题解 | #统计每个学校各难度的用户平均刷题数#

select
    t1.university,
    t3.difficult_level,
    round(count(t1.answer_cnt)/count(distinct t1.answer_cnt),4) avg_answer_cnt
from
    user_profile t1
left join
    question_practice_detail t2
on 
    t1.device_id = t2.device_id
left join
    question_detail t3
on 
    t2.question_id = t3.question_id
where 
    t3.question_id is not null
group by
    t1.university,
    t3.difficult_level