SELECT university,difficult_level,count(qpd.result)/count(DISTINCT u.device_id) as avg_answer_cnt from user_profile as u join question_practice_detail as qpd on u.device_id = qpd.device_id join question_detail as qd on qpd.question_id = qd.question_id group by u.university,qd.difficult_level #第一个点:...