an inquiry

[PeterClemmensen]

Alternatively you can do this. It is not as efficient, but it ensures that every relevant year is actually present for each firm.  

data have;
length code $10;
code='one';
do year=2000 to 2015;
   cashflow=rand('integer', 1000, 10000);
   output;
end;
code='two';
do year=2000 to 2013;
   cashflow=rand('integer', 1000, 10000);
   output;
end;
code='three';
do year=2000 to 2015;
   cashflow=rand('integer', 1000, 10000);
   output;
end;
run;

proc sort data=have;
   by code;
run;

proc transpose data=have out=temp prefix=year;
    by code;
    id year;
    var cashflow;
run;

data want(keep=code year cashflow);
   set temp;
   array years{*} year2000-year2015;
   if nmiss(of years[*])=0 then do i=1 to dim(years);
      cashflow=years[i];
      year=input(compress(vname(years[i]), '0123456789', 'k'), 8.);
      output;
   end;
run;