题解 | 统计每个学校各难度的用户平均刷题数

select distinct
    u.university,
    q2.difficult_level,
    round(
        count(q1.question_id) / count(distinct q1.device_id),
        4
    ) as avg_answer_cnt
from
    question_practice_detail q1
    join question_detail q2 on q1.question_id = q2.question_id
    join user_profile u on u.device_id = q1.device_id
group by
    u.university,
    q2.difficult_level;