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 qpd.device_id),4) as avg_answer_user
 from question_practice_detail as qpd 
 left join user_profile as up
    on qpd.device_id=up.device_id
 left join question_detail as qd 
    on qpd.question_id=qd.question_id
group by university,difficult_level
having university="山东大学"