题解 | #统计每个学校各难度的用户平均刷题数#

select university,
       difficult_level,
       round(count(qpd.question_id) / count(distinct qpd.device_id),4)
       as avg_answer_cnt
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 university, difficult_level
order by university;