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,
    COUNT(qpd.question_id) / COUNT(DISTINCT qpd.device_id) avg_answer_cnt
FROM user_profile up
JOIN question_practice_detail qpd
ON up.device_id = qpd.device_id
JOIN question_detail qd
ON qpd.question_id = qd.question_id
GROUP BY up.university, qd.difficult_level