题解 | #统计每个学校各难度的用户平均刷题数#

select u.university,q.difficult_level,count(q.difficult_level)/count(distinct p.device_id) avg_answer_cnt
from 
user_profile u
inner join 
question_practice_detail p
on u.device_id=p.device_id
left join
question_detail q
on p.question_id = q.question_id
group by difficult_level,university
order by university,difficult_level