题解 | #统计每个用户的平均刷题数#

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