题解 | #统计每个学校各难度的用户平均刷题数#

select u.university, d.difficult_level, round(count(q.question_id)/count(distinct(q.device_id)) ,4) as avg_answer_cnt

from user_profile as u,question_practice_detail as q,question_detail as d

where u.device_id=q.device_id and q.question_id=d.question_id

group by u.university,d.difficult_level