题解 | #统计每个用户的平均刷题数#

select
    up.university,
    qd.difficult_level,
    round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from
    question_detail as qd
    inner join question_practice_detail as qpd on qpd.question_id = qd.question_id
    inner join user_profile as up on qpd.device_id = up.device_id
where up.university ='山东大学'
group by
    qd.difficult_level
order by
    qd.difficult_level;