select #通过筛选及分组的方式计算每个学校的同学平均答题数 b.university, b.difficult_level, sum(b.result_cnt)/count(distinct b.device_id) as avg_answer_cnt from ( select #关联每个同学所属的学校 a.device_id, a.difficult_level, a.result_cnt, up.university from( select #计算每位同学在不同难度下的答题数; qpd.device_id, qd.difficult_level, count(qpd.result...