题解 | 统计每个学校各难度的用户平均刷题数

select
    u.university,
    qd.difficult_level,
    round(
        count(qpd.question_id) / count(distinct u.device_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 qd on qpd.question_id = qd.question_id
group by
    u.university,
    qd.difficult_level
ORDER BY
    u.university ASC,
    qd.difficult_level ASC;