题解 | #统计每个学校各难度的用户平均刷题数#

select 
    up.university,
    qd.difficult_level,
    count(qpd.question_id)/count(distinct up.device_id) #注意，问题数量可有重复，但设备（即用户）不能重复
from 
  user_profile as up
    inner join
  question_practice_detail as qpd on up.device_id = qpd.device_id 
    inner join 
  question_detail as qd on qpd.question_id = qd.question_id
group by up.university , qd.difficult_level
order by up.university asc