I have been researching the best way to calculate median values using
SQL.  Most of the threads I come across include a semi-canned response
from Joe Celko which I have some questions about.

First, Celko says,

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I adapted this SQL to my situation but, unless I am mistaken, it
returns both the left and right statistical medians.  For the life of
me, I can't figure out how to get it to return only the left or right
statistical median.  Luckily, I don't need those specifically.

Celko continues...

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There is a certain "air code" quality to this section (it appears to be
missing a GROUP BY clause and maybe some parenthesis) but again, unless
I am mistaken, the ORed section is redundant (i.e. the section
preceding the OR is identical, aside from some non-critical reordering,
to the section following the OR).

So, finally, my question.  Has anyone used a method similar to this for
computing medians?  Is there anything subtly (or grossly, I suppose)
wrong with the following SQL for calculating  a Financial Median?

SELECT AVG(P1.weight)
      FROM Parts AS P1, Parts AS P2
   GROUP BY P1.weight
HAVING SUM(CASE WHEN P2.weight <= P1.weight
         THEN 1 ELSE 0 END)
   AND SUM(CASE WHEN P2.weight >= P1.weight
         THEN 1 ELSE 0 END)
I think other formulas (such as AVG(DISTINCT P1.weight) could be
substituted in the SELECT clause depending on which "form" of median
was desired but I am more concerned about whether I will always get the
right "starting" records for the median calculation than with the
specific formula for the median calculation.

JLS (jlspub***@hotmail.com) writes:

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On SQL 2005 you can use row_number(). Here is a query which builds on
an idea from SQL Server Itzik Ben-Gan:

   SELECT listlen, datatype,
          cnt = MIN (cnt), median = avg(tookms), avgms = MIN(avgms)
   FROM   (SELECT listlen, datatype, optype, method, calltype,
                  cnt   = COUNT(*)    OVER (PARTITION BY listlen, datatype),
                  avgms = AVG(tookms) OVER (PARTITION BY listlen, datatype),
                  tookms,
                 rowno  = row_number() OVER (PARTITION BY listlen, datatype
                                             ORDER BY tookms)
         FROM   timings) AS x
   WHERE  rowno IN (cnt/2 + 1, (cnt + 1)/2)

tookms is the value I'm computing the median of. The idea is that if
the number of elements are odd, outer WHERE clause will only find one
row, and if there is an even number it will hit two rows, and the median
is the average of those two rows.

Thanks for giving me another take on how this could be done.
Unfortunately, I haven't migrated to SQL 2005 yet.  However, since you
brought it up, I understand that SQL 2005 permits user-defined
aggregates.  Could the Median function be implemented as a user-defined
aggregate?  If it can, that would be great because I wouldn't have to
"re-code" my median logic for various grouping scenarios.  From reading
BOL, I get the impression that user-defined aggregate functions only
"see" the data one record at a time and in unsorted order so the
calculation would require a fair amount of behind the scenes storage
space for large result sets but it seems like it might work.  Any
thoughts?

Erland Sommarskog wrote:

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JLS (jlspub***@hotmail.com) writes:
I have not looked into UDAs much myself, but my impression is the same as
yours. As long as the data comes running, you can only accumulate, and then
at end you can start computing. Which means that if you are ask to asked the
median on 100 million values, you will eat quite some memory.

You are correct, UDAggs in SQL 2005 only "see" one row at a time.  The
internal state could work around that, but unfortunately it too is limited
in size so you can't just store them all and compute the value at the end.

Row_number can be used to see the rows in order, and I'd recommend you use
that to build your own solution in 2005.

For previous releases, the techniques published by others as seen in this
newsgroup are what I recommend.

Thanks,

Conor
SQL Server Query Processor Team

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