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Re: st: eliminate oppositie positive and negative values


From   sara borelli <saraborelli77@yahoo.it>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: eliminate oppositie positive and negative values
Date   Fri, 30 Mar 2007 00:53:57 +0200 (CEST)

thank you very much for your help!
sara

--- Sergiy Radyakin <Radyakin@aoek.uni-hannover.de> ha
scritto:

> 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: "Ronán 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 Márta 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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> >
> > *
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