题解 | #统计每个用户的平均刷题数#

select up.university,qd.difficult_level,format(count(*) / count(distinct qpd.device_id),4) as avg_answer_cnt
from user_profile up
         inner join question_practice_detail qpd on up.device_id = qpd.device_id
         left join question_detail qd on qpd.question_id = qd.question_id
where university = '山东大学'
group by up.university,qd.difficult_level;