Question

我想通过插值两个最近邻居之间的值来计算一个值。我有一个子查询，返回邻居的值及其相对距离，以两列两个元素的形式。

让我们说：

(select ... as value, ... as distance 
 from [get some neighbours by distance] limit 2) as sub

如何通过线性插值计算点的值？是否可以在单个查询中执行此操作？

示例：我的点在距离1处有一个值为10的邻居A，在距离4处有一个值为20的邻居B.该函数应为我的点返回值10 * 4 + 20 * 1 / 5 = 12

我尝试了明显的方法

select sum(value * (sum(distance)-distance)) / sum(distance)

将失败，因为您无法使用group子句中的group子句。使用另一个子查询返回总和也是不可能的，因为那样我就无法同时转发各个值。

Answer 1

如果你真的想要介于point之间，那么有一种内置的方式（但不是聚合函数）：

SELECT center(box(x.mypoint,y.mypoint))
FROM   ([get some neighbours by distance] order by value limit 1) x
      ,([get some neighbours by distance] order by value offset 1 limit 1) y;

如果你想要平均距离：

SELECT avg(x.distance)
FROM   ([get some neighbours by distance] order by value limit 2) as x

请参阅手册中的geometrical function和aggregate functions。

编辑：

对于添加的示例，查询可能如下所示：

SELECT (x.value * 4 + y.value) / 5 AS result
FROM   ([get some neighbours by distance] order by value limit 1) x
      ,([get some neighbours by distance] order by value offset 1 limit 1) y;

我添加了遗失()以获得您期望的结果！

或者，我最后一次刺伤它：

SELECT y.x, y.x[1], (y.x[1] * 4 + y.x[2]) / 5 AS result
FROM  (
    SELECT ARRAY(
        SELECT value FROM tbl WHERE [some condition] ORDER BY value LIMIT 2
        ) x
    ) y

如果您提供了完整的查询和表格定义，那么会更容易。

Answer 2

这是一个丑陋的黑客（基于滥用的CTE;）。它的症结在于

value1 * distance2 + value2 * distance1

可以通过除以distance1 * distance2，重写为

value1/distance1 + value2/distance2

因此，产品（或部门）可以留在他们的行内。在求和之后，乘以（distance1 * distance2）将结果重新调整为所需的输出。对两个以上邻居的推广留给读者练习.YMMV

DROP TABLE tmp.points;
CREATE TABLE tmp.points
    ( pname VARCHAR NOT NULL PRIMARY KEY
    , distance INTEGER NOT NULL
    , value INTEGER
    );

INSERT INTO tmp.points(pname, distance, value) VALUES
  ( 'A' , 1, 10 )
,  ( 'B' , 4, 20 )
,  ( 'C' , 10 , 1)
,  ( 'D' , 11 , 2)
  ;
WITH RECURSIVE twin AS (
    select 1::INTEGER AS zrank
    , p0.pname AS zname
    , p0.distance AS dist
    , p0.value AS val
    , p0.distance* p0.value AS prod
    , p0.value::float / p0.distance AS frac
    FROM tmp.points p0
    WHERE NOT EXISTS ( SELECT * FROM tmp.points px
        WHERE px.distance < p0.distance)
    UNION
    select 1+twin.zrank AS zrank
    , p1.pname AS zname
    , p1.distance AS dist
    , p1.value AS val
    , p1.distance* p1.value AS prod
    , p1.value::float / p1.distance AS frac
    FROM tmp.points p1, twin
    WHERE p1.distance > twin.dist
    AND NOT EXISTS ( SELECT * FROM tmp.points px
        WHERE px.distance > twin.dist
        AND px.distance < p1.distance
        )   
    )
-- SELECT * from twin ;
SELECT min(zname) AS name1, max(zname) AS name2
    , MIN(dist) * max(dist) *SUM(frac) / SUM(dist) AS score
    FROM twin
    WHERE zrank <=2
    ;

结果：

CREATE TABLE
INSERT 0 4
 name1 | name2 | score 
-------+-------+-------
 A     | B     |    12

更新：这个有点更干净...仍未处理关系（需要在外部查询中使用窗口函数或LIMIT 1子句）

WITH RECURSIVE twin AS (
    select 1::INTEGER AS zrank
    , p0.pname AS name1
    , p0.pname AS name2
    , p0.distance AS dist
    FROM tmp.points p0
    WHERE NOT EXISTS ( SELECT * FROM tmp.points px
        WHERE px.distance < p0.distance)
    UNION
    select 1+twin.zrank AS zrank
    , twin.name1 AS name1
    , p1.pname AS name2
    , p1.distance AS dist
    FROM tmp.points p1, twin
    WHERE p1.distance > twin.dist
    AND NOT EXISTS ( SELECT * FROM tmp.points px
        WHERE px.distance > twin.dist
        AND px.distance < p1.distance
        )
    )
SELECT twin.name1, twin.name2
    , (p1.distance * p2.value + p2.distance * p1.value) / (p1.distance+p2.distance)  AS score
    FROM twin
    JOIN tmp.points p1 ON (p1.pname = twin.name1)
    JOIN tmp.points p2 ON (p2.pname = twin.name2)
    WHERE twin.zrank =2
    ;

单个查询中两个值之间的插值

2 个答案:

编辑：