基于集合的方法来删除包含的点

Question

我有这个数据：

IF OBJECT_ID('tempdb..#temp') IS NOT NULL
    DROP TABLE #temp
CREATE TABLE #temp
    (
      Id INT IDENTITY(1, 1) ,
      X FLOAT NOT NULL ,
      Y FLOAT NOT NULL
    )

INSERT INTO #temp (X, Y) VALUES (0, 0)
INSERT INTO #temp (X, Y) VALUES (0, 1)
INSERT INTO #temp (X, Y) VALUES (0, 2)
INSERT INTO #temp (X, Y) VALUES (0.5, 1)
INSERT INTO #temp (X, Y) VALUES (1, 1)
INSERT INTO #temp (X, Y) VALUES (1, 2)
INSERT INTO #temp (X, Y) VALUES (1.5, 0.5)
INSERT INTO #temp (X, Y) VALUES (2, 0)
INSERT INTO #temp (X, Y) VALUES (2, 1)

我想删除包含在其他点中的点，例如：

(0, 1)
(1, 1)
(1.5, 0.5)

获取定义仅由垂直线和水平线组成的外部多边形的最外点，没有冗余（例如（0, 1）是冗余点）。这可以通过 SQL Server 2014 中基于集合的 TSQL 方法来实现吗？

PS:

数据散点图如下：

我想删除圈出的点。最终，我在外边界之后（绘制为红线）。希望这能让它更清楚。

Answer 1

我相信这可能有效。它似乎可以提供您的测试数据。有点粗糙。如果你的实际数据很大，一些 SELECT MIN 和 SELECT MAX 可能可以提前计算。

SELECT * 
-- uncomment this to delete the desired points
-- DELETE #temp
FROM #temp t
WHERE 
(
    -- Internal points
    (
            ( X > (SELECT MIN(X) FROM #temp) AND X < (SELECT MAX(X) FROM #temp) )
        AND ( Y > (SELECT MIN(Y) FROM #temp) AND Y < (SELECT MAX(Y) FROM #temp) )
    )
    -- Exceptions (points with nothing strictly outside them) [Don't want to lose (1,1)]
    AND EXISTS (SELECT * FROM #temp WHERE X > t.X AND Y > t.Y)
)
OR
    -- redundant edge points [(0,1) would be included as an "exception"]
(
    ( (t.X = (SELECT MIN(X) FROM #temp) OR t.X = (SELECT MAX(X) FROM #temp)) 
        AND EXISTS (SELECT * FROM #temp WHERE X = t.X AND Y > t.Y) 
        AND EXISTS (SELECT * FROM #temp WHERE X = t.X AND Y < t.Y)  )
    OR
    ( (t.Y = (SELECT MIN(Y) FROM #temp) OR t.Y = (SELECT MAX(Y) FROM #temp)) 
        AND EXISTS (SELECT * FROM #temp WHERE Y = t.Y AND X > t.X) 
        AND EXISTS (SELECT * FROM #temp WHERE Y = t.Y AND X < t.X)  )
)

基于集合的方法来删除包含的点

set based approach to remove contained points

tsql

sql-server

sql-server-2014