题解 | #统计每个学校各难度的用户平均刷题数#

    ,t3.difficult_level
    ,round(count(t2.question_id)/count(distinct t2.device_id),4) as avg_answer_cnt
  from user_profile t1 
  join question_practice_detail t2 
    on t1.device_id = t2.device_id
  join question_detail t3 
    on t2.question_id = t3.question_id
group by t1.university
    ,t3.difficult_level