题解 | #统计每个用户的平均刷题数#

select university,difficult_level,round(count(*)/count(distinct(t2.device_id)),4)avg_answer_cnt
 from question_detail t1 
      right join 
      question_practice_detail t2 on (t1.question_id=t2.question_id)
      left join
      user_profile t3 on (t2.device_id=t3.device_id)
 group by university,difficult_level
 having university='山东大学'