Ian,
Okay, I'm curious, I did a search on amazon and found Oracle PL/SQL 101 and
Oracle SQL: 101 Frequently Asked Questions. Which of these (or is it a
different one?) are you talking about here?
The reason I ask is, I have done tech-edit (which includes code testing) for
Oracle Press, although not on the PL/SQL book. And when I have done
technical edits, I have insisted on either removing the incorrect code or
replacing it with working code.
If the Oracle Press book was not edited well, a) I'm surprised but b) I will
get in touch with them to have them review it again for the next edition.
Rachel
>From: "MacGregor, Ian A." <[EMAIL PROTECTED]>
>Reply-To: [EMAIL PROTECTED]
>To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
>Subject: RE: Calculating The Median: Error Discovered in Oracle SQL 101
>Co de
>Date: Sun, 01 Jul 2001 08:50:23 -0800
>
>I was 99.999999% that one doesn't toss duplicates when computing a median;
>thanks to the link Jared has posted, I'm now 100% confident. The Oracle
>SQL 101 code will correctly compute a median only if the values in the
>column it is applied against are unique, not including nulls. Median
>calculations are almost always performed against populations which have
>duplicate values.
>
>It's been stated before, "All books have errors". Oracle SQL 101 should
>not be condemned to the trash for the error, but the mistake should
>enforce the rule that code posted here or written in a book by must be
>thoroughly tested before being employed. This rule especially includes
>the code I post.
>
>Ian MacGregor
>Stanford Linear Accelerator Center
>[EMAIL PROTECTED]
>
>
>-----Original Message-----
>Sent: Sunday, July 01, 2001 12:20 AM
>To: Multiple recipients of list ORACLE-L
>Code
>
>
>
>Try this link:
>
>http://math.about.com/science/math/library/howto/htmean.htm
>
>Jared
>
>On Saturday 30 June 2001 19:00, MacGregor, Ian A. wrote:
> > I ran the code from Oracle SQL 101 which Jared posted modifying it to
>find
> > the median of the sal column on that table
> >
> > SQL> select
> > 2 rownum,
> > 3 sal
> > 4 from (
> > 5 select sal
> > 6 from scott.emp
> > 7 where sal is not null
> > 8 union
> > 9 select 1 from dual where 1=2
> > 10 )
> > 11 group by sal, rownum
> > 12 having rownum >= (
> > 13 select decode( mod(total_freq,2),
> > 14 1,trunc(total_freq/2 + 1),
> > 15 0,trunc(total_freq/2)
> > 16 )
> > 17 from (
> > 18 select count(*) total_freq
> > 19 from scott.emp
> > 20 where sal is not null
> > 21 )
> > 22 )
> > 23 and rownum <= (
> > 24 select decode( mod(total_freq,2),
> > 25 1,trunc(total_freq/2 + 1),
> > 26 0,trunc(total_freq/2 + 1)
> > 27 )
> > 28 from (
> > 29 select count(*) total_freq
> > 30 from scott.emp
> > 31 where sal is not null
> > 32 )
> > 33 )
> > 34 /
> >
> > values
> > averaged
> > ROWNUM in median
> > ---------- ----------
> > 7 1600
> > 8 2450
> > ----------
> > Median 2025
> >
> >
>---------------------------------------------------------------------------
> >----------- This answer is different from the result of the code I posted
> > which uses the new analytical functions.
> >
> > select
> > case
> > when mod(number_salaried,2) = 0 then
> > (select sum(sal)/2 from(select sal, row_number()
> > over ( order by sal) as salrank
> > from scott.emp)
> > where salrank = number_salaried/2
> > or salrank = number_salaried/2 +1)
> > else
> > (select sal from(select sal, row_number()
> > over ( order by sal) as salrank
> > from scott.emp)
> > where salrank = ceil(number_salaried/2))
> > end median
> > from (select sal,rank() over (order by sal) as rk from scott.emp),
> > (select count(sal) number_salaried from scott.emp)
> > where rk = 1
> > /
> >
> > MEDIAN
> > ---------
> > 1550
> >
>---------------------------------------------------------------------------
> >--------------------- Emp is a 14 row table . The median should be the
> > average of the seventh and eighth values. I cleared the computes and
> > columns and ran the first part of the SQL 101 code
> >
> > clear computes
> > utes cleared
> > select
> > rownum,
> > sal
> > from (
> > select sal
> > from scott.emp
> > where sal is not null
> > union
> > select 1 from dual where 1=2
> > )
> > group by sal, rownum
> > /
> >
> > ROWNUM SAL
> > ------ ----------
> > 1 800
> > 2 950
> > 3 1100
> > 4 1250
> > 5 1300
> > 6 1500
> > 7 1600
> > 8 2450
> > 9 2850
> > 10 2975
> > 11 3000
> >
> > ROWNUM SAL
> > ------ ----------
> > 12 5000
> >
> > and also ran the part of my code which corresponded. I changed my code
> > slightly so the salrank column would print.
> >
> > SQL> select salrank, sal from(select sal, row_number()
> > 2 over ( order by sal) as salrank
> > 3 from scott.emp)
> > 4 /
> >
> > SALRANK SAL
> > ---------- ----------
> > 1 800
> > 2 950
> > 3 1100
> > 4 1250
> > 5 1250
> > 6 1300
> > 7 1500
> > 8 1600
> > 9 2450
> > 10 2850
> > 11 2975
> >
> > SALRANK SAL
> > ---------- ----------
> > 12 3000
> > 13 3000
> > 14 5000
> >
> > 14 rows selected.
> >
> >
>---------------------------------------------------------------------------
> >------------------- The reason for the different answers is now
>apparent.
> > The SQL 101 code is tossing duplicate records. It's been a long time
> > since my stats classes, but I'm about 99.999999% confident you don't
>purge
> > duplicates when computing a median. But even if I'm wrong about this,
>the
> > SQL 101 code has reduced the set to 12 members, but it is still
>computing
> > the median as if there were 14 members; that is, it is taking the
>average
> > of the 7th and 8th values and not the average of the 6th and 7th.
> >
> > I hope there was a caveat in SQL 101 book stating the code only worked
> > against columns with unique values, not including nulls.
> >
> > Ian MacGregor
> > Stanford Linear Accelerator Center
> > [EMAIL PROTECTED]
>--
>Please see the official ORACLE-L FAQ: http://www.orafaq.com
>--
>Author: Jared Still
> INET: [EMAIL PROTECTED]
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