题解 | #统计每个用户的平均刷题数#

select university, 
        difficult_level, 
        count(q.device_id)/count(distinct q.device_id) as avg_answer_cnt
from question_practice_detail as q,
    question_detail as qd,
    user_profile as u
where q.device_id=u.device_id and q.question_id=qd.question_id and university='山东大学'
group by difficult_level