Re: st: Re: bysort problem

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

Is there a guarantee, that var2+c(epsfloat)*obs will be a unique value (will 
not overlap with the next value)? It is true in the example, but it is not a 
general case.

consider var1=0.5 and var2 equal to
c(epsfloat)
c(epsfloat)
c(epsfloat)
c(epsfloat)

This seems to be also a solution:
replace var2=var2+uniform()/100000

simply add a small random number (noise) to var2. Chances that there will be 
coincidences are also small, but they may not be ignored.



----- Original Message ----- 
From: "Austin Nichols" <austinnichols@gmail.com>
To: <statalist@hsphsun2.harvard.edu>
Sent: Monday, February 26, 2007 5:51 PM
Subject: Re: st: Re: bysort problem



> Nikolaos A. Patsopoulos--
> No loop should be necessary, as this example demonstrates.  My guess
> is that this exercise is also not necessary, but as your ultimate
> desired outcome is not explicit, I cannot know if replacing values is
> actually a good idea for you.
>
> input var1 var2
> .145 .14
> .145 .15
> .145 .15
> .167 .15
> 1.89 .15
> 1.89 .16
> 1.89 .16
> end
> egen oldgroup=group(var1 var2)
> bys var1 var2: gen obs=_n
> replace var2=var2+c(epsfloat)*obs
> egen newgroup=group(var1 var2)
> li, noo clean
>
>
> On 2/26/07, Nikolaos A. Patsopoulos <npatsop@cc.uoi.gr> wrote:
> > The purpose of the code is to eliminate duplicate observations of var1 &
> > var2 (combination).
> > In the firest pass the algorithm fixes duplicates but new ones might
> > come-up, so it should run as long as noone is left. The correction is
> > too small for the real data (the ones here are dummy ones just to test
> > the code) and the possibility of duplicates very small but still present.
> >
> > var2 is changed not var1. This was a mistake I made on earlier e-mail.
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