管道卡住-堆棧溢出沒有其他解決方案

Question

（更新）我正在構建一個模塊來分發基於代理的模型，其思想是將模型拆分到多個進程中，然后當代理到達邊界時，將它們傳遞給處理該區域的處理器。 我可以建立進程並在不進行通信的情況下工作，但是無法使數據通過管道傳遞並更新其他處理器上的模型段。

我已經在stackoverflow上嘗試了解決方案，並構建了該模型的簡單版本。 一旦將模型對象放入管道，模型就會掛起（它適用於python標准數據類型）。 簡單版本只是來回傳遞代理。

 from pathos.multiprocessing import ProcessPool
 from pathos.helpers import mp
 import copy



 class TestAgent:

 "Agent Class-- Schedule iterates through each agent and \
  executes step function"

 def __init__(self, unique_id, model):
      self.unique_id = unique_id
      self.model = model
      self.type = "agent"

 def step(self):

       pass 
       #print ('     ', self.unique_id, "I have stepped")

class TestModel:

   "Model Class iterates through schedule and executes step function for \
   each agent"

   def __init__(self):
       self.schedule = []
       self.pipe = None
       self.process = None

       for i in range(1000):
           a = TestAgent(i, self)
           self.schedule.append(a)


   def step(self):


       for a in self.schedule:
           a.step()

if __name__ == '__main__':

   pool = ProcessPool(nodes=2)

   #create instance of model
   test_model = TestModel()
   #create copies of model to be run on 2 processors
   test1 = copy.deepcopy(test_model)
   #clear schedule
   test1.schedule = []
   #Put in only half the schedule
   for i in range(0,500):
       test1.schedule.append(test_model.schedule[i])  
   #Give process tracker number
   test1.process = 1
   #repeat for other processor
   test2= copy.deepcopy(test_model)
   test2.schedule = []
   for i in range(500,1000):
       test2.schedule.append(test_model.schedule[i])
   test2.process = 2

   #create pipe
   end1, end2 = mp.Pipe()

   #Main run function for each process
   def run(model, pipe):

      for i in range(5):
          print (model.process)#, [a.unique_id for a in model.schedule])
          model.step() # IT HANGS AFTER INITIAL STEP
          print ("send")
          pipe.send(model.schedule)
          print ("closed")
          sched = pipe.recv()
          print ("received")
          model.schedule = sched



   pool.map(run, [test1, test2], [end1,end2])

代理應切換處理器並執行其打印功能。 （我的下一個問題將是同步處理器，以使它們保持在每個步驟上，但一次執行一次。）

Answer 1

我知道了。 我超出了python中的管道緩沖區限制（8192）。 如果代理擁有模型的副本作為屬性，則尤其如此。 下面是上面代碼的有效版本，該版本一次通過代理。 它使用Pympler來獲取所有代理的大小。

from pathos.multiprocessing import ProcessPool
from pathos.helpers import mp
import copy

# do a blocking map on the chosen function

class TestAgent:

"Agent Class-- Schedule iterates through each agent and \
executes step function"

   def __init__(self, unique_id, model):
       self.unique_id = unique_id
       self.type = "agent"

   def step(self):
       pass 


class TestModel:

   "Model Class iterates through schedule and executes step function for \
   each agent"

   def __init__(self):
       from pympler import asizeof 

       self.schedule = []
       self.pipe = None
       self.process = None
       self.size = asizeof.asizeof


       for i in range(1000):
           a = TestAgent(i, self)
           self.schedule.append(a)


   def step(self):


       for a in self.schedule:
           a.step()

if __name__ == '__main__':

   pool = ProcessPool(nodes=2)

   #create instance of model
   test_model = TestModel()
   #create copies of model to be run on 2 processors
   test1 = copy.deepcopy(test_model)
   #clear schedule
   test1.schedule = []
   #Put in only half the schedule
   for i in range(0,500):
       test1.schedule.append(test_model.schedule[i])  
   #Give process tracker number
   test1.process = 1
   #repeat for other processor
   test2= copy.deepcopy(test_model)
   test2.schedule = []
   for i in range(500,1000):
       test2.schedule.append(test_model.schedule[i])
   test2.process = 2

   #create pipe
   end1, end2 = mp.Pipe()

   #Main run function for each process
   def run(model, pipe):

      for i in range(5):
        agents = []
        print (model.process, model.size(model.schedule) ) 
        model.step() # IT HANGS AFTER INITIAL STEP
        #agent_num = list(model.schedule._agents.keys())
        for agent in model.schedule[:]:
            model.schedule.remove(agent)
            pipe.send(agent)
            agent = pipe.recv()
            agents.append(agent)
        print (model.process, "all agents received")
        for agent in agents: 
            model.schedule.append(agent)

        print (model.process, len(model.schedule))



   pool.map(run, [test1, test2], [end1,end2])

Mike McKerns和Thomas Moreau-謝謝您的幫助，使我走上了正確的道路。