题解 | #统计每个用户的平均刷题数#

select t.university,t2.difficult_level,
round(count(t1.question_id)/count(distinct t1.device_id),4) as avg_answer_cnt
from user_profile t
    inner join question_practice_detail t1 on t.device_id= t1.device_id
    inner join question_detail t2 on t1.question_id = t2.question_id
where t.university = "山东大学" -- 只比上一题加入了这个条件
group by t.university,t2.difficult_level
order by t.university desc 

只比上一题，增加了一个山东大学的筛选条件。