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 qpd.device_id) as avg_answer_cnt 
from user_profile as up  join question_practice_detail as  qpd on up.device_id = qpd.device_id
left join question_detail as qd on qpd.question_id = qd.question_id

group by university , difficult_level