题解 | #统计每个用户的平均刷题数#

select university, difficult_level, count(answer_cnt)/count(distinct u.device_id) avg_answer_cnt
from question_practice_detail qpd
left join question_detail qd on qpd.question_id = qd.question_id
left join user_profile u on qpd.device_id = u.device_id
where university = '山东大学' and answer_cnt > 0
group by difficult_level