Question

我正在尝试用Python编写一个数值方案来模拟网络上的偏微分方程。

我使用networkx来实现网络结构。现在我想使用Multiprocessing在给定的时间尺度上迭代每个边缘上相同的数值方案。每条边都有一个矩阵E，E[n,:]就是n时的状态。

我试过这个：

def Scheme(G):
   NUM_WORKERS = 4
   for n in range(0,M): #time iteration
      with multiprocessing.Pool(processes=NUM_WORKERS) as pool:
          for edge in G.edges: #iteration edges
              results = pool.map_async(Single_Scheme,(edge,n))
              results.wait()

   return G

此处Single_Scheme是边缘上的算法，它在时间n + 1更新状态。

之前的代码会返回OSError: [Errno 24] Too many open files。

为什么我收到此错误？

编辑：我尝试添加所有必要的细节。

Single_Scheme函数定义如下：

def Single_Scheme(graph, edge, time):
    V1=edge[0]
    V2=edge[1]
    if not V1 in W:
        l=dt/graph[V1][V2]['dx']
        V=Speed(graph,V1,V2,n)
        if V1 in S:
            graph[V1][V2]['ev_matrix'][n+1,0] = graph[V1][V2]['ev_matrix'][n,0]*(1 - l*V[0])
    else: 
        graph[V1][V2]['ev_matrix'][n+1,0] = graph[V1][V2]['ev_matrix'][n,0]*(1 - l*V[0])
        for V0 in list(graph.pred[V1]):
            flux_prec = Speed(graph,V0,V1,n)[-1]
            graph[V1][V2]['ev_matrix'][n+1,0]= graph[V1][V2]['ev_matrix'][n+1,0] + l*P[V0,V1,V2]*graph[V0][V1]['ev_matrix'][n,-1]*flux_prec
    graph[V1][V2]['ev_matrix'][n+1,1:graph[V1][V2]['N']] = graph[V1][V2]['ev_matrix'][n,1:graph[V1][V2]['N']] - l*(V[1:graph[V1][V2]['N']]*graph[V1][V2]['ev_matrix'][n,1:graph[V1][V2]['N']] - V[0:graph[V1][V2]['N']-1]*graph[V1][V2]['ev_matrix'][n,0:graph[V1][V2]['N']-1])

    return graph[V1][V2]['ev_matrix'][n+1,:]

此处graph[V1][V2]['ev_matrix][n,:]是时间步(V1,V2)边缘n的状态。

进入主要部分时，它按以下方式调用：

%matplotlib inline

import multiprocessing
import networkx as nx
import numpy as np
from networkx.algorithms import topological_sort


G=nx.DiGraph()
G.add_nodes_from([0,1,2,3,4])
S=[1,2]
W=[3,4]
G.add_edges_from([(1,0),(2,0),(0,3),(0,4)])
G[1][0]['lenght']=1
G[2][0]['lenght']=1
G[0][3]['lenght']=1
G[0][4]['lenght']=1
n_nodes=nx.number_of_nodes(G)
P=np.zeros((n_nodes,n_nodes,n_nodes))
P[1,0,3]=0.5 
P[1,0,4]=0.5


for node in list(topological_sort(G)):
   for next_n in list(G.succ[node]):
       G[node][next_n]['ev_matrix']= np.zeros((51,1001))


G[1][0]['ev_matrix'][0,4*int(np.floor(G[1][0]['N']/10)):6*int(np.floor(G[1][0]['N']/10))] = 1
G[2][0]['ev_matrix'][0,2*int(np.floor(G[2][0]['N']/4)):3*int(np.floor(G[2][0]['N']/4))]=1
G=Scheme(G)

数值方案中的多处理

0 个答案: