题解 | 统计每个用户的平均刷题数

select
    university,
    difficult_level,
    avg_answer_cnt
from
    (
        select
            pro.university as university,
            qdet.difficult_level as difficult_level,
            count(1) / count(distinct pro.device_id) as avg_answer_cnt
        from
            user_profile pro
            join question_practice_detail pdet on pro.device_id = pdet.device_id
            join question_detail qdet on qdet.question_id = pdet.question_id
        group by
            pro.university, qdet.difficult_level
    ) temp
where university = '山东大学'

根据23题，直接进行过滤操作。