I have the following dataset
data have;
input date_of_birth;
datalines;
19600101
190002..
200002..
199812..
;
run;
I want to write a program that will create a new dataset, that replaces ‘..’ by the last day of the respective month.
Note: I also want to account for leap years. Therefore, for the 2nd value (190002..), I would want this to read 19000229 because this is a Leap Year
Appreciate all help!
Read First 6 chars only as a char, convert it to a date using input and correct informat and then use INTNX to set it to the end of the month with the alignment option.
Data want;
input date $6.;
date_end = intnx('month', input(date, yymm6.), 0, 'e');
format date_end yymmdd8.;
cards;
.....
Read First 6 chars only as a char, convert it to a date using input and correct informat and then use INTNX to set it to the end of the month with the alignment option.
Data want;
input date $6.;
date_end = intnx('month', input(date, yymm6.), 0, 'e');
format date_end yymmdd8.;
cards;
.....
data have;
infile datalines missover;
input date_of_birth $10.;
datalines;
19600101
190002
200002
199812
;
run;
data want;
set have;
month=substr(date_of_birth,5,2);
year=substr(date_of_birth,1,4);
day=substr(date_of_birth,7,2);
length date date_imp $50.;
if day='' then do;
date=cats(strip(year),strip(put(input(month, best.),z2.)));
if month="12" then date_imp=compress(put((mdy(1,1,input(year,best.)+1)-1), is8601da.),'-');
else date_imp=compress(put((mdy(input(month,best.)+1,1,input(year,best.))-1),is8601da.),'-');
end;
else do;
date=compress(put(mdy(input(month,best.),input(day,best.),input(year,best.)),is8601da.),'-');
date_imp=compress(date,'-');
end;
run;
data have; input cdate_of_birth $; format date_of_birth date9.; cdate_of_birth=compress(cdate_of_birth,,'kd'); if length(cdate_of_birth) lt 8 then date_of_birth= intnx('month',input(catt(cdate_of_birth,'01'),yymmdd8.),0,'e'); else date_of_birth=input(cdate_of_birth,yymmdd8.); datalines; 19600101 190002.. 200002.. 199812.. ;
Art, CEO, AnalystFinder.com
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