题解 | #统计每个用户的平均刷题数#

select u.university, d.difficult_level, count(q.question_id) / count(distinct q.device_id) as "avg_answer_cnt"
from user_profile u
         left join question_practice_detail q on u.device_id = q.device_id
         left join question_detail d on q.question_id = d.question_id
where u.university = '山东大学'
group by d.difficult_level;