题解 | #统计每个学校各难度的用户平均刷题数#

select
    university,
    difficult_level,
    count(qp.question_id)/count(distinct qp.device_id) avg_answer_cnt
from
    question_practice_detail qp
left join user_profile u on u.device_id = qp.device_id
left join question_detail qd on qd.question_id=qp.question_id
group by
    university,
    difficult_level;