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。
根据上面提到的,下面是适用于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
使用
th = ThreadWithReturnValue(target=requests.get, args=('http://www.google.com',))
th.start()
response = th.join()
response.status_code # => 200
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)
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
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。而是检查状态。参见代码末尾如何使用它的示例。
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()