Extracting the date, month and year from a variable

[Gliby Anna]

Dear all

How can I extract the date, month and year from a date variable in STATA

Below is an example of my dataset

* Example generated by -dataex-. To install: ssc install dataex
clear
input str10 Q2_9a
"2018-05-07"
"2011-03-07"
"2018-10-15"
"1900-06-01"
"2018-01-01"
"2014-05-13"
"2014-10-23"
"2017-05-16"
"2018-01-01"
"2014-05-12"
""          
"2018-01-01"
"2014-05-14"
""          
"2013-10-03"
"2016-10-27"
"2008-04-03"
"2014-05-13"
"2015-12-24"
"2017-09-14"
"2012-05-07"
"2014-07-09"
"2011-03-01"
"2016-09-01"
"2018-01-01"
"2018-05-04"
"2014-08-01"
"2010-10-11"
"2018-05-07"
"2018-04-16"
"1993-04-27"
"2015-05-07"
"2012-04-16"
"2018-05-14"
"2018-05-11"
"2016-05-07"
"2015-08-04"
"2005-11-01"
"2017-10-06"
"1900-06-01"
"2017-06-12"
"2011-05-25"
"2015-05-07"
"2014-12-10"
"2014-10-23"
"2016-12-02"
"1900-06-01"
"2008-01-10"
"2005-08-31"
"2005-05-17"
"2006-12-31"
"2010-10-12"
"2000-10-16"
"2011-07-01"
""          
"2005-06-01"
"2016-01-19"
"1900-06-01"
"2016-05-19"
"2015-05-07"
"2015-05-07"
"2008-06-30"
"1998-07-07"
"1985-01-04"
""          
""          
"2004-03-23"
"2008-06-30"
"2018-02-05"
"2014-12-04"
"2014-08-19"
"2003-10-15"
"1999-07-24"
"2015-05-07"
"2015-05-29"
"2015-05-05"
"2015-05-05"
"2014-06-20"
"2015-05-05"
"2001-12-21"
"2009-09-09"
"2015-08-05"
"2018-03-19"
"2017-02-13"
"2011-07-15"
"2018-03-19"
"1985-01-01"
"2007-01-01"
"2015-05-04"
"1900-06-01"
"2015-01-13"
""          
"1981-01-01"
"2015-05-06"
"2017-03-31"
"2017-05-01"
"2015-11-04"
"2016-01-01"
"2015-06-25"
"2016-08-25"
end

Thank you

[Carole J. Wilson]

See the help file for -split-. You’ll want something like:

split Q2_9a, parse("-") gen(part)
rename part1 year
rename part2 month
rename part3 date

[Joseph Coveney]

It's not actually a date variable in Stata. It's a string variable in Stata.

For your case, you'd use the substr() function. Or split.

split Q2_9a, generate(d) parse(-)
rename d1 year
rename d2 month
rename d3 day

If it were a Stata date, you could

help datetime

for the functions to extract the components.

[Nick Cox]

If you use split you will need the destring option to make these results easier to use. Otherwise use daily() as here.

* Example generated by -dataex-. To install: ssc install dataex
clear
input str10 Q2_9a
"2018-05-07"
"2011-03-07"
"2018-10-15"
"1900-06-01"
"2018-01-01"
"2014-05-13"
"2014-10-23"
"2017-05-16"
"2018-01-01"
"2014-05-12"
end 

split Q2_9a, parse(-) destring 

rename (Q2_9a?) (year month day)

list 

     +---------------------------------+
     |      Q2_9a   year   month   day |
     |---------------------------------|
  1. | 2018-05-07   2018       5     7 |
  2. | 2011-03-07   2011       3     7 |
  3. | 2018-10-15   2018      10    15 |
  4. | 1900-06-01   1900       6     1 |
  5. | 2018-01-01   2018       1     1 |
     |---------------------------------|
  6. | 2014-05-13   2014       5    13 |
  7. | 2014-10-23   2014      10    23 |
  8. | 2017-05-16   2017       5    16 |
  9. | 2018-01-01   2018       1     1 |
 10. | 2014-05-12   2014       5    12 |
     +---------------------------------+

gen ddate = daily(Q2_9a, "YMD")

format ddate %td

list 

     +---------------------------------------------+
     |      Q2_9a   year   month   day       ddate |
     |---------------------------------------------|
  1. | 2018-05-07   2018       5     7   07may2018 |
  2. | 2011-03-07   2011       3     7   07mar2011 |
  3. | 2018-10-15   2018      10    15   15oct2018 |
  4. | 1900-06-01   1900       6     1   01jun1900 |
  5. | 2018-01-01   2018       1     1   01jan2018 |
     |---------------------------------------------|
  6. | 2014-05-13   2014       5    13   13may2014 |
  7. | 2014-10-23   2014      10    23   23oct2014 |
  8. | 2017-05-16   2017       5    16   16may2017 |
  9. | 2018-01-01   2018       1     1   01jan2018 |
 10. | 2014-05-12   2014       5    12   12may2014 |
     +---------------------------------------------+