networkx：如何设置自定义成本函数？

Question

我正在关注 networkx 文档 ( 1 )，我想为成本函数（例如node_del_cost和node_ins_cost）设置不同的惩罚。比方说，我想通过三点惩罚删除/插入节点。

到目前为止，我已经创建了两个无向图，它们因标记节点 C（更新代码）而异。

import networkx as nx

G=nx.Graph()
G.add_nodes_from([("A", {'label':'CDKN1A'}), ("B", {'label':'CUL4A'}), 
    ("C", {'label':'RB1'})])

G.add_edges_from([("A","B"), ("A","C")])

H=nx.Graph()
H.add_nodes_from([("A", {'label':'CDKN1A'}), ("B", {'label':'CUL4A'}),
    ("C", {'label':'AKT'})])
H.add_edges_from([("A","B"), ("A","C")])

# arguments
# node_match – a function that returns True if node n1 in G1 and n2 in G2 should be considered equal during matching.
# ignored if node_subst_cost is specified
def node_match(node1, node2):
    return node1['label']==node2['label']

# node_subst_cost - a function that returns the costs of node substitution
# overrides node_match if specified.
def node_subst_cost(node1, node2): 
    return node1['label']==node2['label']

# node_del_cost - a function that returns the costs of node deletion
# if node_del_cost is not specified then default node deletion cost of 1 is used.
def node_del_cost(node1):
    return node1['label']==3    

# node_ins_cost - a function that returns the costs of node insertion
# if node_ins_cost is not specified then default node insertion cost of 1 is used.
def node_ins_cost(node2):
    return node2['label']==3    

paths, cost = nx.optimal_edit_paths(G, H, node_match=None, edge_match=None, 
    node_subst_cost=node_subst_cost, node_del_cost=node_del_cost, node_ins_cost=node_ins_cost, 
    edge_subst_cost=None, edge_del_cost=None, edge_ins_cost=None, 
    upper_bound=None)

# length of the path
print(len(paths))

# optimal edit path cost (graph edit distance).
print(cost)

这给了我2.0一个最佳路径成本和路径7.0长度。但是，我不完全理解为什么，因为我将惩罚设置为 3.0，因此预计编辑距离为 3。

谢谢你的建议！

奥尔哈

Answer 1

如文档中所述，当您将node_subst_cost函数作为参数传递时，它会忽略node_match函数并对任何替换操作应用成本，即使节点相等。因此，我建议您首先评估node_subst_cost功能中的节点相等性，然后相应地应用成本：

def node_subst_cost(node1, node2):
    # check if the nodes are equal, if yes then apply no cost, else apply 3
    if node1['label'] == node2['label']:
        return 0
    return 3


def node_del_cost(node):
    return 3  # here you apply the cost for node deletion


def node_ins_cost(node):
    return 3  # here you apply the cost for node insertion


paths, cost = nx.optimal_edit_paths(
    G,
    H,
    node_subst_cost=node_subst_cost,
    node_del_cost=node_del_cost,
    node_ins_cost=node_ins_cost
)

print(cost)  # which will return 3.0

您也可以对边缘操作执行相同的操作。