题解 | #截至当月的练题情况#

截至当月的练题情况

https://www.nowcoder.com/practice/80ed62a097564fad880b967678e460fc

1.首先找到每个设备id每个月的练题情况

select device_id, date_format(event_date, '%Y-%m') as ym, count(*) counts
            from question_practice_detail
            group by device_id, date_format(event_date, '%Y-%m')                

​ 2.使用窗口函数分别求出总练题数,三个月总共的练题数和月份数量,每个人的练题总数

select device_id,
             ym,
             sum(counts) over (partition by device_id order by ym)                   sum_cnt,
             sum(counts) over (order by ym)                                          total_cnt,
             sum(counts) over (partition by device_id order by ym ROWS 2 preceding ) avg3_,
             count(*) over (partition by device_id order by ym ROWS 2 preceding )    avg3_count


#        sum(counts) over (partition by device_id order by ym ),
      from (select device_id, date_format(event_date, '%Y-%m') as ym, count(*) counts
            from question_practice_detail
            group by device_id, date_format(event_date, '%Y-%m')) diyc

​ 3.在最外层计算最近三个练题月的月平均练题数

select device_id,
       ym,
       sum_cnt,
       avg3_ / avg3_count as avg3_cnt,
       total_cnt
from (select device_id,
             ym,
             sum(counts) over (partition by device_id order by ym)                   sum_cnt,
             sum(counts) over (order by ym)                                          total_cnt,
             sum(counts) over (partition by device_id order by ym ROWS 2 preceding ) avg3_,
             count(*) over (partition by device_id order by ym ROWS 2 preceding )    avg3_count


#        sum(counts) over (partition by device_id order by ym ),
      from (select device_id, date_format(event_date, '%Y-%m') as ym, count(*) counts
            from question_practice_detail
            group by device_id, date_format(event_date, '%Y-%m')) diyc) d;

全部评论
跟着思路做了变,简单易懂, sum(counts) over (partition by device_id order by ym ROWS 2 preceding ) avg3_, count(*) over (partition by device_id order by ym ROWS 2 preceding ) avg3_count 后来思考了下,其实这里楼主就可以直接用avg()就可以得出结果,无需后面再套一层了。 - avg(counts) over (PARTITION by device_id order by ym rows 2 preceding)
点赞 回复 分享
发布于 2022-07-19 14:57

相关推荐

不愿透露姓名的神秘牛友
06-12 18:53
点赞 评论 收藏
分享
程序员牛肉:主要是因为小厂的资金本来就很吃紧,所以更喜欢有实习经历的同学。来了就能上手。 而大厂因为钱多,实习生一天三四百的就不算事。所以愿意培养你,在面试的时候也就不在乎你有没有实习(除非是同级别大厂的实习。) 按照你的简历来看,同质化太严重了。项目也很烂大街。 要么换项目,要么考研。 你现在选择工作的话,前景不是很好了。
点赞 评论 收藏
分享
评论
6
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务