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.question_id) / count(distinct qpd.device_id),
        4
    ) as avg_answer_cnt
from
    question_practice_detail as qpd
     join user_profile as up on up.device_id = qpd.device_id
     join question_detail as qd on qd.question_id = qpd.question_id
group by
    university,
    difficult_level