NetworkX：在图上构建简单的流程研究

Question

如果你考虑下图：

from __future__ import division
import networkx as nx
import matplotlib.pyplot as plt
import numpy as np
from numpy.linalg import inv

G = nx.Graph()

pos={1:(2,3),2:(0,0),3:(6,0)}
G.add_nodes_from(pos.keys())
nx.set_node_attributes(G, 'coord', pos)

PE={1:0,2:60,3:40}
nx.set_node_attributes(G,'PE',PE)
q={1:100,2:0,3:0}
nx.set_node_attributes(G,'q',q)

G.add_edge(1,2)
G.add_edge(1,3)
G.add_edge(2,3)

import math
lengths={}
inv_lengths={}
for edge in G.edges():
    startnode=edge[0]
    endnode=edge[1]
    lengths[edge]=round(math.sqrt(((pos[endnode][1]-pos[startnode][1])**2)+
                                      ((pos[endnode][0]-pos[startnode][0])**2)),2)
    inv_lengths[edge]=round(1/lengths[edge],3)
nx.set_edge_attributes(G, 'length', lengths)
nx.set_edge_attributes(G, 'inv_length', inv_lengths) 
nx.draw(G,pos,node_size=1000,node_color='r',with_labels=True)
nx.draw_networkx_edge_labels(G,pos)
plt.show()

以及以下流程问题：

其中 1 是仅供应节点，而 2 和 3 是仅需求节点，为什么以下解决方案会产生通过每条边的奇怪流量值？似乎 q1=100 甚至都没有考虑，我希望 L2 有 flow=0.

m=nx.laplacian_matrix(G,weight='inv_length')
a=m.todense()

flow={}
res2=np.dot(a,b) #No inverse is required: x=ab
res2=[round(item,3) for sublist in res2.tolist() for item in sublist]
print res2
for i,e in enumerate(G.edges()):
   flow[e]=res2[i]

b=[]
for i,v in enumerate(PE.values()):
    b.append(v)
res2=np.dot(a,b) #No inverse is required: x=ab
res2=[round(item,3) for sublist in res2.tolist() for item in sublist]
print res2

#res2=[-24.62, 19.96, 4.66]

Answer 1

我冒昧地以更简单的方式计算边长：

from scipy.spatial.distance import euclidean
lengths = {}
inv_lengths = {}
for edge in G.edges():
    startnode = edge[0]
    endnode = edge[1]
    d = euclidean(pos[startnode], pos[endnode])
    lengths[edge] = d
    inv_lengths[edge] = 1/d

这就是我实现矩阵方程的方式 :

E = np.array([[0], 
              [60], 
              [40]], dtype=np.float64)

L1 = lengths[(1, 2)]
L2 = lengths[(2, 3)]
L3 = lengths[(1, 3)]

L = np.array([[1/L1 + 1/L3,       -1/L1,       -1/L3],
              [      -1/L1, 1/L1 + 1/L2,       -1/L2],
              [      -1/L3,       -1/L2, 1/L2 + 1/L3]], dtype=np.float64)

qLE = np.dot(L, E)

以上代码产生的结果（大致）与您的相同：

In [55]: np.set_printoptions(precision=2)

In [56]: qLE
Out[56]: 
array([[-24.64],
       [ 19.97],
       [  4.67]])

综上所述，我认为这不是编程问题。也许你应该修改流模型...

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