题解 | #统计每个用户的平均刷题数#

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