按时间序列填充空数据答案

【问题标题】：filling the null data by time series按时间序列填充空数据
【发布时间】：2018-07-24 02:22:31
【问题描述】：

我有一张这样的桌子

ts                    item          infoA         infoB         
2018-02-03 12:00:00   A             null          null          
2018-02-03 12:01:00   null          A1            null          
2018-02-03 12:02:00   A             null          null          
2018-02-03 12:03:00   null          null          null          
2018-02-03 12:04:00   null          A2            null           
2018-02-03 12:05:00   null          null          null         
2018-02-03 12:06:00   B             null          null         
2018-02-03 12:07:00   null          null          B1         
2018-02-03 12:08:00   null          null          null

我只想在相关项目上用时间序列填充空数据

ts                    item          infoA         infoB         
2018-02-03 12:00:00   A             null          null          
2018-02-03 12:01:00   A             A1            null          
2018-02-03 12:02:00   A             A1            null          
2018-02-03 12:03:00   A             A1            null          
2018-02-03 12:04:00   A             A2            null           
2018-02-03 12:05:00   A             A2            null         
2018-02-03 12:06:00   B             null          null         
2018-02-03 12:07:00   B             null          B1         
2018-02-03 12:08:00   B             null          B1

我发现了一个 AGGREGATE 函数 GapFill() 从 this 使用该功能，我可以从

select t1.ts, t1.item, t2.infoA, t3.infoB 
from 
(select ts,gapfill(item) OVER (ORDER BY ts)) t1 
LEFT JOIN (select ts,gapfill(infoA) OVER (ORDER BY ts) as infoA) on (t1.ts = t2.ts and t1.item='A') t2 
LEFT JOIN (select ts,gapfill(infoB) OVER (ORDER BY ts) as infoB) on (t1.ts = t3.ts and t1.item='B') t3

如果我有很多列，如何简化查询。

【问题讨论】：

标签： sql postgresql

【解决方案1】：

您想要的是lag() 上的ignore nulls 选项。但是 Postgres 还不支持它（目前）。

也许最简单的方法是关联子查询：

select t.ts,
       coalesce(item,
                (select t2.item
                 from t t2
                 where t2.ts < t.ts and t2.item is not null
                 order by t2.ts desc
                 fetch first 1 row only
                )
               ) as item,
       coalesce(itemA,
                (select t2.itemA
                 from t t2
                 where t2.ts < t.ts and t2.itemA is not null
                 order by t2.ts desc
                 fetch first 1 row only
                )
               ) as itemA,
       coalesce(itemB,
                (select t2.itemB
                 from t t2
                 where t2.ts < t.ts and t2.itemB is not null
                 order by t2.ts desc
                 fetch first 1 row only
                )
               ) as itemB
from t;

如果您知道这些值是单调递增或递减的，您可以使用max() 或min()。

使用窗口函数的另一种方法使用相同的想法。通过累积计数来识别具有相同值的行组。然后将值分布在行上：

select t.ts,
           max(item) over (partition by grp_item) as item,
           max(itemA) over (partition by grp_item) as itemA,
           max(itemB) over (partition by grp_item) as itemB
from (select t.*,
                    count(item) over (order by ts) as grp_item,
                    count(itemA) over (order by ts) as grp_itemA,
                    count(itemB) over (order by ts) as grp_itemB
         from t
       ) t;

【讨论】：