楼上的SQL步骤也可以用以下程序代替，得出相同结果:

自己写的很麻烦的程序,,,唉...望楼主参考
data new(drop=x11-x33);
merge a(in=ina)  b(rename=(x1=x11 x2=x22  x3=x33));            
by id;
if ina then do;
x1=sum(x1,x11);
x2=sum(x2,x22);
x3=sum(x3,x33);
end;
else do;
x1=x11;
x2=x22;
x3=x33;
end;
run;

学习了，不同的解决思路~~

Here are a couple of ways to do it.

data a;
input id $ x1 x2 x3;
cards;
1      8     20   3
3      3     0     8
4      2     4     0
;
data b;
input id $ x1 x2 x3;
datalines;
1    0     7    3
4    9    0     5
5    8    5     9
;

data want(drop=x11 x22 x33);
  merge a b(rename=(x1=x11 x2=x22 x3=x33));
  by id;
  x1=sum(x1,x11);x2=sum(x2,x22);x3=sum(x3,x33);
  run;

proc print;run;

proc sql;
  select coalesce(a.id,b.id) ,sum(a.x1,b.x1) as x1 ,sum(a.x2,b.x2) as x2,sum(a.x3,b.x3) as x3
  from a
  full join b on a.id=b.id
  order by 1
  ;
  quit;