NetworkX Graph对象的“同构"比较，而不是默认的“地址"比较 [英] 'Isomorphic' comparison of NetworkX Graph objects instead of the default 'address' comparison

问题描述

我想使用NetworkX Graph 对象作为Python dict 中的键.但是，我不希望默认行为进行比较(即按对象的地址).相反，我希望同构图被称为 dict 中相同元素的键.

I would like to use NetworkX Graph objects as keys in a Python dict. However, I do not want the default behavior for comparison (i.e., by the address of the object). Instead, I would like isomorphic graphs to refer to be keys to the same elements in the dict.

此行为是否已在某处实现?我找不到这个方向的任何信息.

Is this behavior already implemented somewhere? I could not find any information in this direction.

如果我必须自己实施，以下评估是否现实?

If I have to implement it myself, is the following assessment realistic?

• 将 networkx.Graph 包装在一个类中.
• 定义 __ eq __ ，以使其调用 is_isomorphic .
• 以某种方式定义 __ hash __ (欢迎提出建议).

• Wrap networkx.Graph in a class.
• Define __eq__ such that it calls is_isomorphic.
• Define __hash__ somehow (suggestions welcomed).

我认为我必须使此包装图不可变，因为:

I think that I would have to make this wrapped Graph immutable, because:

如果类定义了可变对象并实现了 __ eq __()方法，则不应实现 __ hash __()，因为可哈希集合的实现要求键的哈希值是不可变的(如果对象的哈希值更改，它将位于错误的哈希存储桶中).

If a class defines mutable objects and implements an __eq__() method, it should not implement __hash__(), since the implementation of hashable collections requires that a key’s hash value is immutable (if the object’s hash value changes, it will be in the wrong hash bucket).

推荐答案

以下是对networkx图进行子类化并添加您描述的 eq 和 hash 函数的示例.我不确定它是否可以解决您的问题，但应该作为一个开始.

Here is an example of subclassing a networkx Graph and adding a eq and hash function as you describe. I'm not sure if it solves your problem but should be a start.

import networkx as nx

class myGraph(nx.Graph):
    def __eq__(self, other):
        return nx.is_isomorphic(self, other)
    def __hash__(self):
        return hash(tuple(sorted(self.degree().values())))


if __name__ == '__main__':
    G1 = myGraph([(1,2)])
    G2 = myGraph([(2,3)])
    G3 = myGraph([(1,2),(2,3)])
    print G1.__hash__(), G1.edges()
    print G2.__hash__(), G2.edges()
    print G3.__hash__(), G3.edges()
    print G1 == G2
    print G1 == G3
    graphs = {}
    graphs[G1] = 'G1'
    graphs[G2] = 'G2'
    graphs[G3] = 'G3'
    print graphs.items()

输出类似:

3713081631935493181 [(1, 2)]
3713081631935493181 [(2, 3)]
2528504235175490287 [(1, 2), (2, 3)]
True
False
[(<__main__.myGraph object at 0xe47a90>, 'G2'), (<__main__.myGraph object at 0x1643250>, 'G3')]
[aric@hamerkop tmp]$ python gc.py 
3713081631935493181 [(1, 2)]
3713081631935493181 [(2, 3)]
2528504235175490287 [(1, 2), (2, 3)]
True
False
[(<__main__.myGraph object at 0x1fefad0>, 'G2'), (<__main__.myGraph object at 0x27ea290>, 'G3')]

这篇关于NetworkX Graph对象的“同构"比较，而不是默认的“地址"比较的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！