data have;
input a b c;
datalines;
1 . 2
. 3 4
6 7 .
8 9 10
11 12 13
;
proc transpose data=have out=have_trans;
run;

%macro mymacro(dsn=, out=);
data &out;
     set &dsn;
         array col{5};
         do i = 1 to dim(col);
               if col(i) = . then do;
                  do j = i to dim(col)-1;
                     col(j) = col(j+1);
                  end;
           end;
         end;
drop i j;
run;
%mend mymacro;
%mymacro(dsn=have_trans, out=have_out);

proc transpose data=have_out out=results (drop=_name_);
run;

proc iml;
use out.case;
read all var _char_ into caseiml;
do i=1 to nrow(caseiml)-1;
  do j=1 to ncol(caseiml);
    if missing(caseiml[i,j]) then do;
          do k=1 to nrow(caseiml);
            if not missing(caseiml[i+k,j]) then do;
          caseiml[i,j]=caseiml[i+k,j];caseiml[i+k,j]="";
                end;
          end;
    end;
  end;
end;

create out.examm from caseiml;
append from caseiml;

quit;

我尝试用IML做了一下，没成功，关键有个问题，就是如果有连续缺失的就依然有问题了，例如

1 2
. 3
. 4
5 .
6 7

怪我没把问题考虑清楚，如果这个连续问题到晚上还没解决，我采纳楼上仁兄的回答。

submit the macro again
%mymacro(dsn=have_out, out=have_out_1);

then run the proc transform use the new result dataset have_out_1.

If there are five 连续 missing, then submit the macro five times with the same input and output dataset name. To see the result of each time you submit the macro,  you can change the input and output dataset every time as above.

忘了说， 所有的变量必须是数值变量。有＿character＿ 好象不行。你可以看一下 transpose 的结果。

已经用set，missing，merge解决啦；谢谢大家