题解 | #统计每个用户的平均刷题数#

select
    university,
    difficult_level,
    round(
        count(qd.question_id) / count(distinct qpd.device_id),
        4
    ) as avg_answer_cnt
from
    user_profile as up
    inner join question_practice_detail as qpd on up.device_id = qpd.device_id
    inner join question_detail as qd on qd.question_id = qpd.question_id
where university = '山东大学'
group by difficult_level;