Convert date into years

[Marta Oliveira]

Hi, Stata users!
I want to convert data time into years, in order to determine age.
An example of my data is
year yearborn
2000 196010
Any help would be appreciated.

[Clyde Schechter]

Your example does not tell us how these variables are set up in Stata. In the future, please use the -dataex- command to post example data so that we can get an exact replica of your data. (-ssc install dataex-, -help dataex- for instructions on using it.)

So you first have to convert yearborn to a Stata internal format date, and then you have to extract the year from it. I assume that 196010 refers to October 1960.

numdate month_year_of_birth = yearborn, pattern("YM")
gen year_of_birth = year(month_year_of_birth)

[Marta Oliveira]

Thank you, Clyde Schechter for your answer. I just had now availability to try it.
I tried your code
numdate month_year_of_birth = yearborn, pattern("YM") but Stata replied to me "month_year_of_birth unrecognised kind of date-time".
I also tried
numdate yearly month_year_of_birth = yearborn, pattern("YM")
but I just obtain missing values.
Unfortunately I cannot use dataex because the computer that I am utilising cannot have internet connection.

[Marcos Almeida]

Hello Marta,

Please check whether these commands apply to your needs.

. set obs 4
number of observations (_N) was 0, now 4

. input year yearbirth

          year  yearbirth
  1. 2000 196010
  2. 2005 196112
  3. 2013 196103
  4. 2004 196209

. tostring yearbirth, replace
yearbirth was float now str6

. gen yearborn = date(yearbirth, "YM")

. format yearborn %td

. list

     +-----------------------------+
     | year   yearbi~h    yearborn |
     |-----------------------------|
  1. | 2000     196010   01oct1960 |
  2. | 2005     196112   01dec1961 |
  3. | 2013     196103   01mar1961 |
  4. | 2004     196209   01sep1962 |
     +-----------------------------+

. gen yearborn2 = year(yearborn)

. list

     +----------------------------------------+
     | year   yearbi~h    yearborn   yearbo~2 |
     |----------------------------------------|
  1. | 2000     196010   01oct1960       1960 |
  2. | 2005     196112   01dec1961       1961 |
  3. | 2013     196103   01mar1961       1961 |
  4. | 2004     196209   01sep1962       1962 |
     +----------------------------------------+

. gen age = year - yearborn2

. list

     +----------------------------------------------+
     | year   yearbi~h    yearborn   yearbo~2   age |
     |----------------------------------------------|
  1. | 2000     196010   01oct1960       1960    40 |
  2. | 2005     196112   01dec1961       1961    44 |
  3. | 2013     196103   01mar1961       1961    52 |
  4. | 2004     196209   01sep1962       1962    42 |
     +----------------------------------------------+

Best,

Marcos

[Nick Cox]

Note that Clyde had a typo in #2

 numdate month_year_of_birth = yearborn, pattern("YM")

should be

  
 numdate month month_year_of_birth = yearborn, pattern("YM")

[Nick Cox]

Note that if the year and month are in a numeric variable with values like 196010 then

gen byear = floor(yearborn/100)

is a direct one-liner.

Otherwise, Marcos's example helpfully lays out step-by-step logic. It could also be telescoped:

clear 
input year yearbirth
 2000 196010
 2005 196112
 2013 196103
 2004 196209
end 
gen yearborn = year(daily(string(yearbirth), "YM")) 
gen age = year - yearborn
list

     +----------------------------------+
     | year   yearbi~h   yearborn   age |
     |----------------------------------|
  1. | 2000     196010       1960    40 |
  2. | 2005     196112       1961    44 |
  3. | 2013     196103       1961    52 |
  4. | 2004     196209       1962    42 |
     +----------------------------------+