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  • #1

    Extracting year from a date

    Hello,

    First, sorry if this question has already been solved. I researched it but I did not found it.

    I would like to pick up the year of a date. This is an example of what I have:


    Code:
    * Example generated by -dataex-. For more info, type help dataex
clear
input str9 dateintr
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"2-Jul-03" 
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"23-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"12-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"7-Jul-03" 
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"17-Jul-03"
"18-Jul-03"
"18-Jul-03"
"17-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"22-Jun-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"17-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"18-Jul-03"
"19-Jul-03"
"19-Jul-03"
"19-Jul-03"
"19-Jul-03"
end

    After looking around, I think one of the possible solutions is to use date() function, but it is not working as it just gives missing:

    Code:
    gen year=date(dateintr, "DMY")
    Any ideas?
    Tags: None
  • #2
    The reason your solution is not working is that with strings like "19-Jul-03" Stata has no way of knowing whether the 03 refers to 2003, or 1903, or 1803, or... To get the -date()- function to work with these strings, you will have to tell Stata. Assuming all of your dates are in year 2000 or later:
    Code:
    gen date = date(dateintr, "DM20Y")
format date %td
gen year = year(date)
    • 1 like

    Comment

    • #3
      Thanks a lot Clyde Schechter . It works really well.

      Comment

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