如何在 SQL 中获取最新的非负值？

Question

我有以下数据集。

date	orig	dest	value
11-4	S8	B9	-42
10-30	S8	B9	-2999
10-15	S8	B9	959
10-14	S8	B9	480
10-13	S8	B9	0
10-9	S8	B9	503

我怎样才能把它变成这样的东西？

date	orig	dest	value	new value
11-4	S8	B9	-42	959
10-30	S8	B9	-2999	959
10-15	S8	B9	959	959
10-14	S8	B9	480	480
10-13	S8	B9	0	503
10-9	S8	B9	503	503

这是我用过的，不正确：

CASE WHEN value <= 0
     THEN last_value(value) over (PARTITION BY orig, dest ORDER BY date DESC ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
     ELSE value 
END as new_value

这里的目标是不要用最近的最大值替换

Answer 1

这应该在 Redshift 中完成：

SELECT date, orig, dest, value
     , max(value) OVER (PARTITION BY orig, dest, grp) AS new_value
FROM  (
   SELECT date, orig, dest, value
        , count(value > 0 OR NULL) OVER (PARTITION BY orig, dest ORDER BY date) AS grp
   FROM   tbl
   ) sub

解释：

• Retrieve last known value for each column of a row

关于count(expression OR NULL)，见：

• For absolute performance, is SUM faster or COUNT?

不幸的是，Redshift 没有跟上对 Postgres 的改进，并且缺少聚合的 FILTER 子句（自 Postgres 9.4 起！），这会更快、更优雅：

SELECT date, orig, dest, value
     , max(value) OVER (PARTITION BY orig, dest, grp) AS new_value
FROM  (
   SELECT date, orig, dest, value
        , count(*) FILTER (WHERE value > 0) OVER (PARTITION BY orig, dest ORDER BY date) AS grp
   FROM   tbl
   ) sub

见：

• Get max value from a window of rows as new column for all rows
• Aggregate columns with additional (distinct) filters