题解 | #统计每个学校各难度的用户平均刷题数#

select
u.university,
l.difficult_level,
round(count(q.question_id)/count(distinct(q.device_id)), 4) avg_answer_cnt
from user_profile u
join question_practice_detail q on q.device_id = u.device_id
join question_detail l on l.question_id = q.question_id
group by u.university, l.difficult_level;
group by 不要错过