题解 | #统计每个学校各难度的用户平均刷题数#

select 
a.university,
c.difficult_level,
-- 平均刷题数
round(count(b.question_id)/count(distinct (b.device_id)),4) as avg_answer_cnt 

from question_practice_detail b 
left join 
user_profile a 
on a.device_id=b.device_id
left join question_detail c 
on b.question_id=c.question_id

group by a.university,
c.difficult_level;