 concurrent.futures
模块中添加了 Python 3.2。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。
此 api 的一个好处是向 Executor
提交任务会返回一个 Future
对象,该对象将以您提交的可调用对象的返回值完成。
这使得附加 queue
对象变得不必要,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
如果未传入,这将使用默认模块线程池执行程序。
用法和之前很相似:
@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 包装成带有 asyncio.wrap_future
的 asyncio.Future
。这使得它可以轻松地与协程一起工作:
result = await asyncio.wrap_future(long_task(10))
如果您不需要访问底层 concurrent.Future
对象,则可以在装饰器中包含包装:
_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,每当您需要将 cpu 密集型或阻塞代码从事件循环线程中推送出去时,您可以将其放入修饰函数中:
@threadpool
def some_long_calculation():
...
# this will suspend while the function is executed on a threadpool
result = await some_long_calculation()
AttributeError: 'module' object has no attribute 'Lock'
这似乎是从第 y = long_task(10)
行发出的...想法?
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为多线程环境:
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
from queue import Queue
。
我发现的大多数答案都很长,需要熟悉其他模块或高级 python 功能,除非他们已经熟悉答案所涉及的所有内容,否则会让某人感到困惑。
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:我想显着简化事情,所以我创建了一个 ThreadWithResult
类并让它继承自 threading.Thread
。 __init__
中的嵌套函数 function
调用我们要保存其值的线程函数,并在线程完成执行后将该嵌套函数的结果保存为实例属性 self.result
。
创建 this 的实例与创建 threading.Thread
的实例相同。将要在新线程上运行的函数传递给 target
参数,并将函数可能需要的任何参数传递给 args
参数,并将任何关键字参数传递给 kwargs
参数。
例如
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!我包含了 time
和 random
模块来模拟线程的行为,但它们不是实现 original question 中要求的功能所必需的。
我知道在提出问题后我正在回答这个问题,但我希望这可以帮助更多的人!
编辑:我创建 save-thread-result
PyPI package 是为了让您可以访问上面的相同代码并在项目之间重复使用它 (GitHub code is here)。 PyPI 包完全扩展了 threading.Thread
类,因此您也可以在 ThreadWithResult
类上设置您在 threading.thread
上设置的任何属性!
上面的原始答案涵盖了这个子类背后的主要思想,但有关更多信息,请参阅 more detailed explanation (from the module docstring) here。
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)
Stable
Development Status 之后那!在我这样做之后我会在这里更新答案:)
Parris / kindall 的 answer join
/return
答案移植到 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
请注意,Thread
类在 Python 3 中的实现方式不同。
我偷了kindall的答案并稍微清理了一下。
关键部分是将 *args 和 **kwargs 添加到 join() 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
更新的答案如下
这是我最受欢迎的答案,因此我决定使用可在 py2 和 py3 上运行的代码进行更新。
此外,我看到这个问题的许多答案表明对 Thread.join() 缺乏理解。有些完全无法处理 timeout
参数。但是,当您有 (1) 一个可以返回 None
的目标函数并且 (2) 您还将 timeout
arg 传递给 join() 时,您还应该注意一个极端情况。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
_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))
您能确定我们在 TEST 4 中可能遇到的极端情况吗?
问题是我们希望 giveMe() 返回 None(参见测试 2),但我们也希望 join() 在超时时返回 None。
returned is 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。但在现实世界的实例中(目标可能合法地返回 None 或其他东西),我们希望明确检查发生了什么。
以下是解决这种极端情况的方法:
# 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()
target
、args
和 kwargs
参数作为成员变量保存在类中,则可以避免访问线程类的私有变量。
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())
out_queue1
,您将需要循环 out_queue1.get()
并捕获 Queue.Empty 异常:ret = [] ; try: ; while True; ret.append(out_queue1.get(block=False)) ; except Queue.Empty: ; pass
。用于模拟换行符的分号。
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。它也是非阻塞的。您改为检查状态。请参阅代码末尾的如何使用它的示例。
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()
我正在使用这个包装器,它可以轻松地将任何函数转换为在 Thread
中运行 - 处理它的返回值或异常。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
使用示例
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.")
