题解 | #统计每个用户的平均刷题数#

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