这是一个丑陋的黑客行为（基于滥用的CTE；）。关键是
value1 * distance2 + value2 * distance1
可以通过除以distance1 * distance2来重写为
value1/distance1 + value2/distance2
因此，产品（或部门）可以留在其行内。求和后，乘以（distance1 * distance2）可将结果重新缩放为所需的输出。读者可以将对两个以上邻居的归纳作为练习。
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 有点 干净…关联仍未处理（为此需要外部函数中的窗口函数或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
;
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