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AW: AW: st: data management


From   <[email protected]>
To   <[email protected]>
Subject   AW: AW: st: data management
Date   Mon, 7 Dec 2009 19:47:25 +0100

Thanks Uli, Martin, and Nick,
I applied Ulis solution and it works well. 
Thanks a lot!
Best,
Simon

-----Ursprüngliche Nachricht-----
Von: [email protected]
[mailto:[email protected]] Im Auftrag von Ulrich Kohler
Gesendet: Montag, 7. Dezember 2009 16:45
An: [email protected]
Betreff: Re: AW: st: data management

Ups. yes.

You can do it by hand, then

. gen index = _n
. reshape long Var, i(index)
. by index Var, sort: gen countdif = 1 if _n==1
. by index: replace countdif = sum(countdif)
. by index: replace countdif = coundif[_N]
. reshape long 
 





Am Montag, den 07.12.2009, 16:02 +0100 schrieb Martin Weiss:
> <> 
> 
> 
> 
> *************
> egen, group()
> *************
> 
> does not support -by-, though...
> 
> 
> HTH
> Martin
> 
> 
> -----Ursprüngliche Nachricht-----
> Von: [email protected]
[mailto:[email protected]] Im Auftrag von Ulrich Kohler
> Gesendet: Montag, 7. Dezember 2009 16:04
> An: [email protected]
> Betreff: Re: st: data management
> 
> I would use egen-group in the long format for this:
> 
> . gen index = _n
> . reshape long Var, i(index)
> . egen countdif = group(Var), by(index)
> . reshape long 
> 
> Many regards
> Uli
> 
> Am Montag, den 07.12.2009, 15:56 +0100 schrieb [email protected]:
> > Dear all,
> > I have a quick data management question. I would like to count the
number of different values over several variables within a row and save the
result in a new variable.
> > 
> > For example: 
> > Var1 Var2 Var3 Var4 Var5 Var6 Var7 Varx
> > 1         2      3        9       20     1       1      ...
> > 
> > In the example above, the new variable would holds the value 5, since
Var1, Var6 and Var7 all hold the same value (put differently, 5 different
values can be count over var1 - var7). 
> > 
> > I tried the following command: egen byte countdif = diff(var1 var2 var3
var4 var5 var6 var7 varx). However, the new variable is only a dummy (0/1). 
> > 
> > I bet there is a very simple solution ? however I am too blind to see
it?
> > 
> > I appreciate any help and suggestions.
> > Thanks
> > Simon
> > 
> > 
> > 
> > *
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