converting character date value to year4.

[sheren_deep1]

Hi, 

I have a dataset in which the birth date variable is in character format and some of the values do not have the birth month and birth date. Hence, I need to create a new variable in my dataset and just extract the birth year alone. Here is the snippet of the dataset. 

Could anyone please help me on how to create a new variable for the birth month alone. 

Thanks. 

[DJongman · Accepted solution]

Hi,

You could use the function SUBSTR if all vaues have the same format / layout.

SUBSTR Function

[Kurt_Bremser]

Use the scan() function to retrieve the third "word" (separated by slashes), and the input() function to convert the string to numeric.

[Riteshdell]

Hi,

data have;
length DOB $32.;
input DOB $;
informat DOB $32.;
datalines;
01/01/1980
01/12/1960
02/01/2019
31/12/2018
;
run;

proc contents data=work.have;
run;

data want;
set have;
new_year=year(input(DOB,ddmmyy10.));
run; 

[andreas_lds]

---

@Riteshdell wrote:

Hi,

data have;
length DOB $32.;
input DOB $;
informat DOB $32.;
datalines;
01/01/1980
01/12/1960
02/01/2019
31/12/2018
;
run;

proc contents data=work.have;
run;

data want;
set have;
new_year=year(input(DOB,ddmmyy10.));
run; 

---

Does not work with the data provided by @sheren_deep1 - if month and day are zero (see first line of the data posted) the format ddmmyy is unabble to create a date, so that year() returns missing.

Using scan+input (as suggested by @Kurt_Bremser), seems be the easiest way to solve the problem:

data want;
   set have;
   new_year = input(scan(DOB, 3, '/'), 4.);
run;

[sheren_deep1]

Hi, 

Thanks all for the help. I manage to do it with the substr function.