题解 | #统计每个学校的答过题的用户的平均答题数#

select 
    university,
    count(t1.device_id)/count(distinct t1.device_id) as avg_answer_cnt
from question_practice_detail as t1
left join(
    select 
        device_id,
        university
    from user_profile
) as t2
on t1.device_id = t2.device_id
where t1.device_id is not null
group by university;