题解 | 统计每个用户的平均刷题数

select
t1.university,t3.difficult_level,count(t2.question_id)/count(distinct t2.device_id) as avg_answer_cnt 
from 
user_profile as t1 
inner join 
question_practice_detail as t2
on t1.device_id=t2.device_id and t1.university='山东大学'
inner join 
question_detail as t3
on t2.question_id=t3.question_id
group by university,difficult_level