题解 | 统计每个用户的平均刷题数

select
    university,
    difficult_level,
    count(t2.question_id) / count(distinct t2.device_id) as avg_answer_cnt
from 
    question_practice_detail t2
left join
    user_profile t1
on
    t1.device_id = t2.device_id
left join
    question_detail t3
on 
    t2.question_id = t3.question_id

where
    t1.university = '山东大学'
group by t3.difficult_level

count(t2.question_id) / count(distinct t2.device_id) as avg_answer_cnt

注：用户在不同难度下的平均答题题目数，同一个难度的总答题数 / 去重（distinct）的人数。