networkx中的定向加权平衡树导入和最短路径

时间:2012-11-24 22:01:33

标签: python networkx

我有一个平衡的树,分支因子2和高度100,每个边都有一个由文本文件给出的权重,如下所示:

 73 41
 52 40 09
 26 53 06 34
 etc etc until row nr 99

即:从节点0到1的边权重是73,从0到2是41,从1到3是52,等等。

我希望找到从树的根到末尾的最短路径(具有相应的边权重和)。据我所知,这可以通过将所有边权重乘以-1并使用Networkx中的Dijkstra算法来完成。

  1. 算法选择是否正确?
  2. 如何“轻松”将此数据集导入Networkx图形对象?
  3. PS:这是项目Euler Problem 67,找到数字三角形的最大总和。我已经通过memoization递归解决了问题,但我想尝试用Networkx解决它包。

2 个答案:

答案 0 :(得分:4)

  

算法选择是否正确?

是。您可以使用正权重,并调用nx.dijkstra_predecessor_and_distance以从根节点0开始获取最短路径。


  

如何“轻松”将此数据集导入Networkx图形对象?

import networkx as nx
import matplotlib.pyplot as plt

def flatline(iterable):
    for line in iterable:
        for val in line.split():
            yield float(val)

with open(filename, 'r') as f:
    G = nx.balanced_tree(r = 2, h = 100, create_using = nx.DiGraph())
    for (a, b), val in zip(G.edges(), flatline(f)):
        G[a][b]['weight'] = val

# print(G.edges(data = True))

pred, distance = nx.dijkstra_predecessor_and_distance(G, 0)

# Find leaf whose distance from `0` is smallest
min_dist, leaf = min((distance[node], node) 
                     for node, degree in G.out_degree_iter()
                     if degree == 0)
nx.draw(G)
plt.show()

答案 1 :(得分:3)

我不确定我是否完全理解输入格式。但类似的东西应该有效:

from itertools import count
import networkx as nx
adj ="""73 41
52 40 09
26 53 06 34"""
G = nx.Graph()
target = 0
for source,line in zip(count(),adj.split('\n')):
    for weight in line.split():
        target += 1
        print source,target,weight
        G.add_edge(source,target,weight=float(weight))
# now call shortest path with weight="weight" and source=0