题解 | 统计每个用户的平均刷题数

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