题解 | #统计每个学校各难度的用户平均刷题数#

# 计算每个学校用户不同难度下的用户平均答题题目数情况

select 
    university,
    difficult_level,
    count(q.question_id)/count(distinct u.device_id)
from question_practice_detail q
join user_profile u on u.device_id = q.device_id
join question_detail qd on q.question_id = qd.question_id
group by university, difficult_level

;