题解 | #统计每个学校各难度的用户平均刷题数#

select
    university, 
    difficult_level,
    round(count(question_practice_detail.question_id) / count(distinct(question_practice_detail.device_id)), 4) as 'avg_answer_cnt'
    
from 
    user_profile
inner join
    question_practice_detail
on 
    user_profile.device_id = question_practice_detail.device_id
inner join
    question_detail
on
    question_detail.question_id = question_practice_detail.question_id  
group by
    university, difficult_level

思路：

1、先分组，按照学校、不同难度

2、三表内联， 没有null的情况

3、求 用户平均答题量 = 答题总数 （question_practice_detail.question_id）/ 答题人数 （question_practice_detail.device_id 去重一下）

round(count(question_practice_detail.question_id) / count(distinct(question_practice_detail.device_id)), 4)