使用 LangChain 和 Neo4j 的 Graph RAG
Adam Cowley
Manager, Developer Education at Neo4j
列出以下文本中的人物:
{text}
此文本中的人物有:
* Sampson - Capulet 家仆。
* Gregory - Capulet 家另一位仆人。
* Abram - Montague 家仆人。
* Balthasar - Montague 家仆人。
列出以下文本中的人物:
{text}
此文本列出的人物:
* Sampson - Capulet 家仆
* Gregory - Capulet 家仆
* Abram - Montague 家仆
* Benvolio - Montague 家仆
examples = graph.query("""
MATCH (c:Character)-[:BELONGS_TO]->(f:Family)
RETURN c.id AS id, c.name AS name, f.name AS belongsToFamily
""")
for character in example:
print(character)
{"id": "romeo-montague", "name": "Romeo", "family": "Montague"},{"id": "juliet-capulet", "name": "Juliet", "family": "Capulet"},{"id": "sampson", "name": "Sampson", "family": "Capulet"},...
SystemMessagePromptTemplate.from_template(""" 提取与以下实体对应的信息。忽略未包含在下列清单中的人物:{examples}""",partial_variables={"examples": [ dict(e) for e in examples ]})





MATCH path = (c:Character {id: "romeo-capulet"})-[r]-(inCommon:Character),(inCommon)-[r2]-(other:Character)RETURN c.id AS id, c.name AS name, other.name AS otherName, other.id AS otherId,c.name = other.name AS shareName,EXISTS { (c)-[:BELONGS_TO]->(:Family)<-[:BELONGS_TO]-(other) } AS belongToFamily,count(*) as nodesInCommon,collect(path) AS pathsORDER BY nodesInCommon DESC LIMIT 10

MATCH (:Character {id: 'romeo-montague'})-[:SIMILAR_TO*0..1]-(character),
(character)-[:HAS_LINE]->(line),
(line)-[:SPOKEN_TO]->(target)
RETURN character, line
使用 LangChain 和 Neo4j 的 Graph RAG