题解 | #统计每个用户的平均刷题数#

select t1.university,t3.difficult_level,count(t3.difficult_level)/count(DISTINCT t1.device_id)
from user_profile as t1
inner join question_practice_detail as t2 
on t1.device_id = t2.device_id and university = "山东大学"
inner join question_detail as t3
on t2.question_id = t3.question_id
group by t3.difficult_level