题解 | 统计每个学校各难度的用户平均刷题数

select
    up.university,
    qd.difficult_level,
    round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from
    question_practice_detail as qpd
    inner join user_profile as up on qpd.device_id = up.device_id
    inner join question_detail as qd on qpd.question_id = qd.question_id
group by
    up.university,
    qd.difficult_level

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