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Hello Friends,
I have character date in YYMM format.(Column Name EXPDATE)
e.g. '2004' which means April 2020
'2104' which means April 2021
'2204' which means April 2022
'2304' which means April 2023
How to convert it into date format ?
Being a beginner in SAS I googled the same and use below mentioned logic
format exp date9.;
Exp = intnx('month', input(put(EXPDATE,4.), yymmn4.), 1)-1;
;
Which gives me output 30Apr1920 instead of 2020 and similarly for all remaining values where YY value is above 20.
Please help me to resolve the same.
Thanks,
Prasad Devasthali
Business Analyst,HDFC Bank
Accepted Solutions
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Can't test ist right now, but maybe setting YEARCUTOFF solves the issue.
See YEARCUTOFF= System Option :: SAS(R) 9.4 System Options: Reference, Fourth Edition
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Can't test ist right now, but maybe setting YEARCUTOFF solves the issue.
See YEARCUTOFF= System Option :: SAS(R) 9.4 System Options: Reference, Fourth Edition
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Thanks andreas_lds..Thanks
RW9...Your solutions resolved my problem...!!
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Your problem is YEARCUTOFF I expect it is the same as mine which defaults to 1920 set it to something greater than your largest 2 digit year; Also you don't need PUT(EXPDATE,4.) as EXPDATE is already character, or so you say. If you want the last day of the month EXP you can advance by 0 and use E for the alignment option. This it looks like you know how INTNX works.
data exp;
expdate = '2004';
x = put(EXPDATE,4.);
dt = input(put(EXPDATE,4.), yymmn4.);
put x= dt=date9.;
Exp = intnx('month',input(EXPDATE,yymmn4.),1)-1;
ExpLastday = intnx('month',input(EXPDATE,yymmn4.),0,'e');
put (exp expl:) (=date9.);
run;
28 expdate = '2004';
29 x = put(EXPDATE,4.);
30 dt = input(put(EXPDATE,4.), yymmn4.);
31 put x= dt=date9.;
32 Exp = intnx('month',input(EXPDATE,yymmn4.),1)-1;
33 ExpLastday = intnx('month',input(EXPDATE,yymmn4.),0,'e');
34 put (exp expl:) (=date9.);
35 run;
x=2004 dt=01APR2020
Exp=30APR2020 ExpLastday=30APR2020
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Hi,
data tmp;
infile datalines;
input text_date $;
num_date=input("20"||text_date,yymmn6.);
format num_date date9.;
datalines;
2004
2104
2204
2304
;
run;
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Note sure why you are getting the wrong century, but it is not hard to add the century back in. Is the existing variable character or numeric? You text says it is character but your example code is treating it as numeric. If it is numeric then convert it to text using the Z4. format to preserve any leading zeros.
data tmp;
infile datalines;
input text_date $4.;
num_date1=input(text_date,yymmn4.);
num_date2=input(cats('20',text_date),yymmn6.);
num_date3=input(cats(text_date,'01'),yymmdd6.);
num_date4=input(cats('20',text_date,'01'),yymmdd8.);
format num_date: yymmdd10.;
put (_all_) (=);
datalines;
2004
2104
2204
2304
;
run;
text_date=2004 num_date1=2020-04-01 num_date2=2020-04-01 num_date3=2020-04-01 num_date4=2020-04-01
text_date=2104 num_date1=2021-04-01 num_date2=2021-04-01 num_date3=2021-04-01 num_date4=2021-04-01
text_date=2204 num_date1=2022-04-01 num_date2=2022-04-01 num_date3=2022-04-01 num_date4=2022-04-01
text_date=2304 num_date1=2023-04-01 num_date2=2023-04-01 num_date3=2023-04-01 num_date4=2023-04-01
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Personal rant: Anyone still writing 2 digit years deserves many lashes with a wet spaghetti noodle or similar chastisement.
Of course I may be too sensitive after Y2K reprogramming and storing data from the early 60's.
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I reckon you're preachin' to the choir. I doubt that @ImPrasad is responsible for the date entry.
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Very true data_null_;!!!Why should I write such dates in such nightmare formats...Anyways Cheers ...Happy Weekend Guyss!!