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AW: st: AW: Counting months over several years in a variable
From
"Martin Weiss" <[email protected]>
To
<[email protected]>
Subject
AW: st: AW: Counting months over several years in a variable
Date
Sat, 29 May 2010 16:50:46 +0200
<>
Ok, this kind of problem can be solved via -egen, group()-
*************
clear*
inp int Year byte month
2008 8
2008 8
2008 8
2008 9
2008 9
2008 10
2008 10
2008 11
2008 11
2008 12
2009 1
2009 1
2009 2
2009 3
2009 3
end
egen myvar=group(Year month)
list, sepby(Year month) noo
*************
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: [email protected] [mailto:[email protected]] Im Auftrag von George Chioran
Gesendet: Samstag, 29. Mai 2010 16:35
An: [email protected]
Betreff: Re: st: AW: Counting months over several years in a variable
Sorry, i forgot to mention that there are in total over 100 000 observations, over a period of 1,5 years, so myvar can only take 18 distinct values at maximum.
There are a lot of duplicates in the database like this:
Year month mvar
2008 8 1
2008 81
200881
2008 9 2
200892
200810 3
2008103
2008114
2008 11 4
2008 12 5
20091 6
200916
2009 2 7
200938
2009 3 8
… … …
----- Original Message ----
From: Martin Weiss <[email protected]>
To: [email protected]
Sent: Sat, May 29, 2010 3:51:52 PM
Subject: st: AW: Counting months over several years in a variable
<>
So this code -generate-s "myvar" across the entire dataset. Make sure it does what you want:
*************
clear*
inp int Year byte month
2008 8
2008 9
2008 10
2008 11
2008 12
2009 1
2009 2
2009 3
end
gen myvar=_n
list, noo
*************
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: [email protected] [mailto:[email protected]] Im Auftrag von George Chioran
Gesendet: Samstag, 29. Mai 2010 15:07
An: [email protected]
Betreff: st: Counting months over several years in a variable
Dear statalisters,
I want to count the
months in several years. I have variables year and month, and want to obtain
variable mvar. My database looks like this.
Year month mvar
2008 8 1
2008 9 2
2008 10 3
2008 11 4
2008 12 5
2009 1 6
2009 2 7
2009 3 8
… … …
Thank you in advance
for your help,
George
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