data a;
  set a;
  obs = _N_;
run;

proc sql noprint;
   create table c(drop = obs) as
     select a.name,b.v1, b.v2, a.v3, a.v4, a.obs
     from a left join b
     on a.name = b.name
    order by obs ;
quit;
   
  

代码跑出来的结果是这样的：

name	v1	v2	v3	v4
hirs1681	1	32	23	23
hirs1681	2	22	23	23
hirs1681	3	1	23	23
hirs1681	3	1	443	21
hirs1681	2	22	443	21
hirs1681	1	32	443	21
hirs1681	3	1	21	22
hirs1681	1	32	21	22
hirs1681	2	22	21	22
orissa			4	1
Purepixe	4	34	34	34
virtuoso	5	56	33	34
virtuoso	8	1	33	34
ndokcepl	6	64	21	22
frontend	7	32	24	1
virtuoso	5	56	3	4
virtuoso	8	1	3	4

不是我说顺序合并。
我感觉那种double set可能会管用，但自己还不会写这种code

我没有仔细看书，尝试下列代码：
data a;
  set a;
  obs = _N_;
run;

proc sql noprint;
   create table c as
     select a.name,b.v1, b.v2, a.v3, a.v4, a.obs
     from a left join b
     on a.name = b.name
    order by obs ;
quit;

data c(drop = obs);
  set c; by obs;
  if first.obs then output;
run;

结果，好像还是不对

好吧，仔细研究了一下，双set也不好实现。
data a;
  set a;
  obs = _N_;
run;

proc sort data =a; by name; run;

data a;
        set a; by name;
        if first.name then vobs = 0;
    vobs + 1;
run;

proc sort data = b; by name; run;
data b;
        set b; by name;
        if first.name then vobs = 0;
    vobs + 1;
run;

proc sql noprint;
   create table c as
     select a.name,b.v1, b.v2, a.v3, a.v4, a.obs
     from a left join b
     on a.name = b.name and a.vobs = b.vobs
    order by obs ;
quit;

data c(drop = obs);
  set c; by obs;
  if first.obs then output;
run;

把a先去掉观测4，然后merge，再加上观测4，再恢复顺序，也可以实现

用merge实现：

data a;
  set a;
  obs = _N_;
  if name = 'orissa' then obs1 = 999;
  else obs1 = obs;
run;

proc sort data = a ; by obs1; run;


data c;
  merge b a;
  drop obs1;
run;

proc sort data = c out= d(drop = obs); by obs; run;

利用双set过程实现：

data a(index=(namev = (name vobs)));
  set a;
  length namelst $2000;
  retain namelst;
  namelst = catx('', namelst, name);
  vobs = count(namelst, name, 't');
  drop namelst;
run;

data b(index=(namev = (name vobs)));
  set b;
  length namelst $2000;
  retain namelst;
  namelst = catx('', namelst, name);
  vobs = count(namelst, name, 't');
  drop namelst;
run;

data c;
  retain name variable1-variable4;
  set a ;
  set b key = namev;
  if _iorc_ eq 0 then do;
     _iorc_=0;
     _error_=0;
  end;
  else do;
    _iorc_=0;
    _error_=0;
    variable1 = .;
    variable2 = .;
  end;
  drop vobs;
run;

谢谢您的回复，程序完美运行。

谢谢您的帮助！！！！

不客气。你有什么经典的笑话吗？来一个？谢谢！

帮我看看这个问题吧，多谢多谢！https://bbs.pinggu.org/thread-3604873-1-1.html