Re: Median with standard SQL

From: Jess Likens <likens_at_students.uiuc.edu>
Date: Wed, 11 Jul 2001 18:35:00 -0500
Message-ID: <gk537.748$oz3.10473_at_vixen.cso.uiuc.edu>


Well, seeing as how I turned in the homework on which this was a question prior to posting to the newsgroup, I think perhaps your assumption was a bit uncalled for, no? I already had a solution, but it is (in my opinion) unwieldy and inelegant, so I was simply looking for some other possibilities.

-Jess

"Alan" <alanshein_at_erols.com> wrote in message news:9ihmd2$1jn$1_at_bob.news.rcn.net...
> Your professor at the University of Illinois probably wants you to really
> think about this one and try lots of different things. Finding a
 ready-made
> answer in a newsgroup is not the best way to go in the long term. It is
 much
> better to try to figure it out on your own, so here is a hint:
>
> The answer is constrained in the problem.
>
> "Jess Likens" <likens_at_students.uiuc.edu> wrote in message
> news:wPT27.633$oz3.8465_at_vixen.cso.uiuc.edu...
> > Well, I don't think that works for both even and odd numbers of rows.
 And
 I
> > don't think it works if there are multiple rows with identical values.
 But
> > thanks.
> >
> > -Jess
> >
> > "Aakash Bordia" <a_bordia_at_hotmail.com> wrote in message
> > news:9ida78$vi6$1_at_stlnews.stl.ibm.com...
> > > Hint....
> > > select column from table O where (select count(*) from table where
> > > column<O.column)=(select count(*)/2 from table)
> > > Thanks
> > > Aakash
> > >
> > > "Jess Likens" <likens_at_students.uiuc.edu> wrote in message
> > > news:yqm27.364$oz3.4759_at_vixen.cso.uiuc.edu...
> > > > Does anyone have a method to get the median of a relation using only
 the
> > > > following SQL commands: SELECT, DISTINCT, WHERE, FROM, LIKE,
 EXISTS,
 IN,
> > > > UNION, INTERSECT, EXCEPT/MINUS, ANY, ALL, COUNT, SUM, AVG, MAX, MIN,
 GROUP
> > > > BY, HAVING, ORDER BY, JOIN? I can't seem to come up with anything.
 The
> > > > median of n elements occurs at (n+1)/2 for an odd number of elements
 and
 it
> > > > is the average of n/2 and n/2 and (n+1)/2. Any help would be much
> > > > appreciated.
> > > >
> > > > -Jess
> > > >
> > > >
> > >
> > >
> >
> >
>
>
Received on Thu Jul 12 2001 - 01:35:00 CEST

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