用于检查缺失日期的 sas 宏代码

Question

如果输入日期在下面给出，那么日期输出应该是给定的

我希望 sas 宏代码检查输入中缺少的日期

    1.-MMM-YYYY = 01-MMM-YYYY
    --YYYY    = 01-JAN-YYYY 
    DD--YYYY  = DD-JAN-YYYY 
    DD-MMM-   = DD-MMM-0000

Answer 1

我创建了一个数据集，其中包含缺少日、月和年的所有可能情况。下面的数据步骤解决了所有可能性以及如何处理它们以按照您请求的格式创建结果变量 new_bdate。

data have; 
  input birthdate $20.; 
   datalines4;
  -JAN-1975 
  --1977 
  02--1978 
  03-JAN- 
  01-JAN-1980
  03--
  -JAN-
;;;;
run;

data haha2(Keep = birthdate new_bdate);
  set have;  
   format day $2. month $3. Year $4. New_bdate $11.;
   call scan(birthdate, 1, First_Pos, First_Length);
   First_Word = substrn(birthdate, First_Pos, First_Length);
   call scan(birthdate, -1, Last_Pos, Last_Length);
   Last_Word = substrn(birthdate, Last_Pos, Last_Length);
   if length(birthdate) = 11 then length1 = 'Y';
  else length1 = 'N';

   if length1 = 'Y' then do;
      Day = substr(birthdate,1,2);
      Month = substr(birthdate,4,3);
      Year = substr(birthdate,8,4);
   end;
   else if length1 = 'N' then do;
     if (First_Pos = '1' and Last_Length = '2')  then do 
              Day = First_Word;
              Month = 'JAN';
              Year = '0000';
        end;
      else if (First_Pos = '1' and Last_Length = '3')  then do 
              Day = First_Word;
              Month = Last_Word;
              Year = '0000';
        end;
     else if (First_Pos = '1' and Last_Length = '4')  then do 
              Day = First_Word;
              Month = 'JAN';
              Year = Last_Word;
        end;
   if (First_Pos = '2' and Last_Length = '3')  then do 
              Day = '01';
              Month = First_Word;
              Year = '0000';
        end;
      else if (First_Pos = '2' and Last_Length = '4')  then do 
              Day = '01';
              Month = First_Word;
              Year = Last_Word;
        end;
     else if (First_Pos = '3' and Last_Length = '4')  then do 
              Day = '01';
              Month = 'JAN';
              Year = Last_Word;
        end;
    end;
New_bdate = Day||'-'||Month||'-'||Year;
run;

Answer 2

Peeyush， 如果您在问题中输入的列表是完整的，那么下面的代码段是获得您想要的结果的简单方法。我放了一个宏和非宏的 if-then 语句来处理它。

 data inputss;
       input in  $  ;
       cards;
    -MMM-YYYY 
    --YYYY    
    DD--YYYY  
    DD-MMM-   
    ;

data result;
   set inputs;
if in = '-MMM-YYY' then Date_Output = '01-MMM-YYYY';
else if in = '--YYYY' then Date_Output  = '01-JAN-YYYY' ;
else if in = 'DD--YYYY' then Date_Output  = 'DD-JAN-YYYY' ;
else if in = 'DD-MMM-' then Date_Output   = 'DD-MMM-0000';
run;



%macro checkmiss;
    %if  in = '-MMM-YYY' %THEN Date_Output = '01-MMM-YYYY';
    %else %if  in = '--YYYY' %THEN Date_Output  = '01-JAN-YYYY' ;
    %else %if  in = 'DD--YYYY' %THEN Date_Output  = 'DD-JAN-YYYY' ;
    %else %if  in = 'DD-MMM-' %THEN Date_Output   = 'DD-MMM-0000';
run;
%mend checkmiss;
%checkmiss;