题解 | 统计每个学校各难度的用户平均刷题数

select u.university, qq.difficult_level,
round(count(q.question_id)/count(distinct q.device_id),4) as avg_answer_cnt
from user_profile u 
 join question_practice_detail q
on u.device_id = q.device_id
join question_detail qq
on q.question_id = qq.question_id

group by u.university, qq.difficult_level

平均题数=总题数/总（去重后）人数！！

不是直接count（question_id）也不是直接avg(question_id)