Question

假设我有这张表：

select * from window_test;

 k | v
---+---
 a | 1
 a | 2
 b | 3
 a | 4

最终我想得到：

 k | min_v | max_v
---+-------+-------
 a | 1     | 2
 b | 3     | 3
 a | 4     | 4

但我会很高兴得到这个（因为我可以使用distinct轻松过滤它）：

 k | min_v | max_v
---+-------+-------
 a | 1     | 2
 a | 1     | 2
 b | 3     | 3
 a | 4     | 4

使用PostgreSQL 9.1+窗口函数可以实现这一点吗？我试图了解我是否可以使用单独的分区来查看此示例中k=a的第一次和最后一次出现（按v排序）。

Answer 1

这将使用样本数据返回您想要的结果。不确定它是否适用于现实世界的数据：

select k, 
       min(v) over (partition by group_nr) as min_v,
       max(v) over (partition by group_nr) as max_v
from (
    select *,
           sum(group_flag) over (order by v,k) as group_nr
    from (
    select *,
           case
              when lag(k) over (order by v) = k then null
              else 1
            end as group_flag
    from window_test
    ) t1
) t2
order by min_v;

我遗漏了DISTINCT。

Answer 2

编辑：我想出了以下查询 - 根本没有窗口函数：

WITH RECURSIVE tree AS (
  SELECT k, v, ''::text as next_k, 0 as next_v, 0 AS level FROM window_test
  UNION ALL
  SELECT c.k, c.v, t.k, t.v + level, t.level + 1
    FROM tree t JOIN window_test c ON c.k = t.k AND c.v + 1 = t.v),
partitions AS (
  SELECT t.k, t.v, t.next_k,
         coalesce(nullif(t.next_v, 0), t.v) AS next_v, t.level
    FROM tree t
   WHERE NOT EXISTS (SELECT 1 FROM tree WHERE next_k = t.k AND next_v = t.v))
SELECT min(k) AS k, v AS min_v, max(next_v) AS max_v
  FROM partitions p
 GROUP BY v
 ORDER BY 2;

我现在提供了两个有效的查询，我希望其中一个可以帮助你。

对于此变体，

SQL Fiddle。

如何实现这一目标的另一种方法是使用支持序列。

创建支持序列：
```
CREATE SEQUENCE wt_rank START WITH 1;
```

查询：

WITH source AS (
  SELECT k, v,
         coalesce(lag(k) OVER (ORDER BY v), k) AS prev_k
    FROM window_test
    CROSS JOIN (SELECT setval('wt_rank', 1)) AS ri),
ranking AS (
  SELECT k, v, prev_k,
         CASE WHEN k = prev_k THEN currval('wt_rank')
              ELSE nextval('wt_rank') END AS rank
    FROM source)
SELECT r.k, min(s.v) AS min_v, max(s.v) AS max_v
    FROM ranking r
    JOIN source s ON r.v = s.v
   GROUP BY r.rank, r.k
   ORDER BY 2;

Answer 3

这不会为您完成工作，无需窗口，分区或合并。它只是使用传统的SQL技巧通过自联接找到最近的元组，并在差异上找到最小值：

SELECT k, min(v), max(v) FROM (
    SELECT k, v, v + min(d) lim FROM (
        SELECT x.*, y.k n, y.v - x.v d FROM window_test x
        LEFT JOIN window_test y ON x.k <> y.k AND y.v - x.v > 0) 
    z GROUP BY k, v, n)
w GROUP BY k, lim ORDER BY 2;

我认为这可能是一个更关键的问题。解决方案，但我不确定它的效率。

窗口函数和更多“本地”聚合

3 个答案: