题解 | #统计每个学校各难度的用户平均刷题数#

select u.university,qd.difficult_level,
round(count(qpd.device_id)/(count(distinct qpd.device_id)),4) 
as avg_answer_cnt
from (user_profile u inner join 
question_practice_detail qpd on u.device_id = qpd.device_id) 
inner join question_detail qd on qpd.question_id = qd.question_id
group by u.university,qd.difficult_level;