SAS Invalid Leap Year Yymmdd8 Release Date

Question

I am reading some raw data that have a couple of bad dates. In particular, someone hit February 29th during a non-leap year. For instance:

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

The client would like this to return by February 28th. Is there a quick / efficient way to do this? for example, the equivalent:

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

Any suggestions gratefully received!

Answer 1

I would set the dates a little more carefully. here is one way. NTN.

 %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. */ 

Answer 2

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 ***/ 

You can incorporate this into your load anyway.

Answer 3

A slightly modified version (I think) of the answer above. It avoids error messages, although with "??" 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; 

Answer 4

This is ugly and there are notes about conversions, but it works and does not cause errors.

 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