题解 | #统计每个用户的平均刷题数#

select
    university,
    difficult_level,
    COUNT(q1.question_id) / COUNT(distinct q1.device_id) as avg_answer_cnt
FROM
    question_practice_detail as q1
    inner join user_profile as q2
        on q1.device_id = q2.device_id and q2.university = '山东大学'
    left join question_detail as q3
        on q1.question_id = q3.question_id
GROUP BY q3.difficult_level;