题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    t1.university,
    t3.difficult_level,
    ROUND(COUNT(t2.question_id)/COUNT(DISTINCT(t2.device_id)),4) AS avg_answer_cnt
FROM user_profile AS t1
    JOIN question_practice_detail AS t2 ON t1.device_id=t2.device_id
    JOIN question_detail AS t3 ON t2.question_id=t3.question_id
GROUP BY t1.university,t3.difficult_level;