题解 | #统计每个用户的平均刷题数#

select 
  t1.university,
  t3.difficult_level,
  round(count(t2.question_id)/count(distinct(t2.device_id)),4) as avg_answer_cnt
from 
  user_profile as t1,
  question_practice_detail as t2,
  question_detail as t3
where t1.university = "山东大学" 
      and t1.device_id = t2.device_id
      and t2.question_id = t3.question_id
group by t3.difficult_level