[Webinar] Streamline your web hosting managementRegister Today

x
?
Solved

Find Median using SQL Server 2000

Posted on 2012-03-28
9
Medium Priority
?
362 Views
Last Modified: 2012-04-11
I have Max() working as in following query and want to find Mean & Median.....

select 'Max' as Func, Max(b.Proc_Cnt) Procedures
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF'
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b

How do I convert above to get Median?

Thanks,
Ashok
0
Comment
Question by:Ashok
  • 5
  • 4
9 Comments
 
LVL 9

Expert Comment

by:rajeevnandanmishra
ID: 37778959
Hi,
MEAN can be calculated by using AVG function like:
select 'Max' as Func, Max(b.Proc_Cnt) Procedures
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF'
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b
0
 
LVL 9

Accepted Solution

by:
rajeevnandanmishra earned 2000 total points
ID: 37779005
And Median can be found by using below query:
select 'Median' as Func,
(SELECT top 1 Proc_Cnt FROM
(select top 50 percent Proc_Cnt
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF'
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b ORDER BY Proc_Cnt ) c ORDER BY 1 DESC )
+
(SELECT top 1 Proc_Cnt FROM
(select top 50 percent Proc_Cnt
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF'
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b ORDER BY Proc_Cnt DESC) c ORDER BY 1 )/2


Though it could be better if you put your inner processed result in a temp table and then run the query on that temp table.
0
 
LVL 13

Author Comment

by:Ashok
ID: 37779422
MEAN can be calculated by using AVG function like:

I do not see the use of AVG function in above.

I will try the median when I go to work tomorrow morning.

Thanks,
Ashok
0
Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 
LVL 13

Author Comment

by:Ashok
ID: 37779426
Do I just change SUM to

select 'Avg' as Func, Avg(b.Proc_Cnt) Procedures

in my Max() query?

Ashok
0
 
LVL 9

Expert Comment

by:rajeevnandanmishra
ID: 37779446
My mistake. Yes,just change the MAX to AVG in your query and that will provide you the MEAN.
0
 
LVL 13

Author Closing Comment

by:Ashok
ID: 37781931
Thanks a lot
0
 
LVL 13

Author Comment

by:Ashok
ID: 37782653
by the way, this did not work for my data.
Median was out of range (not between Min & Max).

Ashok
0
 
LVL 9

Expert Comment

by:rajeevnandanmishra
ID: 37783561
Hi,
Logically it should work. But could be some mistake by me. Can you put the result of below two queries:
SELECT top 1 Proc_Cnt FROM 
(select top 50 percent Proc_Cnt 
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no 
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR 
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF' 
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b ORDER BY Proc_Cnt ) c ORDER BY 1 DESC 

SELECT top 1 Proc_Cnt FROM 
(select top 50 percent Proc_Cnt 
from (
select a.Stu_Code, sum(a.proc_cnt) Proc_Cnt
from (
select s.stu_code, t.chart_no, t.last_date, count(*) proc_cnt
from Treatment  t, Student s, Patient p
where s.stu_code = t.provider_id_credited
and p.chart_no = t.chart_no 
and t.last_date between '6/30/2009' and '7/1/2010'
and s.clinic_location like 'GRP%'
 and t.njds_code <> '00000'
 and (t.njds_code like 'D%' or t.njds_code like '0%')
and t.njds_code NOT BETWEEN '90000' and '90100'
and (t.chart_no Not Like 'M%' and t.chart_no Not Like 'F%' and t.chart_no <> '292021')
and (t.Activity Not in ('CP','CS','NS') OR t.Activity is Null)
and ((p.group_assignment IS NULL) OR 
     (p.group_assignment <> 'MTHN' AND p.group_assignment <> 'NRTF' 
  AND p.group_assignment <> 'SOM' AND p.group_assignment <> 'GWY'))
group by s.stu_code, t.chart_no, t.last_date) a
group by a.stu_code
) b ORDER BY Proc_Cnt DESC) c ORDER BY 1

Open in new window

0
 
LVL 13

Author Comment

by:Ashok
ID: 37833883
Proc_Cnt
321

(1 row(s) affected)

Proc_Cnt
322

(1 row(s) affected)
0

Featured Post

Get expert help—faster!

Need expert help—fast? Use the Help Bell for personalized assistance getting answers to your important questions.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

What if you have to shut down the entire Citrix infrastructure for hardware maintenance, software upgrades or "the unknown"? I developed this plan for "the unknown" and hope that it helps you as well. This article explains how to properly shut down …
An alternative to the "For XML" way of pivoting and concatenating result sets into strings, and an easy introduction to "common table expressions" (CTEs). Being someone who is always looking for alternatives to "work your data", I came across this …
Via a live example, show how to extract information from SQL Server on Database, Connection and Server properties
Via a live example, show how to backup a database, simulate a failure backup the tail of the database transaction log and perform the restore.
Suggested Courses

612 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question