| Date: | Wed, 10 Sep 2008 10:11:18 +0200 |
| Reply-To: | Spousta Jan <JSpousta@CSAS.CZ> |
| Sender: | "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU> |
| From: | Spousta Jan <JSpousta@CSAS.CZ> |
| Subject: | Re: Selecting the top 3% of Service Users |
|
| In-Reply-To: | A<04B618C2F0D4574098D8B3BDDC1B3B2F020C8477@bc-mail-03.bci.buttecounty.net> |
| Content-Type: | text/plain; charset="iso-8859-2" |
Hi Kevin,
Try this (the first part is only for testing purposes).
* create a fake dataset.
INPUT PROGRAM.
- loop ident = 1 to 1000.
- compute service_minutes = uniform(100).
- end case.
- end loop.
- END FILE.
END INPUT PROGRAM.
form ident (f4).
execute.
**** the solution ****.
* create percentiles.
RANK service_minutes (D) /NTILES (100) into percentiles .
* select top 3%.
select if percentiles <= 3.
execute.
Best regards,
Jan
-----Original Message-----
From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Secrist, Kevin
Sent: Wednesday, September 10, 2008 12:18 AM
To: SPSSX-L@LISTSERV.UGA.EDU
Subject: Selecting the top 3% of Service Users
Hi SPSS Guru's,
I have a question that is probably very simple, but here it goes. I need to select the top 3% of service users based on service minutes in a mental health department. The Population is all clients. What would the code for this be?
Thanks
Kevin Secrist, Administrative Analyst, Associate
Butte County Department of Behavioral Health
Administrative Services Division
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