题解 | #统计每个学校各难度的用户平均刷题数#

select 
    t1.university, 
    t2.difficult_level, 
    (count(t2.question_id) / count(distinct t2.device_id)) avg_answer_cnt
from user_profile t1,
(
    select t1.device_id, t2.question_id, t2.difficult_level
    from question_practice_detail t1, question_detail t2
    where t1.question_id = t2.question_id
) t2
where t1.device_id = t2.device_id
group by t1.university, t2.difficult_level