>>>
>>>>
Scott, Nick
thanks for your help. I believe I was not very explicit in my email. Nick's solution works fine but it takes the same value for B in 1999 and multiplies it by C for previous years. What I should have mentioned in my email is that I want the new value of B created for each of the years before and after 1999 to be multiplied by C.
Scott, your solution works but only for the year 1998. I have tried gsort -year and that is doing the job.
Many thanks
Dev
>>> [email protected] 07/12/04 05:00pm >>>
How about this:
while b == . {
replace b=b[_n+1]*c if b==. & year <1999
replace b=b[_n-1]*c if b==. & year >1999
}
Scott
----- Original Message -----
From: Dev Vencappa <[email protected]>
Date: Monday, July 12, 2004 10:19 am
Subject: Re: st: filling previous missing observations repeatedly.
> >>
> >>
> >>
> Stata users, I have the following data set
>
> year B C Country
> 1995 . x Belgium
> 1996 . x Belgium
> 1997 . x Belgium
> 1998 . x Belgium
> 1999 x x Belgium
> 2000 . x Belgium
> 2001 . x Belgium
> 2002 . x Belgium
>
> Variable B has data for only 1999 and missing for other years. I
> want to extrapolate missing years using variable C, which is some
> kind of multiplicative scalar that varies over the years. Writing
> a command like the following seems to work fine to fill missing
> values after year 1999:
>
> replace B=B[_n-1]*C if B==. & year>1999 & country=="Belgium"
>
> However, the following does not seem to work (or at least fills
> the missing value for the year 1998 only):
>
> replace B=B[_n+1]*C if B==. & year<1999
>
> I believe this is because Stata works only forward and not
> backward. Is there a way to ask Stata to work backwards starting
> with the value from the year 1999 and repeat the command for prior
> years? I have tried the following loop command but that does not
> seem to work either (this command does not even fill in the
> missing value for the year 1998, for obvious reasons)
>
> while B[_n+1]!=. & B==. & country=="Belgium"{
> replace B=B[_n+1]*C
> }
>
> I think I am missing a syntax of Stata here.
>
> Many thanks
> Dev
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