Re: st: eliminate oppositie positive and negative values

["Sergiy Radyakin" <Radyakin@aoek.uni-hannover.de>]

clear
input num personid date charge proc xm
406 124 2001 -5274 121 -1050
409 124 2001 -5274 121 -1050
409 124 2001 -5274 121 -1050
408 124 2001 5274 121 1050
451 124 2001 5274 121 1050
453 124 2001 5274 121 1768
end

l

gen 
whatever_id=string(personid)+"#"+string(date)+"#"+string(proc)+"#"+string(abs(xm))
gen abs_charge=abs(charge)

gen posit=1 if charge>0
gen negat=1 if charge<0

sort whatever_id abs_charge charge
by whatever_id abs_charge:egen npos=count(posit)
by whatever_id abs_charge:egen nneg=count(negat)

by whatever_id abs_charge:gen ndel=npos*(npos<=nneg)+nneg*(nneg<npos)
gen flag=0
by whatever_id abs_charge:replace flag=1 if _n<=ndel | _n>_N-ndel
drop if flag
drop abs_charge posit negat npos nneg ndel flag whatever_id
l




Sergiy



----- Original Message ----- 
From: "Michael Blasnik" <michael.blasnik@verizon.net>
To: <statalist@hsphsun2.harvard.edu>
Sent: Thursday, March 29, 2007 2:42 PM
Subject: Re: st: eliminate oppositie positive and negative values



> Woops, I copied your own syntax errors below.  Clearly, yoiu can't have a 
> variable named abs(xm) and the code should be changed to absxm.  Also, I 
> would combine replace todrop line in with the gen todrop line as one 
> logical condition, since it should be faster that way, especially if you 
> are working with a large dataset and many times though the loop.
>
> Michael
>
> ----- Original Message ----- 
> From: "Michael Blasnik" <michael.blasnik@verizon.net>
> To: <statalist@hsphsun2.harvard.edu>
> Sent: Thursday, March 29, 2007 8:32 AM
> Subject: Re: st: eliminate oppositie positive and negative values
>
>
>
> > That wouldn't work because there could be offsetting charges and non 
> > offsetting charges on the same date so the sum wouldn't be zero.
> >
> > Although there may be better ways to do this,  to repeat the loop until 
> > there is nothing to change you can  use while loop that continues based 
> > on a test.
> >
> > gen abscharge=abs(charge)
> > gen abs(xm)=abs(xm)
> > local go=1
> > while `go' {
> > bysort personid proc date abscharge abs(xm)(charge) :
> > gen byte todrop=charge!=charge[_n-1] & _n>1
> > bysort personid proc date abscharge absxm (charge) :
> > replace todrop=1 if charge!=charge[_n+1] & _n<_N
> > qui count if todrop==1
> > if r(N)==0 local go=0
> > drop if todrop==1
> > drop todrop
> > }
> >
> >
> > Michael Blasnik
> >
> > ----- Original Message ----- 
> > From: "Ron�n Conroy" <rconroy@rcsi.ie>
> > To: <statalist@hsphsun2.harvard.edu>
> > Sent: Thursday, March 29, 2007 5:11 AM
> > Subject: Re: st: eliminate oppositie positive and negative values
> >
> >
> >
> > > On 29 M�rta 2007, at 02:41, sara borelli wrote:
> > >
> > > > I need to drop the negatives and the positives with
> > > > the same date abs(charge) abs(xm) proc personid. As
> > > > suggested by Michael Blasnik in a previous e-mail I
> > > > runned the following:
> > > doesn't that mean that the charges for a given date should sum to zero?
> > >
> > > If that's the case then can't you just sum the charges within each 
> > > person within each date and check if they are zero?
> > *
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> > *   http://www.ats.ucla.edu/stat/stata/
> *
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