题解 | #统计每个用户的平均刷题数#

select
    "山东大学" as university,
    difficult_level,
    count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt
from
    question_practice_detail as qpd
    inner join user_profile as up on up.device_id = qpd.device_id
    and up.university = "山东大学"
    inner join question_detail as qd on qd.question_id = qpd.question_id
group by
    difficult_level