题解 | #统计每个用户的平均刷题数#

select up.university university, qd.difficult_level difficult_level, count(qpd.question_id) / count(distinct qpd.device_id) avg_answer_cnt from question_practice_detail qpd

inner join user_profile up
    on up.device_id = qpd.device_id and up.university = '山东大学'
    
inner join question_detail qd
    on qd.question_id = qpd.question_id 
group by qd.difficult_level