得到所有连续的行相差一定值？

小编典典

得到所有连续的行相差一定值？

sql

我正在努力做到这一点，因为它涉及比较连续的行。我正在尝试将相差一定数值的值进行分组。例如，让我们说我有这个表：

CREATE TABLE #TEMP (A int, B int)

-- Sample table
INSERT INTO #TEMP VALUES 
(3,1), 
(3,2), 
(3,3),
(3,4),
(5,1),
(6,1),
(7,2),
(8,3),
(8,4),
(8,5),
(8,6)

SELECT * FROM #TEMP

DROP TABLE #TEMP

让我们说我必须对所有相差1且具有相同A值的值进行分组。然后，我试图获得如下输出：

A B GroupNo
3 1 1
3 2 1
3 3 1
3 4 1
5 1 2
6 1 3
7 2 4
8 3 5
8 4 5
8 5 5
8 6 5

(3,1) (3,2) (3,3) (3,4)并且(8,3) (8,4) (8,5) (8,6)由于它们的值相差1而被归入同一组。我将首先展示我的尝试：

CREATE TABLE #TEMP (A int, B int)

-- Sample table
INSERT INTO #TEMP VALUES 
(3,1), (3,2), (3,3), (3,4), (5,1), (6,1), (7,2),
(8,3), (8,4), (8,5), (8,6)

-- Assign row numbers and perform a left join
-- so that we can compare consecutive rows
SELECT ROW_NUMBER() OVER (ORDER BY A ASC) ID, * 
INTO #TEMP2
FROM #TEMP

;WITH CTE AS
(
    SELECT X.A XA, X.B XB, Y.A YA, Y.B YB
    FROM #TEMP2 X
    LEFT JOIN #TEMP2 Y
    ON X.ID = Y.ID - 1
    WHERE X.A = Y.A AND
    X.B = Y.B - 1
)
SELECT XA, XB
INTO #GROUPS
FROM CTE
UNION 
SELECT YA, YB
FROM CTE
ORDER BY XA ASC

-- Finally assign group numbers
SELECT X.XA, X.XB, Y.GID
FROM #GROUPS X
INNER JOIN
(SELECT XA, ROW_NUMBER() OVER (ORDER BY XA ASC) GID
    FROM #GROUPS Y
    GROUP BY XA
) Y
ON X.XA = Y.XA

DROP TABLE #TEMP
DROP TABLE #TEMP2
DROP TABLE #GROUPS

我将在一个大表（大约3000万行）上执行此操作，因此我希望有一个更好的方法可以对任意值执行此操作（例如，不仅可以相差1，还可以是2或3，稍后合并到程序中）。关于我的方法是否没有错误以及是否可以改进的任何建议？

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2021-04-22

共1个答案

小编典典

declare @Diff int = 1

;with C as
(
  select A, 
         B,
         row_number() over(partition by A order by B) as rn
  from #TEMP
),
R as
(
  select C.A,
         C.B,
         1 as G,
         C.rn
  from C
  where C.rn = 1
  union all
  select C.A,
         C.B,
         G + case when C.B-R.B <= @Diff 
               then 0
               else 1
             end,
         C.rn
  from C
    inner join R
       on R.rn + 1 = C.rn and
          R.A = C.A       
)
select A,
       B,
       dense_rank() over(order by A, G) as G
from R
order by A, G

2021-04-22