topic Re: Sas year comparision in SAS Programming

Sas year comparision

Singla14 — Wed, 06 Sep 2017 09:26:09 GMT

, I have input file having year ( let's say 2015 to 2017 YYYY) .
I am trying to fetch system year ( year(today())
And then compare year from input file to fetched year.
If matches , those rows should come in output.

If comparison for years is not working ..Have tried reading input year in chat..Integer both..But nothing working.
Can some one please help..Urgent

Re: Sas year comparision

PeterClemmensen — Wed, 06 Sep 2017 09:32:34 GMT

show us some code 🙂

Re: Sas year comparision

RW9 — Wed, 06 Sep 2017 09:43:59 GMT

Show us some code and show us some test data data in the form of a datastep.

E.g.:

data have;
  input year;
datalines;
2013
2017
;
run;

Re: Sas year comparision

alexal — Wed, 06 Sep 2017 11:32:14 GMT

@Singla14,

Something like this?

data years;
  input year;
datalines;
2013
2014
2015
2016
2017
;
run;
proc print data=years; where year=year(today()); run;

Re: Sas year comparision

Singla14 — Thu, 07 Sep 2017 01:45:09 GMT






Input file is of 150..
Year is starting at 36th..
A0214494      FLDNIW0000414600DRIN 201401010....
A0214494      FLDNIW0000104200CRIN 201401010..
A0330094      FLDQIW0000104200CRSD 201401010..


SAS code:

DATA INFLE1;
DATERUN = TODAY();
FINAL_YEAR=YEAR(DATERUN);
FORMAT FINAL_YEAR YEAR4. ;
RUN ;
PROC PRINT DATA=INFLE1;

DATA INFLE2;
INFILE DD01;
INPUT
     @36 YEARX       4.
     @1 RECORD  $CHAR150.;
FORMAT YEARX YEAR4.;
RUN;
PROC PRINT DATA=INFLE2;

DATA OUTP1;
SET INFLE1 INFLE2;
IF  YEARX = FINAL_YEAR THEN OUTPUT OUTP1 ;
RUN;
DATA FINAL1;
SET OUTP1 ;
FILE DD01O;
PUT @1 RECORD $CHAR150.;
RUN;

YEARX = FINAL_YEAR is not working ... i have tried other formatting options too... like took input year in Char as well.. but 0 records in output..

I think issue is ..sas 's internal format.. 2017 would mean..1965 when i am moving it into YEAR4....

Re: Sas year comparision

andreas_lds — Thu, 07 Sep 2017 08:06:41 GMT

I suggest you start reading the SAS documentation, because your code has so many errors resulting from missing knowledge.

You don't need INFLE1, you can use the functions directly when createding infle2.
A SET statement with multiple datasets appends the datasets.
Using a format designed for variables containing dates on a variable containing just the year-part of a date, will lead to wrong results. You don't need formats to solve the problem.

data INFLE2;
infile DD01;
input 
     @36 YearX 4.
     @1 Record  $char150.;

   if YearX = year(today());
run;