个人认为题面应该更新为, 不同学校, 不同难度, 回答过问题的同学的平均回答次数
然后就是清除聚合函数都是干嘛的, 就比如avg(字段)
求得是把所有这个字段的值加起来然后除以总数据条数
select
university,
difficult_level,
count(q.question_id) / count(distinct u.device_id) as avg_answer_cnt
from user_profile as u
inner join question_practice_detail as q
on u.device_id = q.device_id
inner join question_detail as d
on q.question_id = d.question_id
group by university, difficult_level;