题解 | #统计每个学校各难度的用户平均刷题数#

select 
    u.university,
    qd.difficult_level,
    count(q.question_id) / count(distinct q.device_id) as avg_answer_cnt

from 
    user_profile as u,
    question_practice_detail as q,
    question_detail as qd
    
where
    u.device_id = q.device_id
    and q.question_id = qd.question_id
    
group by
    university, difficult_level