Python多线程与子线程返回值获取的两种主流方案

方案一：继承Thread类封装

import threading
class MyThread(threading.Thread):
 def __init__(self,func,args=()):
  super(MyThread,self).__init__()
  self.func = func
  self.args = args
 def run(self):
  self.result = self.func(*self.args)
 def get_result(self):
  try:
   return self.result # 如果子线程不使用join方法，此处可能会报没有self.result的错误
  except Exception:
   return None
def foo(a,b,c):
 time.sleep(1)
 print a*2,b*2,c*2,
 return a*2,b*2,c*2
st = time.time()
li = []
for i in xrange(4):
 t = MyThread(foo,args=(i,i+1,i+2))
 li.append(t)
 t.start()
for t in li:
 t.join() # 一定要join，不然主线程比子线程跑的快，会拿不到结果
 print t.get_result()
et = time.time()
print et - st

0 2 4 (0, 2, 4)
4 6 8 2 4 6 (2, 4, 6)
(4, 6, 8)
6 8 10 (6, 8, 10)
1.00200009346

方案二：使用Queue队列传递数据

import sys 
import threading 
import Queue 
q = Queue.Queue() 
def worker1(x, y): 
 func_name = sys._getframe().f_code.co_name 
 print "%s run ..." % func_name 
 q.put((x + y, func_name)) 
def worker2(x, y): 
 func_name = sys._getframe().f_code.co_name 
 print "%s run ...." % func_name 
 q.put((x - y, func_name)) 
if __name__ == '__main__': 
 result = list() 
 t1 = threading.Thread(target=worker1, name='thread1', args=(10, 5, )) 
 t2 = threading.Thread(target=worker2, name='thread2', args=(20, 1, )) 
 print '-' * 50 
 t1.start() 
 t2.start() 
 t1.join() 
 t2.join() 
 while not q.empty(): 
  result.append(q.get()) 
 for item in result: 
  if item[1] == worker1.__name__: 
   print "%s 's return value is : %s" % (item[1], item[0]) 
  elif item[1] == worker2.__name__: 
   print "%s 's return value is : %s" % (item[1], item[0])