Question

我太困惑了，无法提出正确的方法：

我有这个有向图：

并有两个字典，分别显示出场和入场分数

graph_to = {'a':{'b':2,'c':3},'b':{'a':1,'d':4}}
graph_from = {'a':{'b':1},'b':{'a':2},'c':{'a':3},'d':{'b':4}}

例如，在graph_to中，节点a进入得分为2的节点b，然后进入得分为3的节点c。并在graph_from节点a中收到节点b的得分1。

我想创建undirected graph以便对两个节点之间的分数求和。它应该成为这本字典：

graph = {
    'a':{'b':3,'c':3},
    'b':{'a':3,'d':4},
    'c':{'a':3},
    'd':{'b':4}
}

Answer 1

您可以尝试制作collections.defaultdict()个对象collections.Counter()，并在迭代两个图格时求和边计数：

    Process current = Process.GetCurrentProcess();
    // Enumerate through all the process resources on the share
    // local computer that the specified process name.
    foreach (Process process in
                     Process.GetProcessesByName(current.ProcessName))
      {
         if (process.Id != current.Id)

           {
                  NativeMethods.SetForegroundWindow( process.MainWindowHandle);
                  NativeMethods.ShowWindow(process.MainWindowHandle,
                           WindowShowStyle.Restore);
                  break;
           }
      }

哪个给：

from collections import defaultdict
from collections import Counter
from pprint import pprint

graph_to = {'a':{'b':2,'c':3},'b':{'a':1,'d':4}}
graph_from = {'a':{'b':1},'b':{'a':2},'c':{'a':3},'d':{'b':4}}

undirected_graph = defaultdict(Counter)

def sum_edges(graph, result):
    for node, edges in graph.items():
        for edge in edges:
            result[node][edge] += edges[edge]

sum_edges(graph_to, undirected_graph)
sum_edges(graph_from, undirected_graph)

pprint(undirected_graph)

注意：defaultdict(<class 'collections.Counter'>, {'a': Counter({'b': 3, 'c': 3}), 'b': Counter({'d': 4, 'a': 3}), 'c': Counter({'a': 3}), 'd': Counter({'b': 4})})和Counter是defaultdict的子类，因此您可以将它们与普通词典一样对待。

如果您确实想要最终无向图中的普通字典，则可以使用以下两种dict理解之一：

dict

此外，您还可以在此处使用dict.update()重构dict((k, dict(v)) for k, v in undirected_graph.items()) # {'a': {'b': 3, 'c': 3}, 'b': {'a': 3, 'd': 4}, 'c': {'a': 3}, 'd': {'b': 4}} {k: dict(v) for k, v in undirected_graph.items()} # {'a': {'b': 3, 'c': 3}, 'b': {'a': 3, 'd': 4}, 'c': {'a': 3}, 'd': {'b': 4}}：

sum_edges()

Answer 2

我希望我们很乐意掌握一切，这里有简单的逻辑

    out_dict = {}
    for key in graph_to : 
        for sub_key in graph_to[key]:
            if  key in graph_from and sub_key in graph_from[key]:
                    out_dict[key] = {sub_key: graph_to[key][sub_key] + graph_from[key][sub_key]}
            else:
                    out_dict[key].update({sub_key: graph_to[key][sub_key]})
    graph_from.update(out_dict)
    print(graph_from)

输出：

{'a': {'b': 3, 'c': 3}, 'b': {'a': 3, 'd': 4}, 'c': {'a': 3}, 'd': {'b': 4}}

从有向图创建无向图

2 个答案: