Aggregation with every N number of records grouped

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Aggregation with every N number of records grouped

sibir1us

SSC Eights!

Points: 828
More actions
December 10, 2009 at 3:35 am

#135815

I would like to generate an output which shows me the average value of Col2 for every 5 rows and the starting number of the 5 row set (i.e. Col1).
For an example, the output of my query should be 3 rows as follows:
StartingNumber | AvgValue
1 | 9.2
5 | 186.6
10 | 284.25
If there is no 5th row for the last set, then just average the ones that are left (in this case 4 records).
I have a table with the following model and data:
--===== If the test table already exists, drop it
IF OBJECT_ID('TempDB..#Table_1','U') IS NOT NULL
DROP TABLE #Table_1
--===== Create the test table with
CREATE TABLE #Table_1
(
[Col1] [int] NULL,
[Col2] [int] NULL
)
--===== Insert the test data into the test table
INSERT INTO #Table_1
(Col1, Col2)
SELECT '1','1' UNION ALL
SELECT '2','5' UNION ALL
SELECT '3','2' UNION ALL
SELECT '4','6' UNION ALL
SELECT '5','32' UNION ALL
SELECT '6','56' UNION ALL
SELECT '7','2' UNION ALL
SELECT '8','798' UNION ALL
SELECT '9','43' UNION ALL
SELECT '10','34' UNION ALL
SELECT '11','563' UNION ALL
SELECT '12','23' UNION ALL
SELECT '13','465' UNION ALL
SELECT '14','86'
Make everything as simple as possible, but not simpler.
Albert Einstein

Viewing 6 posts - 1 through 5 (of 5 total)

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Mark Cowne One Orange Chip Points: 26952 More actions · Answer 1

Maybe this?

SELECT MIN(Col1) AS StartingNumber,

AVG(Col2*1.0) AS AvgValue

FROM dbo.Table_1

GROUP BY (Col1-1) / 5

ORDER BY MIN(Col1)

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sibir1us SSC Eights! Points: 828 More actions · Answer 2

My question was to get the aggregation for every 5 rows of data, no matter how many rows there are in the table. If I have 503 records in my table, I would respectively have 101 records with aggregated data (where the 101st is just an aggregation of the last 2).

/I have to improve on my question posting technique, and I will. /

I just edited the original post to give a better look, according to the posting standards.

Thank you for your reply.

Make everything as simple as possible, but not simpler.
Albert Einstein

Mitesh Oswal SSCrazy Points: 2798 More actions · Answer 3

--===== If the test table already exists, drop it

IF OBJECT_ID('TempDB..#Table_1','U') IS NOT NULL

DROP TABLE #Table_1

--===== Create the test table with

CREATE TABLE #Table_1

(

[Col1] [int] IDENTITY(1,1),

[Col2] [int] NULL

)

declare @i INT

SELECT @i = 1

WHILE(@i<2)

BEGIN

--===== Insert the test data into the test table

INSERT INTO #Table_1

(Col2)

SELECT '1' UNION ALL

SELECT '5' UNION ALL

SELECT '2' UNION ALL

SELECT '6' UNION ALL

SELECT '32' UNION ALL

SELECT '56' UNION ALL

SELECT '2' UNION ALL

SELECT '798' UNION ALL

SELECT '43' UNION ALL

SELECT '34' UNION ALL

SELECT '563' UNION ALL

SELECT '23' UNION ALL

SELECT '465' UNION ALL

SELECT '86'

select @i = @i + 1

END

SELECT

MIN(COL1) AS ID,

CAST( SUM (Col2)/SUM(1.0) AS DECIMAL(18,2))

FROM #Table_1

GROUP BY

(Col1-1) / 5

Regards,
Mitesh OSwal
+918698619998

Pavel Pawlowski Ten Centuries Points: 1244 More actions · Answer 4

you can use NTILE function with argument equals to CEILING of count of records in the table divided by number specifying number of records in the group.

WITH NTileTable AS (

SELECT

NTile(CONVERT(int, (SELECT CEILING(COUNT(1) / 5.0) FROM #Table_1))) OVER(ORDER BY Col1) NTileCol,

Col1,

Col2

FROM #Table_1

)

SELECT

NTileCol,

Min(Col1) As StartCol,

AVG(Col2) AS AVGCol2

FROM NTileTable

GROUP BY NTileCol

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 5

sibir1us (12/10/2009)
My question was to get the aggregation for every 5 rows of data, no matter how many rows there are in the table. If I have 503 records in my table, I would respectively have 101 records with aggregated data (where the 101st is just an aggregation of the last 2).
/I have to improve on my question posting technique, and I will. /
I just edited the original post to give a better look, according to the posting standards.
Thank you for your reply.

This should show you how it's done, and give you a result:

CREATE TABLE #Table_1

(

[Col1] [int] NULL,

[Col2] [int] NULL

)

--===== Insert the test data into the test table

INSERT INTO #Table_1

(Col1, Col2)

SELECT '1','1' UNION ALL

SELECT '2','5' UNION ALL

SELECT '3','2' UNION ALL

SELECT '4','6' UNION ALL

SELECT '5','32' UNION ALL

SELECT '6','56' UNION ALL

SELECT '7','2' UNION ALL

SELECT '8','798' UNION ALL

SELECT '9','43' UNION ALL

SELECT '10','34' UNION ALL

SELECT '11','563' UNION ALL

SELECT '12','23' UNION ALL

SELECT '13','465' UNION ALL

SELECT '14','86'

SELECT MIN(Col1), SUM(Col2), COUNT(*)

FROM (

SELECT *, CAST((Col1-1)/5 AS INT) AS [Rank]

FROM #Table_1

) d

GROUP BY [Rank]

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

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