小编点评

**目录构建节点字典** 该函数用于构建一个包含匹配节点的字典，用于拼成 Cypher CQL 语句。 ```python def build_nodedict(self, args): """ 构建节点字典 :param args: {'看东西 Sometimes Clear Sometimes Unclear': ['symptom']} :return: 组装成 => {'symptom': '看东西 Sometimes Clear Sometimes Unclear'} """ node_dict = {} for arg, types in args.items(): for type in types: if type not in node_dict: node_dict[type] = [arg] else: node_dict[type].append(arg) return node_dict ``` **构建 Cypher CQL 语句** ```python # 查询症状会导致哪些疾病 sql = [ "MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '{0}' return m.name, r.name, n.name\".format(i) for i in entities] # 查询症状会导致哪些疾病 sql = [ "MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '{0}' return m.name, r.name, n.name\".format(i) for i in entities ] # 查询疾病常用药品－药品别名记得扩充 sql = [ "MATCH (m:Disease)-[r:used_drugs]->(n:Drug) where m.name = '{0}' return m.name, r.name, n.name\".format(i) for i in entities ] # 查询疾病的忌口 sql = [ "MATCH (m:Disease)-[r:noteat_foods]->(n:Foods) where m.name = '{0}' return m.name, r.name, n.name\".format(i) for i in entities ] ``` **测试示例** ```python # 输入question_class = {'args': {'看东西 Sometimes Clear Sometimes Unclear': ['symptom']}, 'question_types': ['symptom_disease']} cql = handler.parser_main(question_class) # 打印CQL语句 print(cql) ``` **输出** ``` [ {'question_type': 'symptom_disease', 'sql': "MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '看东西 Sometimes Clear Sometimes Unclear' return m.name, r.name, n.name"} ] ```

正文

这边只是为了测试，演示效果和思路，实际应用中，可以通过NLP构建CQL
接上一篇的问题分类

question = "请问最近看东西有时候清楚有时候不清楚是怎么回事"
# 最终输出
data = {'args': {'看东西有时候清楚有时候不清楚': ['symptom']}, 'question_types': ['symptom_disease']}

question = "干眼常用药有哪些"
# 最终输出
data = {'args': {'干眼': ['disease']}, 'question_types': ['disease_drug']}

question = "干眼哪些不能吃"
data = {'args': {'干眼': ['disease']}, 'question_types': ['disease_not_food']}

构建节点字典

目的，为了拼CQL，查出符合条件的节点详情

def build_nodedict(self, args):
    """
    构建节点字典
    :param args: {'看东西有时候清楚有时候不清楚': ['symptom']}
    :return: 组装成 => {'symptom': '看东西有时候清楚有时候不清楚'}
    """
    node_dict = {}
    for arg, types in args.items():
        for type in types:
            if type not in node_dict:
                node_dict[type] = [arg]
            else:
                node_dict[type].append(arg)
    return node_dict

# 输入：
{'看东西有时候清楚有时候不清楚': ['symptom']}
# 输出：
{'symptom': ['看东西有时候清楚有时候不清楚']}

构建Cypher CQL语句

# 查询症状会导致哪些疾病
if question_type == 'symptom_disease':
    sql = ["MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '{0}' return m.name, r.name, n.name".format(i) for i in entities]

# 查询症状会导致哪些疾病
if question_type == 'symptom_disease':
    sql = ["MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '{0}' return m.name, r.name, n.name".format(i) for i in entities]

# 查询疾病常用药品－药品别名记得扩充
if question_type == 'disease_drug':
    sql = ["MATCH (m:Disease)-[r:used_drugs]->(n:Drug) where m.name = '{0}' return m.name, r.name, n.name".format(i) for i in entities]


# 查询疾病的忌口
if question_type == 'disease_not_food':
    sql = ["MATCH (m:Disease)-[r:noteat_foods]->(n:Foods) where m.name = '{0}' return m.name, r.name, n.name".format(i) for i in entities]

node_dict.get('symptom')

Test

if __name__ == '__main__':
    handler = QuestionPaser()
    question_class = {'args': {'看东西有时候清楚有时候不清楚': ['symptom']}, 'question_types': ['symptom_disease']}
    cql = handler.parser_main(question_class)
    print(cql)

输出：

# 输入
question_class = {'args': {'看东西有时候清楚有时候不清楚': ['symptom']}, 'question_types': ['symptom_disease']}
# 输出
[{'question_type': 'symptom_disease', 'sql': ["MATCH (m:Disease)-[r:has_symptom]->(n:Symptom) where n.name = '看东西有时候清楚有时候不清楚' return m.name, r.name, n.name"]}]

# 输入：
question_class = {'args': {'干眼': ['disease']}, 'question_types': ['disease_drug']}
# 输出： 
[{'question_type': 'disease_drug', 'sql': ["MATCH (m:Disease)-[r:used_drugs]->(n:Drug) where m.name = '干眼' return m.name, r.name, n.name"]}]

# 输入：
question_class = {'args': {'干眼': ['disease']}, 'question_types': ['disease_not_food']}
# 输出：
[{'question_type': 'disease_not_food', 'sql': ["MATCH (m:Disease)-[r:noteat_foods]->(n:Foods) where m.name = '干眼' return m.name, r.name, n.name"]}]

后面根据 生成的 CQL语句，查询出知识图谱中对应的数据，

源代码地址：https://gitee.com/VipSoft/VipQA