Calculate Median!!!

May 29, 2007

I need to calculate Median on each calculated result from the query below. There is one Median function available in SQL2K but it is not working. Can anyone help me in this regard.

------------------------------------------------------------------------
SELECT INS.Code As [code],INS.FinYr as [YEAR], ' ' As [FIRE BUSINESS], (INSRev.FI_NetPremLessIns / INSRev.FI_GrPremium) * 100 As [Rention Ratio],
(INSRev.FI_NetClaimPaid/INSRev.FI_AdjNetPremium)*100 As [Claim Ratio], ((INSRev.FI_AgencyCommPaid+INSRev.FI_ReInsCommPaid+INSRev.FI_MgmtExpenses+INSRev.FI_OthExpenses)/INSRev.FI_AdjNetPremium)*100 As [Expense Ratio],
((INSRev.FI_NetClaimPaid/INSRev.FI_AdjNetPremium)*100)+(((INSRev.FI_AgencyCommPaid+INSRev.FI_ReInsCommPaid+INSRev.FI_MgmtExpe nses+INSRev.FI_OthExpenses)/INSRev.FI_AdjNetPremium)*100) As [Combine Ratio],
(INSRev.FI_ClosingBal/INSRev.FI_NetClaimPaid) As [Unexpired Risk Reserve to Net Claim(x)],(INSRev.FI_MgmtExpenses/INSRev.FI_AdjNetPremium)*100 As [Management Expenses to Adj. Net Premium],
(INSRev.FI_AgencyCommPaid/INSRev.FI_AdjNetPremium)*100 As [Agency Commissioned to Adj. Net Premium]

FROM
(InsuranceGen As INS LEFT JOIN INSURANCEGen As INS1 ON (INS1.FinYr=INS.FinYr-1 AND INS.Code=INS1.Code))
LEFT JOIN INSRevGen as INSRev ON (INS.Code=INSRev.Code AND INS.FinYr=INSRev.FinYr) WHERE INS.Code IN ('ABC1','ABC2','ABC3') AND INS.FinYr=2005 ORDER BY INS.Code, INS.FinYr
----------------------------------------------------------------------

Thanks

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Calculate Median LTM

Apr 22, 2008

Using SQL Server 2005, I have the following query to calculate the median sales of each quarter over the past 5 years:


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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Calculate Median Value

Dec 15, 2005

I am converting a report created using Crystal Reports 10 to Reporting Services.  The report contains a list of items with dollar values.  The original report displays both the Average and Median value.  I can easily ( using avg(Field1.Value!) ) determine the average but cannot find a function to determine the median.

How can I add the Median to the report?

 

Thanks in advance,

Greg

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Using Report Builder 3.0 To Calculate Median?

Sep 14, 2015

I'm using custom code in an expression to calculate the median of a column. It works fine up to a point. Like if there are 35 rows in the result set (or up to some number) but when it gets bigger results, like 42 rows or more it doesn't work, the median is -1. This is the custom code I'm using (found online).

I don't see anything limiting the count, but it comes back with a -1 median so I think that means the count is not > 0.

I have a column in the report called arrival_to_complete which is like: 53 min

I create a column with expression: =MAX(Code.AddValue(Val(Fields!Arrival_to_complete.Value))) to fill the array using just the number part of the column. Then in the report I have 'Median =' <expression>, where the expression is: =Code.GetMedian() I run the report with 2 parameters a begin date and an end date. I don't see where any of this should be limiting the median calculation so I don't get why it works sometimes and not other times.

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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Calculate Median Of Difference In Days Between Records

Jan 24, 2006

I have a table of sample data

Samples(sample_no, sample_date..)

I have no idea how to do the following in sql server or if its even possible:

1. Calculate the difference in days between all samples.
2. Select the median result

Any trick to get this done would be really helpful

thanks,

DB

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Calculate Median Of Difference In Days Between Records

Jan 24, 2006

I have a table of sample data

Samples(sample_no, sample_date..)

