题解 | #统计每个学校各难度的用户平均刷题数#

select
    university,
    difficult_level,
    count(q.question_id) / count(distinct (u.device_id)) as avg_answer_cnt 
from
    user_profile as u
    join question_practice_detail as q on u.device_id = q.device_id
    join question_detail as d on q.question_id = d.question_id
group by
    university,
    difficult_level