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。
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()
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
这是我根据@Kindall的回答创建的版本。
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做的:
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)
GuySoft的想法很棒,但我认为对象不一定要从Thread继承,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