I have no idea how to do the following in sql server or if its even possible:

1. Calculate the difference in days between all samples.
2. Select the median result

Any trick to get this done would be really helpful

thanks,

DB

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Better Median??

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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New To SQL, Need To Get The Median Value

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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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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Jan 28, 2006

I've noticed that SQL Server (and other DBMSs I've looked at) doesn't seem to have a built-in function for finding the median of a range of numbers.

Gack!

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Sep 28, 2006

I have a long term need for a median function, I was wondering if anyone has or knows of some code or user defined functions somewhere that would do this. Ideally you could use it just liket the rest of the aggregate functions like AVG, etc.

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Finding Median Value

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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Sep 7, 2007

Im trying to find a funtion for median's if there is one. can anyone help?

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Feb 24, 2006

I have a table (cars) with 3 fields:VIN, Class, sell_price101, sports, 10000102, sports, 11000103, luxury, 9000104, sports, 11000105, sports, 11000106, luxury, 5000107, sports, 11000108, sports, 11000109, luxury, 9000i need to write a query that WITHOUT USING A FUNCTION will return themedian selling price for each class of car. result should look like:Class, Med_Priceluxury, 9000sports, 11000thanks to all u SQLers

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The Median Function In SQL Server

Jul 20, 2005

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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:


MEDIAN( [Test Fact].[Loan Number].Members, [Measures].[Loan Amount])

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

Thanks!

- Grant

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Median Function In T-SQL For SS2005?

Oct 13, 2006

Hello!

I have been trying to find a T-SQL function that would calculates a Median statistical value for me. I am runnnig on SS 2005? Any examples of using this function would be greatly appreciated.

Thanks for any responses!





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Analysis :: Median Calculation Using MDX

May 28, 2015

 I have a fact table fct_line_details having two columns mtid, productivity

  mtid     productivity
  1           400
1           200
1          600
2          700
3         900

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

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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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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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Apr 22, 2004

Folks,

I have a an sql table having 10,000 Rows.

The table has 5 columns. They are S1,S2,S3,S4 and M

I need the best way to calculate the MEDIAN for each row and store it in M. i.e. M=median(S1,S2,S3,S4)

I've come across articles where they've calculated medians on table rows, not table columns. But my requirement involves table columns. I guess transposing the columns into rows and then calculating median should be possible, but if I do that for 10000 rows using a cursor, then it would take a loooooong time.

Please suggest the best way to counter this

Thanx

Kiran

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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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Oct 27, 2015

I need to get the median for the 10 columns in my table. For the sake of example, I've reduced it to 2 columns. 

The code below works perfectly if i compute for only 1 column and certainly doesn't for multiple columns.

Is there a way to better handle the median computation in one pass, if multiple columns are involved?

DECLARE @tbl as table (id_n int, col1 int, col2 int)
insert into @tbl
values (1, 1, 2), (1,3, 4), (1, 5, 7), (2, 4, 7), (2, 7, 7), (2, 3, 5), (2,5, 5), (3, 1, 2), (3, 3, 5), (3, NULL,11)
select *
from @tbl
order by id_n, col1

[Code] .....

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T-SQL (SS2K8) :: How To Reduce Execution Time Of Median Calculations

Apr 7, 2015

I wrote a procedure to calculate median:

============================
ALTER proc [dbo].[sp_CalculateMedianTimeInDepartmentMinutes]

@StartDate date
,@EndDate date
as
--== Check if count is even or odd
declare @modulo int
select @modulo = (Select COUNT(*)%2 from ED_data where AdmitDateTime between @StartDate and @EndDate )
--=== Get Median

[Code] ....

My fellow developer is using this code to calcuate a madians in many columns (see below). The problem is that it takes about 2 minutes to execute this code. Is there a way to reduce the time of execution?

I attach also a sample of the view

==============
ALTER PROCEDURE [dbo].[sp_ED_Measures]
@StartDate date,
@EndDate date,
@Hospital varchar(5)
AS
BEGIN
SET NOCOUNT ON;

[Code] ......

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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.

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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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Hi all,
 
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