python - 在 neo4j 中创建关系的有效方法

Question

我有一个 Neo4j 数据库，其中填充了数千个节点，但没有定义任何关系。我有一个包含节点之间关系的文件，因此我想在数据库中创建的这些节点之间创建关系。我目前的做法是:

from py2neo import NodeSelector,Graph,Node,Relationship
graph = Graph('http://127.0.0.1:7474/db/data')
tx = graph.begin()
selector = NodeSelector(graph)
with open("file","r") as relations:
    for line in relations:
        line_split=line.split(";")
        node1 = selector.select("Node",unique_name=line_split[0]).first()
        node2 = selector.select("Node",unique_name=line_split[1]).first()
        rs = Relationship(node1,"Relates to",node2)
        tx.create(rs)
tx.commit()

当前的方法需要两次数据库查询才能获取节点以形成关系+创建关系。鉴于数据库中当前存在节点，是否有更有效的方法？

Answer 1

您可以在填充关系时使用某种形式的节点缓存:

from py2neo import NodeSelector,Graph,Node,Relationship
graph = Graph('http://127.0.0.1:7474/db/data')
tx = graph.begin()
selector = NodeSelector(graph)
node_cache = {}

with open("file","r") as relations:
    for line in relations:
        line_split=line.split(";")

        # Check if we have this node in the cache
        if line_split[0] in node_cache:
            node1 = node_cache[line_split[0]]
        else:
            # Query and store for later
            node1 = selector.select("Node",unique_name=line_split[0]).first()
            node_cache[line_split[0]] = node1

        if line_split[1] in node_cache:
            node2 = node_cache[line_split[1]]
        else:
            node2 = selector.select("Node",unique_name=line_split[1]).first()
            node_cache[line_split[1]] = node2

        rs = Relationship(node1,"Relates to",node2)
        tx.create(rs)

tx.commit()

通过上述内容，您只会加载每个节点一次，并且仅当该节点出现在您的输入文件中时。