Calculating Median In SQL

Aug 25, 2004

Hi All,
I have a table that of server names and their execution times that run in to hundreds of thousands of records. What i need is some SQL that gives me the median execution times for each of these different servers. At the moment i have some SQL that only gives me the median for all the records in the table not the median execution time for every different server name. For example my tables looks something like this;

ServerName | ExecTime
-----------------------
server1 | 0.07
server2 | 0.17
server1 | 0.27
server1 | 0.37
server2 | 0.47
server1 | 0.57
server1 | 0.67
server2 | 0.77

My SQL below gives me

ServerName | ExecTime
-----------------------
server1 | 0.37

Where as i want

ServerName | ExecTime
-----------------------
server1 | 0.37
server2 | 0.47

Here is my SQL, hope someone can modify it and thanks in advance.


Code:


SELECT DISTINCT instance, exec_time AS median
FROM (SELECT instance, exec_time
FROM (SELECT TOP 1 exec_time = exec_time * 1.0, instance
FROM (SELECT TOP 50 PERCENT exec_time, instance
FROM llserverlogs
ORDER BY exec_time) sub_a
ORDER BY 1 DESC) sub_1
UNION ALL
SELECT instance, exec_time
FROM (SELECT TOP 1 exec_time = exec_time * 1.0, instance
FROM (SELECT TOP 50 PERCENT exec_time, instance
FROM llserverlogs
ORDER BY exec_time DESC) sub_b
ORDER BY 1) sub_2)

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Calculating Median Value From Measures And Dimensions

May 24, 2007

I am facing some problem in calculating Median

I am trying to calculate the median value using one of the measures and a dimension value.



Time is a measure in my cube and OpId is one of the dimensions.The result is as follows:



opid time median

1 55

2 23

3 23

Total 23



The Time here for Op Id 1 is the aggregation for all the rows whose OpId is 1.I want the median of the values whose OpId is 1 which is not showing at the moment.



What I am getting here is the median for all of the OpId but what I really want is the median for each of the individual Opid's as well.



I am using a calculated field Median with the following expression.



MEDIAN

( [Dim Operation].[Dim Operation].currentmember.children ,[Measures].[Elapsed Time])



Thanks

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Calculating Median Values On Column In A Table

Feb 19, 2012

I need to calculate a median on a column in a table. The code I have is:

Code:
Select gender,
CASE
when gender = 'F' then 'Female'
when gender = 'M' then 'Male'
else 'Unknown'
end as test,
datediff(day, [admit_date], getdate()) as 'datediffcal',
from [tbl_record]
How do I calculate the median on the datediffcal column?

It doesn't matter if the resultset only shows the median result. So if the output shows:

median
15

that's fine. Minimally, I need the median value.

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Aug 13, 2015

I have a database with 1million+ records in and i'm trying to collect the median values of column(2) for all distinct values in column (1)

Example DB:

Column 1 Column 2
978555 500
978555 502
978555 480
978555 490
978324 1111
978324 1102
978311 122
978311 120
978994 804
978320 359

and I need it to display on SELECT as

column 1 column 2
978555 495
978324 1106
978311 121
978994 804
978320 359

Is this possible on 2008 R2?

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Analysis :: Calculating A Rolling Median Over A Period Of 3 Years?

Jun 17, 2015

calculating a rolling median over a period of 3 years.

I already calculate median and I've tried to calculate rolling median over a period of 3 years as below.

 MEDIAN([Date].[Year].CurrentMember.Lag(3):[Date].[Year].CurrentMember,[Measures].[median])

What this does is, it calculates the median of the medians over the period of 3 years. But, what I'm looking for is the overall median of the underlying measure over a period of 3 years.

What I have now:

Year1 - 41,52,73;  Median1 - 52
Year2 - 6,9,12;  Median2- 9
Year3 - 24,68,89; Median3 - 68
Overall Median of 9,52,68 - 52

What I need:

Year1 - 41,52,73;  Median1 - 52
Year2 - 6,9,12;  Median2- 9
Year3 - 24,68,89; Median3 - 68

Overall Median of 41,52,73,6,9,12,24,68,89 is 41 

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Dec 7, 2002

I recently had to use my own little median technique again on a report here at work, and had posted it before, but wasn't sure if anyone had seen it. I have read Celko's and others techniques for generating a median and haven't seen one more efficent.

