从连续行中选择当前和上一个值

Question

我有一个结果集，需要从中提取输出。 结果集如下：

+----------+-------------+----------------+-----------+
| Issue_id | create_date | transtimestamp | Npriority |
+----------+-------------+----------------+-----------+
| 556365   | 2020-02-02  | 1580628330     | 2         |
| 556365   | 2020-02-02  | 1580629317     | 0         |
| 773656   | 2020-02-02  | 1580608104     | 1         |
| 773656   | 2020-02-02  | 1580609625     | 0         |
| 111252   | 2020-02-02  | 1580622323     | 1         |
| 111252   | 2020-02-02  | 1580623399     | 2         |
+----------+-------------+----------------+-----------+

我想从上面生成一个输出：

+----------+-------------+------------+-----------+
| Issue_id | create_date | prev_value | cur_value |
+----------+-------------+------------+-----------+
| 556365   | 2020-02-02  | 2          | 0         |
| 773656   | 2020-02-02  | 1          | 0         |
| 111252   | 2020-02-02  | 1          | 2         |
+----------+-------------+------------+-----------+

我正在使用以下内容，但无法获得所需的结果集：

select  Issue_id,Npriority,lag(Npriority)   over (partition by Issue_id order by trantimestamp) as prev_value FROM (select * from results)

Answer 1

BigQuery 可以通过聚合来做到这一点：

select Issue_id, create_date,
       array_agg(Npriority order by transtimestamp desc limit 1)[ordinal(1)] as latest_value,
       array_agg(Npriority order by transtimestamp desc limit 2)[ordinal(2)] as prev_value
from results r
group by Issue_id create_date;

Answer 2

你可以做一个简单的自连接

select a.Issue_id, a.create_date, b.cur_value as prev_value, a.cur_value
from results a inner join results b on a.Issue_id = b.Issue_id 
and a.transtimestamp   > b.transtimestamp  

Answer 3

你走在正确的轨道上。

with data as (
  select 
  *
  , lag(Npriority,1) over (partition by Issue_id order by transtimestamp asc) as prev_Npriority -- Get previous
  , row_number() over (partition by Issue_id order by transtimestamp desc) as rn -- order by most recent
 from `project.dataset.table`
)
select Issue_id, create_date, prev_Npriority as prev_value, Npriority as cur_value
from data 
where rn = 1 -- filter to most recent