题解 | 统计每个学校各难度的用户平均刷题数

select
university,difficult_level,round(count(question_id)/count(distinct  device_id),4) as avg_answer_cnt
from (
         select q.device_id,
                u.university,
                q.question_id,
                qu.difficult_level
         from question_practice_detail q
                  inner join user_profile u
                             on q.device_id = u.device_id
                  inner join question_detail qu
                            on q.question_id = qu.question_id

         order by u.university,qu.difficult_level

     ) t
group by university,difficult_level