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
一种常见的解决方案是用装饰器来包装函数foo
result = queue.Queue()
def task_wrapper(*args):
result.put(target(*args))
那么整个代码可能是这样的
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
Note
一个重要的问题是返回值可能是无序的。 (事实上，返回值不一定保存到队列中，因为您可以选择任意线程安全的数据结构)
其他回答
join总是返回None，我认为你应该子类化Thread来处理返回代码等。
一种常见的解决方案是用装饰器来包装函数foo
result = queue.Queue()
def task_wrapper(*args):
result.put(target(*args))
那么整个代码可能是这样的
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
Note
一个重要的问题是返回值可能是无序的。 (事实上，返回值不一定保存到队列中，因为您可以选择任意线程安全的数据结构)
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
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
我知道这个线程是旧的....但我也遇到了同样的问题…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()
