使用pytholog定义带症状列表的知识库查询失败，求解决方法

想用pytholog构建一个疾病对应症状列表的知识库，写了如下代码：

import pytholog as pl
diseases_kb = pl.KnowledgeBase()
diseases_kb([
     "disease(diabetes, [frequent_urination, extreme_thirst])",
     "disease(pneumonia, [cough, fever, shortness_of_breath])",
])

但执行查询print(diseases_kb.query(pl.Expr("disease(diabetes, Symptoms)")))时，返回结果是['No']。试过用()、list()替换列表语法都没用，怀疑是pytholog解析时会移除括号并按逗号拆分参数，有没有可行的解决办法？

pytholog基于Prolog实现，而Prolog的事实参数并不支持Python风格的列表直接解析，得换用符合Prolog规范的方式来处理，这里有几种可行方案：

方案1：拆分疾病-症状为独立事实（推荐） #

这是Prolog中存储一对多关系的标准做法，避免列表参数的解析问题，同时更灵活：

import pytholog as pl
diseases_kb = pl.KnowledgeBase()
# 添加疾病-症状对应事实
diseases_kb([
     "disease_symptom(diabetes, frequent_urination)",
     "disease_symptom(diabetes, extreme_thirst)",
     "disease_symptom(pneumonia, cough)",
     "disease_symptom(pneumonia, fever)",
     "disease_symptom(pneumonia, shortness_of_breath)",
     # 定义规则，用findall收集某疾病的所有症状
     "get_disease_symptoms(Disease, Symptoms) :- findall(S, disease_symptom(Disease, S), Symptoms)"
])
# 查询糖尿病的症状
print(diseases_kb.query(pl.Expr("get_disease_symptoms(diabetes, Symptoms)")))

执行后会返回[{'Symptoms': ['frequent_urination', 'extreme_thirst']}]，完全符合需求。

方案2：使用Prolog兼容的列表语法直接定义 #

如果坚持要保留列表形式，确保定义和查询都使用Prolog的列表语法，同时用pl.Expr构造事实：

import pytholog as pl
diseases_kb = pl.KnowledgeBase()
# 用pl.Expr包装每个事实，确保列表被正确解析
diseases_kb([
    pl.Expr("disease(diabetes, [frequent_urination, extreme_thirst])"),
    pl.Expr("disease(pneumonia, [cough, fever, shortness_of_breath])")
])
# 查询
result = diseases_kb.query(pl.Expr("disease(diabetes, Symptoms)"))
print(result)

这种方式能让pytholog正确识别列表参数，返回预期的症状列表。

方案3：手动构造列表参数（调试用） #

如果上述方法仍有问题，可以尝试将列表作为原子字符串传递，再通过Prolog规则转换：

import pytholog as pl
diseases_kb = pl.KnowledgeBase()
diseases_kb([
     "disease(diabetes, 'frequent_urination,extreme_thirst')",
     "disease(pneumonia, 'cough,fever,shortness_of_breath')",
     # 定义规则拆分字符串为列表
     "split_symptoms(Str, List) :- split_string(Str, ',', '', List)",
     "get_disease_symptoms(Disease, Symptoms) :- disease(Disease, SymStr), split_symptoms(SymStr, Symptoms)"
])
# 查询
print(diseases_kb.query(pl.Expr("get_disease_symptoms(diabetes, Symptoms)")))

不过这种方式需要依赖Prolog的字符串处理能力，通用性不如前两种方案。

内容的提问来源于stack exchange，提问作者Elissa