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Re: st: Working with time series operators and loops in If Qualifier


From   Nick Cox <njcoxstata@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Working with time series operators and loops in If Qualifier
Date   Thu, 5 May 2011 18:36:01 +0100

Uunfortunately for you this is indeed fantasy syntax.

You have to break down

L(1/`x').(abc >= `z'/100)

So you first have to create

gen myvar = abc >= `z'/100

and then do something like this

gen true = 1

forval k = 1/`x' {
replace false = 0 if  L`k'.myvar == 0
}

Later

drop myvar true

Nick

On Thu, May 5, 2011 at 4:48 PM,  <Hauptseminar@gmx.de> wrote:

> I am trying to integrate time operators and a loop in a if qualifier.
>
> Here is my code:
>
> forvalues x = 1/10 {
> forvalues z = 1/100 {
>
> gen hrec_`z'_`x' = 0
>
> gen hrec_start_`z'_`x' = 0
>
> replace hrec_`z'_`x' = 1 if L(1/`x').(abc >= `z'/100) & ( vw >= 0 | dvw >=0)
>
> replace hrec_start_`z'_`x' = 1 if hrec_`z'_`x'==1 & L1.hrec_`z'_`x'==0
>
> replace hrec_`z'_`x'=0 if (abc < `z'/100 | vw<=0) & hrec_start_`z'_`x' ~=1
> }
> }
>
> Stata wont let the time operator L(1/`x') work in the if qualifier and I understand the reason for this. Unfortunately, I have no clue how to write the code of this instead.
>
> I want the if qualifier to check if 1 to x lagged values are above a numerial value without typing every single qualification. Do you have any advice how to solve this problem?
>
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