题解 | 统计每个用户的平均刷题数

select 
    u.university,
    qd.difficult_level,
    round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt
from user_profile u
join question_practice_detail qpd
on u.device_id=qpd.device_id
join question_detail qd
on qpd.question_id=qd.question_id
group by u.university,qd.difficult_level
having u.university="山东大学"
order by avg_answer_cnt desc