Does anyone have a better way they can think of? I think this bad boy is pretty short & efficient.

First, if you want to return the middle number or the higher one next to the middle if there is an even number:

SELECT x.Value AS median
FROM Vals x
CROSS JOIN Vals y
GROUP BY x.Value
HAVING SUM(SIGN(x.Value-y.Value)) IN (1,0)

Change the " IN (1,0)" to "IN (-1,0)" to get the lower value if there is an even # of values.

Basically, we are saying compare each number to all possible numbers, and add up values of 1,0 or -1 depending if the first number is less, equal or higher than the second. The number that returns 0 is right in the middle ... If there is no middle, a -1 or 1 is returned. There will never be a 0 and (-1 or 1) at the same time returned.

To get the financial median (avg of the 2 values middle values if there is an even number), you need to encapsulate the results of the above into a subquery, allow for not just (-1,0) but all three (-1,0,1) and then take the AVG of the values returned.

That is,

SELECT Avg(Median) as Median FROM
(
SELECT x.Value AS median
FROM Vals x
CROSS JOIN Vals y
GROUP BY x.Value
HAVING SUM(SIGN(x.Value-y.Value)) IN (1,0,-1)
) A

If there is an even number of values, the lower and higher middle ones are averaged. If there is an odd number, only the middle value is returned and averaged (which of course has no effect).

Most other techniques used several COUNT(*) subqueries which this one avoids.

Critique and enjoy!

- Jeff

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Sep 27, 2007

I have a table that contains the following:
customer (ID of customers)
product (Product description numeric value)
UOM (unit of measure like each or pak)
avgprice (Avg proce that this product and uom was sold)


I need to find the median value for a product, uom. Then I need create a table that shows product,uom,avgprice,median grouped by product and uom
I have been at this for two days with no luck

Thanks in advance for any help




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Gack!

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Mar 12, 2007

Hi,

I have looked all over the web to try to find some very basic / simple explanations of how to get a median value from a group of records in a table but with no luck

the problem i am having is that all the information i find is always centered around getting a median using every single row in the table. except i have groups of data in the table and want to work out a median for each group. the group is identified by 4 different columns (the 5th column is what i want to get the median on but for each group not the entire table) and i want to produce a resulting table that has 1 row for each group and therefore contains the median value for the group instead of the individual numbers that it currently has. e.g. the current table is like this

column1 column2 column3 column 4 column5(median of this)

value 1 value 2 value 3 value 4 1.2
value 1 value 2 value 3 value 4 1.0
value 1 value 2 value 3 value 4 1.5
value 2 value 3 value 4 value 5 0.2
value 2 value 3 value 4 value 5 0.4
etc...

and i need a query to get the results to show like this

column1 column2 column3 column 4 column5

value 1 value 2 value 3 value 4 1.0
value 2 value 3 value 4 value 5 0.3
etc...

This is driving me crazy and i will be very helpful if anyone can help

the statement i need to add it to is:

select pat_demid, pat_lastname, meas_gendate, meas_id, test_gendate, avg(srtot)as meansrtot, avg(sreff)as meansreff, avg(BFRaw)as BF_Rawmean, avg(BFTGV)as BF_TGVmean, count(srtot)as countsrtot, count(sreff)as countsreff
from bodyparametersf
where (srtot is not null) OR (sreff is not null)
group by pat_demid, pat_lastname, meas_gendate, meas_id, test_gendate

thanks very much

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------------------------------------------------------------------------
SELECT INS.Code As [code],INS.FinYr as [YEAR], ' ' As [FIRE BUSINESS], (INSRev.FI_NetPremLessIns / INSRev.FI_GrPremium) * 100 As [Rention Ratio],
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FROM
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----------------------------------------------------------------------

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WITH CompMedian AS

(

SELECT SoldDate, SoldPrice, ROW_NUMBER() OVER(PARTITION BY Convert(Varchar(5),Year(SoldDate)) + Convert(Varchar(5), RIGHT(CAST(100+DATEPART(QQ,SoldDate) AS CHAR(3)), 2)) ORDER BY SoldPrice) AS RowNum, COUNT(*) OVER(PARTITION BY Convert(Varchar(5),Year(SoldDate)) + Convert(Varchar(5), RIGHT(CAST(100+DATEPART(QQ,SoldDate) AS CHAR(3)), 2))) AS Cnt FROM tbl_Orders
WHERE Status = 'Sold'

