Ian:
Steve Adams briefly addressed P1 a few weeks ago in "What is a fib?" on his
Ixora site.
I don't recall the details.
-----Original Message-----
Sent: Monday, July 02, 2001 3:06 PM
To: Multiple recipients of list ORACLE-L
I ran the Celko algorthm against scott.emp. Wow was that fast! Then I ran
it against the 70,000 record table which my alogorthm took 8 seconds to
compute. The statment hung. Armed with "ORACLE 101 Performance Tuning" by
Gaja, Kirti, and a John A. Kostelac, I discovered that the event being
waited for was "direct path read." The explanation states this is usually
due to contention on devices.
I was running this on a test machine the with no other user sql statements
being run. The only conflict must be self-inflicted. I then went to the
v$session_wait table to further clarify the contention, but was not
successful.
V$SESSION_EVENT showed that the total_waits columns was only being
incremented for the "direct path read" event. The state of the wait for
"direct path read" event was always "waited short time"
What does the p1 value of v$session_wait indicate in this case? It does not
appear to be a file_id.
I killed the query after several minutes. The "direct path read" wait
events continued to increment until then.
Ian MacGregor
Stanford Linear Accelerator Center
[EMAIL PROTECTED]
-----Original Message-----
Sent: Monday, July 02, 2001 7:47 AM
To: Multiple recipients of list ORACLE-L
Code
Ian,
Nice stuff!
Since this subject was brought back up, I thought maybe some would be
interested in the following. I've never had a need to calculate a median,
but, I knew Celko's SQL for Smarties had a few variations and examples from
various people, each with caveats. And then there were differences between
what he termed statistical and financial mean, and some other things as
well. Anyway, a google search turned up another Celko solution. And this one
also brings up the concept of weighted median.
Here is his example,
http://www.intelligententerprise.com/db_area/archives/1999/992004/celko.shtm
l, modified by me to use the standard EMP table's SAL column:
SQL> SELECT AVG(DISTINCT x.sal)
2 FROM (SELECT F1.sal
3 FROM emp F1, emp F2
4 GROUP BY F1.empno, F1.sal
5 HAVING SUM(CASE WHEN F2.sal = F1.sal
6 THEN 1 ELSE 0 END)
7 >= ABS(SUM(CASE WHEN F2.sal < F1.sal THEN 1
8 WHEN F2.sal > F1.sal THEN -1
9 ELSE 0 END)))
10 X
11 /
AVG(DISTINCTX.SAL)
------------------
1550
The link above goes into some detail regarding the logic behind the query
and how his query finally reached the form above. I may never need to do a
median, but, this subject has been a good opportunity for learning. I've
tested the above with even, odd, multiple occurences of SAL, null,s etc. It
seems to work, but, everyone have a whack at it if you like.
Regards,
Larry G. Elkins
[EMAIL PROTECTED]
> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of MacGregor,
> Ian A.
> Sent: Saturday, June 30, 2001 9:00 PM
> To: Multiple recipients of list ORACLE-L
> Subject: Calculating The Median: Error Discovered in Oracle SQL 101 Code
>
>
> 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: MacGregor, Ian A.
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