题解 | #统计每个学校的答过题的用户的平均答题数#

select university ,round(count(question_id)/count(distinct a.device_id),4) AS avg_answer_cnt from user_profile AS a join question_practice_detail AS b 
on a.device_id=b.device_id 
group by university 
order by university 

认识到答题和用户数据的对应关系，这道题就迎刃而解了