题解 | 统计每个学校各难度的用户平均刷题数

select
    university,
    difficult_level,
    round(count(qpd.id) / count(distinct u.id), 4) as avg_answer_cnt
from
    user_profile u
join 
    question_practice_detail qpd on u.device_id = qpd.device_id
join 
    question_detail q on q.question_id = qpd.question_id
group by
    u.university,
    difficult_level
order by
    u.university,
    difficult_level