题解 | #统计每个用户的平均刷题数#

select university,difficult_level,round(count(difficult_level)/count(distinct a.device_id),4) as avg_answer_cnt
from user_profile as a, question_practice_detail as b,question_detail as c
where a.device_id=b.device_id and b.question_id=c.question_id and university='山东大学'
group by university,difficult_level;