基于集合的方法来删除包含的点

Question

我有这些数据：

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)

我想删除其他点中包含的点，例如：

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

获得定义外部多边形的最外部点，该外部多边形仅包括没有冗余的垂直和水平线（例如（0,1）是冗余点）。这可以通过SQL Server 2014中基于集合的TSQL方法实现吗？

PS：

数据的散点图如下：

我想删除包围点。最终，我在外边界（绘制为红线）之后。希望这更清楚。

Answer 1

我相信这可行。它似乎可以提供您的测试数据。 有点粗糙。如果您的实际数据很大，可能会提前计算一些SELECT MIN和SELECT MAX。

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)  )
)