topic Re: Unexpected year value with date informat in SAS Programming

Unexpected year value with date informat

Julie4435637 — Mon, 12 Nov 2018 02:54:51 GMT

Hello, I have 3 variables that put together should be a date. The year keeps coming up as 2002 although the value 2018 was used in the statement to read in the value. I have pasted my code and some data below. What am I missing?

data rpt2;
set rpt1(obs=15 keep=dom mth yr);
yr4=(ifc(index(yr, ":") > 0, "2018", yr));
mod_dt_char=cats("'",dom, mth, yr4,"'d"); /* value before turning into date */
mod_dt=input(cats("'",dom, mth, yr4,"'d"),date.); /* convert to date */
format mod_dt mmddyy10.;
run;

mth	dom	yr	yr4	mod_dt_char	mod_dt
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	14	2018	2018	14Feb2018'd	02/14/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	14	2018	2018	14Feb2018'd	02/14/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Feb	14	2018	2018	14Feb2018'd	02/14/2002
Feb	23	2018	2018	23Feb2018'd	02/23/2002
Apr	11	2016	2016	11Apr2016'd	04/11/2002
Apr	11	2016	2016	11Apr2016'd	04/11/2002
Apr	8	2016	2016	8Apr2016'd	04/08/1920
Apr	8	2016	2016	8Apr2016'd	04/08/1920
Apr	11	2016	2016	11Apr2016'd	04/11/2002

Re: Unexpected year value with date informat

Reeza — Mon, 12 Nov 2018 03:37:34 GMT

Yr4 is a character value.
Your ‘dates’ don’t need the quotation mark or d after it. Remove those and try INPUT again.

Re: Unexpected year value with date informat

Reeza — Mon, 12 Nov 2018 03:39:42 GMT

And use DATE9 instead of date, most likely the default is date7, which can mess things up.

Re: Unexpected year value with date informat

Julie4435637 — Mon, 12 Nov 2018 03:50:40 GMT

Thank you it worked - the date10. informat was the right one. I had never had problem before using date. without the length specified.

data rpt3;
set rpt2(keep=mod_dt_char);
mod_dt=input(mod_dt_char, date10.);
format mod_dt mmddyy10.;
run;

mod_dt_char	mod_dt
'23Feb2018'd	02/23/2018
'23Feb2018'd	02/23/2018
'14Feb2018'd	02/14/2018
'23Feb2018'd	02/23/2018
'23Feb2018'd	02/23/2018
'14Feb2018'd	02/14/2018
'23Feb2018'd	02/23/2018
'23Feb2018'd	02/23/2018
'14Feb2018'd	02/14/2018
'23Feb2018'd	02/23/2018
'11Apr2016'd	04/11/2016
'11Apr2016'd	04/11/2016
'8Apr2016'd	04/08/2016
'8Apr2016'd	04/08/2016
'11Apr2016'd	04/11/2016

Re: Unexpected year value with date informat

singhsahab — Mon, 12 Nov 2018 03:53:42 GMT

Hi Julie,

instead of using cats function , you can directly use MDY date function. Below is the sample code

data test;
year='2018';
month='Feb';
day='19';
;
run;


proc format ;
value $datef
 'Feb'=02
 'Apr'=04
 ;
 run;


data want;
set test;
new_date=mdy(input(month,$datef.),input(day,2.),input(year,4.));
format new_date ddmmyy10.;
run;

I created format as per your value available in month. You've to add remaining month.

Thanks.

Re: Unexpected year value with date informat

Nav_Sweeney — Mon, 12 Nov 2018 04:22:08 GMT

Deviates from the main question/topic. Not relevant

Re: Unexpected year value with date informat

Reeza — Mon, 12 Nov 2018 04:22:30 GMT

You don’t need the quotation marks or d. That still holds. You only need those for hardcoded dates in your code.

Re: Unexpected year value with date informat

Reeza — Mon, 12 Nov 2018 04:24:44 GMT

That becomes 4 function calls, 3 x input + MDY + format vs 2, cat + input.

Re: Unexpected year value with date informat

Nav_Sweeney — Mon, 12 Nov 2018 04:26:24 GMT

Absolutely and doesn't keep up with OPs needs in any case IMHO. Your initial response was neat and well pointed.

Re: Unexpected year value with date informat

Kurt_Bremser — Mon, 12 Nov 2018 14:06:34 GMT

The default length for the date. informat is 7.

Since the first 7 characters of your string are

'23feb2

you get a year of 2002. Remove the quotes, and use the date9. informat.
With regards to the default length of the informat, see Maxim 31.