题解 | #统计每个用户的平均刷题数#

SELECT university,difficult_level, COUNT(qp.question_id)/ count(distinct (qp.device_id)) AS avg_answer_cnt FROM question_practice_detail qp join user_profile u ON u.device_id = qp.device_id join question_detail q on q.question_id = qp.question_id WHERE university = "山东大学" group by difficult_level