AND SoldDate >= DATEADD(Year, -5, Convert(DateTime, Convert(Varchar(5),Month(GetDate())) + Convert(Varchar(5), '/1/') + Convert(Varchar(5), YEAR(GetDate()))))

AND SoldDate < DATEADD(Year, 0, Convert(DateTime, Convert(Varchar(5),Month(GetDate())) + Convert(Varchar(5), '/1/') + Convert(Varchar(5), YEAR(GetDate()))))
)

SELECT Convert(Varchar(5),Year(SoldDate)) + Convert(Varchar(5), RIGHT(CAST(100+DATEPART(QQ,SoldDate) AS CHAR(3)), 2)) AS CompDate, AVG(SoldPrice) AS CompMedian

FROM CompMedian

WHERE RowNum IN((Cnt + 1) / 2, (Cnt + 2) / 2)

GROUP BY Convert(Varchar(5),Year(SoldDate)) + Convert(Varchar(5), RIGHT(CAST(100+DATEPART(QQ,SoldDate) AS CHAR(3)), 2))

ORDER BY CompDate;


Now my client would like me to change the query so that each quarter would represent the median for the past 12 months ending with that quarter. I've been looking at this for hours and I'm at a loss. Anyone have any thoughts?

Thanks in advance,
Russ

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How can I add the Median to the report?

 

Thanks in advance,

Greg

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Mar 4, 2008



I used the Median() function in a calculated measure in a cube. All seemed well until one of my users, a statistician, pointed out that the displayed values were incorrect. I investigated this and finally built the simplest cube with three records and the displayed median does not make sense, i.e. equal what a median value should be.

The 3 records have loan amounts as a measure with these values: $102,500, $168,400, and $172,181 and loan number keys of 1, 2, and 3. That's it.

The median should be $168,400 since the number of items is odd and that is the middle value.

But the cube calculated measure displays $170,290.50. It appears to be taking the average of the middle row and the next value.

When I increase the number of records I get the same odd behavior but it not only alternates (of course) with odd and even (since they use different formulats for odd and even) but the odd wrong results alternate within themselves. The apparent calculations for 5 or 7 records are different compared to 3, 9, 11, 13, 15, 17 records. The first set seems to be calculating odd number of items median by averaging the middle value + the value BEFORE the middle value while the second set of odd rows (3, 9, etc) seem to be calculating median by averaging the middle value + the value AFTER the middel value. The even numbers result in the larger of the two middle values being selected instead of the average (financial median) or the lower number (as one poster claimed for statistical median).

My calculated measure MDX is very simple:


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Ideas anyone? Am I missing something? Is this a bug?

Thanks!

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Oct 13, 2006

Hello!

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3         900

I want to calculate the median for each mtid in SSAS . (median for mtid 1=400 )

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Dim values As New System.Collections.ArrayList
Function AddValue(newValue As Decimal)
If values is Nothing Then
values = New System.Collections.ArrayList
End If
values.Add(newValue)

[code]....

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[Code] .....

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============================
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I attach also a sample of the view

==============
ALTER PROCEDURE [dbo].[sp_ED_Measures]
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Skills
SL_ID

1
4
Skill 1
3

2
5
Skill 2
2

2
6
Skill 3
2

3
7
Skill 4
4

3
8
Skill 5
4

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Skill 6
NULL

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Skill 7
NULL

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11
Skill 8
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6
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Skill 9
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7
13
Skill 10
4

8
14
Skill 11
1

9
15
Skill 12
3

9
16
Skill 13
3

10
17
Skill 14
6

11
18
Skill 15
1
Skills Catagory








CATID
Org
Skillcat

1
Org1
Skill Category 1

2
Org2
Skill Category 2

3
Org3
Skill Category 3

4
Org4
Skill Category 4

5
Org5
Skill Category 5

6
Org6
Skill Category 6

7
Org7
Skill Category 7

8
Org8
Skill Category 8

9
Org9
Skill Category 9
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SL_ID
Service_line

1
Service Line 1

2
Service Line 2

3
Service Line 3

4
Service Line 4

5
Service Line 5

6
Service Line 6

7
Service Line 7
 
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