4

我正在阅读一些有几个错误日期的原始数据。具体来说，有人在非闰年输入了“2 月 29 日”。例如：

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

客户希望恢复到 2 月 28 日。是否有快速/有效的方法来执行此操作？例如相当于：

IF iserror(date) 然后 date=date-1; ?

任何建议都感激不尽！

4

4 回答 4

4

我会更加小心地确定日期。这是一种方法。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.
*/
于 2009-02-21T16:43:44.003 回答
3

这是来自 SCONSIG 的一个很好的提示。关联

/******************************************************************/
 /***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 ***/

可以以一种或另一种方式将其合并到您的负载中。

于 2009-02-18T16:53:59.197 回答
1

上述答案的略微修改版本（我认为）。它通过“??”避免了错误消息 在输入函数中。

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;
于 2009-02-20T13:55:23.700 回答
0

这并不漂亮，并且有转换说明，但它可以工作并且不会出错。

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;

输出：

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
于 2009-02-19T21:38:25.853 回答