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Re: st: RE: a question on how to create a complex loop
From
Lim Lee <[email protected]>
To
[email protected]
Subject
Re: st: RE: a question on how to create a complex loop
Date
Wed, 8 Feb 2012 18:34:24 -0500
another way would be to create a binary variable for each criteria
then creating another variable summing the criteria by patient using
egen:
1) gen "x"=1 if diag1==icd?
2) egen "y"=total(x), by(id)
best,
Lim
On Wed, Feb 8, 2012 at 11:22 AM, Reinhardt Jan Dietrich
<[email protected]> wrote:
> You could transpose the data into wide format using reshape (type: help reshape to see the code). Then you will get diagcateg1, diagcateg2, etc., i.e. a variable for each measurement point/occasion.
> Then generate a new variable
> Gen diagcat_count = diagcateg1 + diagcateg2
> usw.
>
> Best
> Jan
>
>
> -----Original Message-----
> From: [email protected] [mailto:[email protected]] On Behalf Of Aluko Hope
> Sent: Mittwoch, 8. Februar 2012 17:01
> To: [email protected]
> Subject: st: a question on how to create a complex loop
>
> Dear Stata list users,
>
> I am a health services researcher who is relatively new to STATA and I
> am trying to count the number of times each patient satisfies a
> particular criteria. I think I need to do this by creating a complex
> loop but I am stuck as to how to go about it.
>
> The data is structured in the following manner:
>
> id diag1 diag2 diag3 visit# diagcateg1
> 1 icd icd icd 3 1
> 1 icd icd icd 3 1
> 1 icd icd icd 3 1
> 2 icd icd icd 4 0
> 2 icd icd icd 4 0
> 2 icd icd icd 4 0
> 2 icd icd icd 4 0
> 3 icd icd icd 2 1
> 3 icd icd icd 2 1
>
> the icd refer to really icd9 and the diagcateg1 is a binary variable
> that already classifies certain icd9 into categories.
>
> in words, what i want to do now is:
> whenever diagcategory is 1, count the number of times per patient (id)
> that there is a diag1 diag2 or diag3 that fit a particular criteria. i
> would be happy with the total number per patient but would ideally
> like to be able to describe the total number per patient per visit.
>
> Please help,
>
> Many thanks in advance,
>
> Aluko Hope
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