tags:

views:

750

answers:

5

I am reading in some raw data that has a couple of bad dates. Specifically, someone has keyed in "29th Feb" on a NON leap year. For example:

data _null_;
input test :yymmdd8.;
format test date9.;
cards;
20270229
run;

The customer would like this to revert to the 28th Feb. Is there a quick / efficient method of doing this? eg an equivalent of:

IF iserror(date) then date=date-1; ?

Any suggestions gratefully received!

+2  A: 

Here is a great tip from SCONSIG. link

/******************************************************************/
 /***TIP00039.SAS                                                ***/
 /***   Leap Year Problem                                        ***/
 /***                                                            ***/
 /***   Most of us know that if the year is divisible by 4 then  ***/
 /***   that year is a Leap Year.  However, if the year is a     ***/
 /***   century year and is NOT divisible by 400 then that       ***/
 /***   century year is NOT A LEAP YEAR.                         ***/
 /***   (ie, 1700, 1800, 1900, 2100, 2200, 2300 are not LEAP     ***/
 /***        YEARS)                                              ***/
 /***                                                            ***/
 /******************************************************************/

 data leapyear;
   do year = 1600 to 2400 by 100;
    date = mdy(02,29,year);        /*** Leap Date                       ***/
    if date = . then do;           /*** If FEB 29th but not a Leap Year ***/
       date = mdy(03,01,year) - 1; /*** Make date March 1st and then    ***/
    end;                           /***   subtract 1 day                ***/
    output;
   end;
   format date mmddyy10.;
 run;

 proc print; run;

 /*** end of sas program - TIP00039 ***/

It is possible to incorporate this into your load in one way or the other.

AFHood
Thanks AFHood - I am trying a different approach at the moment (have posted a new question) but if I can't get rid of the error message that results I think I will use this instead (putting the years into a macro variable). Nice bit of code! thanks..
Bazil
A: 

The following code resolves the issue but unfortunately leaves an error message:

data null;
format test2 date9.;
input test ;
test2=INPUT(PUT(test,8.),yymmdd8.);
if error =1 then do;
error=0;
test2=INPUT(PUT(test-1,8.),yymmdd8.);
end;
put test2=;
cards;
20270229
run;

Bazil
A: 

This isn't pretty and there are conversion notes but it works and doesn't error.

data testit;
   format test2 yymmdd10.;
   input x $8.;
   mod4 = mod(mod((substr(x,1,4)/4),4) * 10,10);   
   if mod4 NE 0 then x = x - 1;
   test2=input(x,yymmdd8.);
   put x= test2=;

cards;
20080229
20090229
20100229
20110229
20120229
20130229
20270229
run;

Output:

x=20080229 test2=2008-02-29
x=20090228 test2=2009-02-28
x=20100228 test2=2010-02-28
x=20110228 test2=2011-02-28
x=20120229 test2=2012-02-29
x=20130228 test2=2013-02-28
x=20270228 test2=2027-02-28
CTKeane
A: 

Slightly modified version (I think) of the answer above. It avoids the error messages though with the "??" in the input function.

data testit;
   format indate yymmdd10.;
   input x $8.;

   indate = input(x, ?? yymmdd8.);
   if indate=. then indate= input(put(x - 1, 8.), ?? yymmdd8.);


    put indate=;

cards;
20080229
20090229
20100229
20110229
20120229
20130229
20270229
run;
Jay Stevens
+1  A: 

I would be a bit more careful fixing dates. here is one way. hth.

%put sysvlong=&sysvlong sysscpl=&sysscpl;           
/* sysvlong=9.02.01M0P020508 sysscpl=W32_VSPRO */

/* read a date both as character(temp) and numeric(date).
   if the numeric date is missing then check if the
   character date ends with "0229," if so, then change it
   to "0228" and see if it is a valid date. 
   If OK, then that is it. otherwise, keep it missing. */
%let FEB29 = 0229; 
%let FEB28 = 0228;
data one;
  drop temp;
  input temp $char8. @1 date ?? yymmdd8.;
  if missing(date) then link fix;
  format date b8601da.;
  put (_all_) (=);
  return;
fix:
  if length(strip(temp))^=8 then return;
  if substr(temp,5) ^= "&FEB29" then return;
  date = input(cat(substr(temp,1,4), "&FEB28"), ?? yymmdd8.);
return;
cards;
20080229  ok
20090229  should be changed to 28th
201XX229  this should be missing
20110229  -> 28
20120229  ok 
20130229  -> 28
20270229  -> 28
;
run;

/* on log
temp=20080229 date=20080229
temp=20090229 date=20090228
temp=201XX229 date=.
temp=20110229 date=20110228
temp=20120229 date=20120229
temp=20130229 date=20130228
temp=20270229 date=20270228
NOTE: The data set WORK.ONE has 7 observations and 1 variables.
*/
Chang Chung