如何在 Python 中从字典创建 igraph 对象

发布于 2024-10-29 13:09:01 字数 348 浏览 6 评论 0原文

我在 Python 程序中使用字典来表示图形。我使用字典的键来表示顶点，使用值来表示每个顶点的相邻节点。字典目前看起来像这样：

{  
   'v1' : ['v2','v3'],
   'v2' : ['v1'],
   'v3' : ['v1','v4'],
   'v4' : ['v3']
    // And so on. 
}

是否有一种简单的方法来创建新的 igraph 对象来自这本词典？如果没有简单的方法，那么下一个最佳选择是什么？

原文

I'm using a dictionary to represent a graph in my Python program. I'm using the keys of the dictionary to represent vertices and the values to represent the adjacent nodes of each vertex. The dictionary currently looks like this:

{  
   'v1' : ['v2','v3'],
   'v2' : ['v1'],
   'v3' : ['v1','v4'],
   'v4' : ['v3']
    // And so on. 
}

Is there a straightforward way to create a new igraph object from this dictionary? If there's no easy way, what's the next-best option?

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旧时光的容颜 2024-11-05 13:09:01

好吧，根据 docs，似乎 igraph 期望将顶点编码为整数。因此，您需要指定从顶点到整数的映射，然后您实际上可以像这样继续操作：

G= {'v1': ['v2', 'v3'], 'v2': ['v1'], 'v3': ['v1', 'v4'], 'v4': ['v3']}
mvi= {'v1': 1, 'v2': 2, 'v3': 3, 'v4': 4}
graph= igraph.Graph(edges= [(mvi[v], mvi[a]) for v in G.keys() for a in G[v]])

Well, according to the docs, it seems that igraph expects vertices encoded as integers. So you need to specify a mapping from your vertices to integers and then you could actually proceed for example like this:

G= {'v1': ['v2', 'v3'], 'v2': ['v1'], 'v3': ['v1', 'v4'], 'v4': ['v3']}
mvi= {'v1': 1, 'v2': 2, 'v3': 3, 'v4': 4}
graph= igraph.Graph(edges= [(mvi[v], mvi[a]) for v in G.keys() for a in G[v]])

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望她远 2024-11-05 13:09:01

我做了类似的事情，也将顶点名称导入到图中：

relations = {'v1': ['v2','v3'], 'v2': ['v1'], 'v3': ['v1','v4']}
g = igraph.Graph()
g.add_vertices(list(set(list(relations.keys()) + list([a for v in relations.values() for a in v]))))
g.add_edges([(v, a) for v in relations.keys() for a in relations[v]])

I did something like this which imports names of vertices to the graph also:

relations = {'v1': ['v2','v3'], 'v2': ['v1'], 'v3': ['v1','v4']}
g = igraph.Graph()
g.add_vertices(list(set(list(relations.keys()) + list([a for v in relations.values() for a in v]))))
g.add_edges([(v, a) for v in relations.keys() for a in relations[v]])

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