题解 | #统计每个学校的答过题的用户的平均答题数#

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