题解 | #统计每个学校的答过题的用户的平均答题数#

select  university,
 count(C.question_id)/count(distinct C.device_id) as avg_answer_cnt
   from (
     select 
     A.university, 
     B.question_id,
     A.device_id
    from 
    user_profile AS A inner join question_practice_detail AS B
    on  A.device_id = B.device_id) as C
    group by C.university
    order by C.university;