
From  "Sergiy Radyakin" <Radyakin@aoek.unihannover.de> 
To  <statalist@hsphsun2.harvard.edu> 
Subject  Re: st: eliminate oppositie positive and negative values 
Date  Thu, 29 Mar 2007 14:49:44 +0200 
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[_n1] & _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:doesn't that mean that the charges for a given date should sum to zero?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 email I runned the following:
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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