Fabian Brenner wrote:
I have the following problem:
Variable "date" has the following form "11301998"
(the format is str8 %8s). I want to create a new
variable called "calculation_date" (format: str6 %9s).
Calculation date should have the form "199811".
(that means changing date from ddmmyyyy to yyyymm)
=========================================================
"11301998" is in format mmddyyyy, but I assume that you
do mean ddmmyyyy as stated elsewhere.
// Testdata:
clear
input str8 date_s
30111998
01022003
24122008
end
// You can use the substr() function to isolate parts of a string
gen str6 ym_s = substr(date_s,5,4) + substr(date_s,3,2)
// but for most purposes date variables are a lot more flexible
// (I assume that you have Stata 10). With -format- you can decide
// the display format:
gen date_dmy = date(date_s,"DMY")
format date_dmy %tdDDNNCCYY
gen date_ym = date_dmy
format date_ym %tdCCYYNN
. list
+----------------------------------------+
| date_s ym_s date_dmy date_ym |
|----------------------------------------|
1. | 30111998 199811 30111998 199811 |
2. | 01022003 200302 01022003 200302 |
3. | 24122008 200812 24122008 200812 |
+----------------------------------------+
You probably would benefit from studying the output from
. help date
Hope this helps
Svend
________________________________________________________
Svend Juul
Institut for Folkesundhed, Afdeling for Epidemiologi
(Institute of Public Health, Department of Epidemiology)
Vennelyst Boulevard 6
DK-8000 Aarhus C, Denmark
Phone, work: +45 8942 6090
Phone, home: +45 8693 7796
Fax: +45 8613 1580
E-mail: [email protected]
_________________________________________________________
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