题解 | #统计每个用户的平均刷题数#

select university,difficult_level,count(qpd.question_id)/count(distinct qpd.device_id) as avg_answer_cnt
from user_profile as u
inner join question_practice_detail as qpd
   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;