chentianrui
169 lines
5.5 KiB
Python
169 lines
5.5 KiB
Python
from neo4j import GraphDatabase
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import pandas as pd
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import json
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class KnowledgeGraphQuerier:
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def __init__(self, uri, user, password):
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self.driver = GraphDatabase.driver(uri, auth=(user, password))
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def close(self):
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self.driver.close()
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def get_node_name(self, node_types, properties):
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"""根据节点类型和可用属性确定节点名称"""
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# 根据用户提供的信息，为不同类型的节点设置名称属性
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priority_map = {
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"ProjectDivisionSet": ["name"],
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"ProjectDivisionItem": ["name"],
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"Quota": ["name"],
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"MainMaterial": ["name"],
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"Equipment": ["name"],
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"MaterialOrEquipment": ["name"],
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"CostSet": ["GUID"],
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"CostItem": ["id"],
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"FeeTableTemplateSet": ["name"],
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"FeeTableTemplateItem": ["name"],
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"FeeCollection": ["name"],
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"FeeScheduleSet": ["name"],
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"FeeScheduleItem": ["name"],
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"Fee": ["name"],
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"ProjectPropertySet": ["name"],
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"ProjectProperty": ["软件名称"],
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}
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# 默认优先级顺序（适用于未明确定义的节点类型）
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default_priority = ["名称", "项目名称", "费用名称", "工程名称", "id", "GUID", "序号", "编码", "代码"]
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# 确定使用哪个优先级列表
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priority_list = None
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for node_type in node_types:
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if node_type in priority_map:
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priority_list = priority_map[node_type]
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break
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if not priority_list:
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priority_list = default_priority
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# 按优先级查找第一个存在的属性
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for prop in priority_list:
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if prop in properties and properties[prop]:
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return properties[prop]
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# 如果没有找到任何匹配的属性，返回第一个非空属性
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for prop, value in properties.items():
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if value:
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return value
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return "无名称"
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def get_simple_hierarchy(self):
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"""获取简洁的图谱层级结构"""
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# 首先获取所有节点和它们的名称
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with self.driver.session() as session:
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result = session.run(
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"""
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MATCH (n)
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RETURN id(n) AS node_id, labels(n) AS node_types, properties(n) AS properties
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"""
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)
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nodes = {}
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for record in result:
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node_id = record["node_id"]
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node_types = record["node_types"]
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properties = record["properties"]
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# 获取节点名称
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name = self.get_node_name(node_types, properties)
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# 以第一个类型作为节点的主要类型
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main_type = node_types[0] if node_types else "Unknown"
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nodes[node_id] = {"id": node_id, "type": main_type, "name": name, "children": []}
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# 获取节点之间的关系
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with self.driver.session() as session:
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result = session.run(
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"""
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MATCH (a)-[r]->(b)
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RETURN id(a) AS source_id, type(r) AS relationship_type, id(b) AS target_id
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"""
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)
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# 构建父子关系
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for record in result:
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source_id = record["source_id"]
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target_id = record["target_id"]
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if source_id in nodes and target_id in nodes:
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nodes[source_id]["children"].append(target_id)
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# 找出根节点（没有父节点的节点）
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all_nodes = set(nodes.keys())
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child_nodes = set()
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for node in nodes.values():
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child_nodes.update(node["children"])
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root_nodes = all_nodes - child_nodes
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# 构建层级结构
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hierarchy = []
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for root_id in root_nodes:
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hierarchy.append(self._build_simple_hierarchy(root_id, nodes))
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# 如果没有根节点，则返回所有节点的平级结构
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if not hierarchy:
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for node_id in nodes:
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node_info = nodes[node_id]
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hierarchy.append({node_info["type"]: node_info["name"]})
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return hierarchy
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def _build_simple_hierarchy(self, node_id, nodes):
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"""递归构建简洁的节点层级结构"""
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node = nodes[node_id]
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node_type = node["type"]
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node_name = node["name"]
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# 创建当前节点的键值对
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result = {node_type: node_name}
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# 如果有子节点，递归处理
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if node["children"]:
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children = []
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for child_id in node["children"]:
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children.append(self._build_simple_hierarchy(child_id, nodes))
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# 将子节点添加到当前节点下
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result["children"] = children
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return result
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def main():
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# 连接信息
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uri = "bolt://172.20.0.145:7687"
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user = "neo4j"
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password = "password"
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querier = KnowledgeGraphQuerier(uri, user, password)
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try:
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# 获取简洁的图谱层级结构
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print("正在构建简洁的图谱层级结构...")
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hierarchy = querier.get_simple_hierarchy()
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# 将层级结构保存为JSON文件
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print("正在保存图谱层级结构到JSON文件...")
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with open("kg_simple_hierarchy.json", "w", encoding="utf-8") as f:
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json.dump(hierarchy, f, ensure_ascii=False, indent=2)
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print("简洁的图谱层级结构已保存到 kg_simple_hierarchy.json")
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finally:
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querier.close()
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if __name__ == "__main__":
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main()
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