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st: RE: histogram


From   Nick Cox <n.j.cox@durham.ac.uk>
To   "'statalist@hsphsun2.harvard.edu'" <statalist@hsphsun2.harvard.edu>
Subject   st: RE: histogram
Date   Thu, 14 Oct 2010 15:03:39 +0100

This is almost the function of -egen, tag()-. 

egen tag = tag(id) 

But with your problem, it is better to go back to first principles:

bysort id (mean) : gen tag = _n == 1 
histogram mean if tag 

The step of sorting within -id- was what you were missing. 

Nick 
n.j.cox@durham.ac.uk 

Schöler, Lisa

I have different observations within one id:

Example

+----------------------------------+
  | id   mean(brand) brand  year   |
  |--------------------------------|
  |  1   5      		7     1999 |
  |  1   5      		5     2000 |
  |  1   5   		3     2001 |
  |  1   5   		5     2002 |
  |  1   .      		.     2003 |
  |--------------------------------|
  |  id   mean(brand) brand  year  |
  |--------------------------------|
  |  2   .      		.     1999 |
  |  2   6      		7     2000 |
  |  2   6   		8     2001 |
  |  2   6   		5     2002 |
  |  2   6      		4     2003 |  
  |--------------------------------|
  |  id   mean(brand) brand  year  |
  |--------------------------------|
  |  2   .      		.     1999 |
  |  2   6      		7     2000 |
  |  2   6   		8     2001 |
  |  2   6   		5     2002 |
  |  2   6      		4     2003 |
   +------------------------------+

Now I need to get a histogram that uses only one observation for each id for the mean(brand). Therefore I wanted to include a dummy=1, so I can use the command 

. histogram if dummy==1

Can anybody help me how I can create a dummy=1 only for one observation per id? I was gonna use 

. by id: gen dup = cond(_N==1,0,_n)

and then

. histogram if dup==1

But the problem here is that I have missings within my id groups, so sometimes my dup per id starts with 2.

If anybody has a better idea than creating a dummy, I am very happy to hear about it.


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