线程模块注意事项
舒适的回报价值&线程函数的异常处理是一种常见的“Pythonic”需求,并且确实应该由 threading
模块提供 - 可能直接在标准 Thread
类中提供。 ThreadPool
简单任务的开销太大 - 3 个管理线程,大量官僚作风。不幸的是,Thread
的布局最初是从 Java 复制而来的——您可以从仍然无用的第一个 (!) 构造函数参数 group
中看到它。
根据所提到的内容,这是适用于 Python3 的更通用的解决方案。
import threading
class ThreadWithReturnValue(threading.Thread):
def __init__(self, *init_args, **init_kwargs):
threading.Thread.__init__(self, *init_args, **init_kwargs)
self._return = None
def run(self):
self._return = self._target(*self._args, **self._kwargs)
def join(self):
threading.Thread.join(self)
return self._return
用法
th = ThreadWithReturnValue(target=requests.get, args=('http://www.google.com',))
th.start()
response = th.join()
response.status_code # => 200
考虑到@iman 对@JakeBiesinger 答案的评论,我将其重新组合为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()
join
总是返回 None
,我认为你应该继承 Thread
来处理返回码等等。
您可以在线程函数的范围之上定义一个可变对象,并将结果添加到其中。 (我还修改了代码以兼容python3)
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)
这将返回 {'world!': 'foo'}
如果您使用函数输入作为结果字典的键,则保证每个唯一输入都会在结果中给出一个条目
将目标定义为
1) 接受参数 q
2) 将任何语句 return foo
替换为 q.put(foo); return
所以一个函数
def func(a):
ans = a * a
return ans
会成为
def func(a, q):
ans = a * a
q.put(ans)
return
然后你会这样继续
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))]
您可以使用函数装饰器/包装器来制作它,这样您就可以将现有函数用作 target
而无需修改它们,但请遵循此基本方案。
results = [ans_q.get() for _ in xrange(len(threads))]
GuySoft 的想法很棒,但我认为对象不一定必须从 Thread 继承,并且 start() 可以从接口中删除:
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
如前所述,多处理池比基本线程慢得多。使用此处某些答案中提出的队列是一种非常有效的选择。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相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)
这是我创建的 @Kindall's answer 版本。
这个版本使得您所要做的就是输入带有参数的命令来创建新线程。
这是用 Python 3.8 制作的:
from threading import Thread
from typing import Any
def test(plug, plug2, plug3):
print(f"hello {plug}")
print(f'I am the second plug : {plug2}')
print(plug3)
return 'I am the return Value!'
def test2(msg):
return f'I am from the second test: {msg}'
def test3():
print('hello world')
def NewThread(com, Returning: bool, *arguments) -> Any:
"""
Will create a new thread for a function/command.
:param com: Command to be Executed
:param arguments: Arguments to be sent to Command
:param Returning: True/False Will this command need to return anything
"""
class NewThreadWorker(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
ntw = NewThreadWorker(target = com, args = (*arguments,))
ntw.start()
if Returning:
return ntw.join()
if __name__ == "__main__":
print(NewThread(test, True, 'hi', 'test', test2('hi')))
NewThread(test3, True)
一种常见的解决方案是用装饰器包装你的函数 foo
result = queue.Queue()
def task_wrapper(*args):
result.put(target(*args))
那么整个代码可能看起来像这样
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
笔记
一个重要的问题是返回值可能是无序。 (其实return value
不一定保存到queue
,因为你可以选择任意的thread-safe数据结构)
Kindall's answer 在 Python3 中
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
这是一个非常古老的问题,但我想分享一个对我有用并帮助我的开发过程的简单解决方案。
这个答案背后的方法是这样一个事实,即“新”目标函数 inner
通过称为闭包的东西将原始函数的结果(通过 __init__
函数传递)分配给包装器的 result
实例属性.
这允许包装类保留调用者随时访问的返回值。
注意:此方法不需要使用 threading.Thread
类的任何损坏的方法或私有方法,尽管没有考虑 yield 函数(OP 没有提到 yield 函数)。
享受!
from threading import Thread as _Thread
class ThreadWrapper:
def __init__(self, target, *args, **kwargs):
self.result = None
self._target = self._build_target_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,您可以使用以下代码运行 pytest
(假设您已安装)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'
最好的方法...定义一个全局变量,然后在线程函数中更改变量。没有东西可以传入或取回
from threading import Thread
# global var
radom_global_var = 5
def function():
global random_global_var
random_global_var += 1
domath = Thread(target=function)
domath.start()
domath.join()
print(random_global_var)
# result: 6
我知道这个线程很旧......但我遇到了同样的问题......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()
不定期副业成功案例分享
threading
,而不是尝试不同的库,加上池大小限制引入了一个额外的潜在问题,这发生在我的案例中。TypeError: __init__() takes from 1 to 6 positional arguments but 7 were given
。有什么办法可以解决吗?join
有一个超时参数,应该一起传递_Thread__target
的事情)。你会让任何试图将你的代码移植到 python 3 的人讨厌你,直到他们弄清楚你做了什么(因为使用了在 2 和 3 之间更改的未记录的特性)。很好地记录您的代码。