题解 | #统计每个学校各难度的用户平均刷题数#

select
    up.university,
    qd.difficult_level,
    count(qpd.question_id) / count(distinct up.device_id) as avg_answer_cnt
from
    question_practice_detail as qpd
    inner join user_profile as up on qpd.device_id = up.device_id
    inner join question_detail as qd on qd.question_id = qpd.question_id
group by
    university,
    difficult_level