题解 | #统计每个用户的平均刷题数#

select
    p.university,
    q.difficult_level,
    count(d.question_id) / count(distinct d.device_id) as avg_answer_cnt
from
    user_profile p
    join question_practice_detail d on d.device_id = p.device_id
    join question_detail q on d.question_id = q.question_id
where p.university = '山东大学'
group by
    p.university,
    q.difficult_level