proc sql noprint;

create table result

        as select * from (

                select distinct a.id,b.id as id_match,a.old_value,b.old_value as new_value

                        from test as a,test as b

                        where b.old_value ne .

                        group by a.id

                        having a.id=b.id or (abs(a.id-b.id)=min(abs(a.id-b.id)) ))

        group by id

        having id_match=min(id_match);

quit;

如果数据比较大，不知这个速度如何

data ffile;
input obs x;
cards;
1 .
2 566
3 .
4 .
5 .
6 .
7 .
8 .
9 .
10 .
11 998
12 .
13 .
14 223
15 .
16 .
17 .
18 87
19 .
20 56
21 .
22 .
23 .
24 .
25 .
;
run;
proc print;run;

data ffile1;
set ffile;
order=_n_;
x_origin=x;
retain ord_1 0;
if missing(x) then ord_1=ord_1+1;else ord_1=0;
run;

proc sort data=ffile1 ;
by  descending order;
run;

data ffile1;
set ffile1;
retain ord_2 0;
if missing(x) then ord_2=ord_2+1;else ord_2=0;
run;

proc sort data=ffile1 ;
by  order;
run;

data ffile2;
set ffile1;
if ord_1<=ord_2 and ord_1^=0 then inter=1;
if ord_1>ord_2  then inter=2;
run;

data ffile3;
set ffile2;
retain value;
if x^=. then value=x;
if inter=1 then x=value;
drop value;
run;

proc sort data=ffile3 ;
by  descending order;
run;

data ffile4;
set ffile3;
retain value;
if x^=. then value=x;
if inter=2 then x=value;
drop value;
run;

data ffile5;
set ffile4;
retain value;
if x^=. then value=x;
if x=. then x=value;
drop value;
run;

proc sort data=ffile5 ;
by  order;
run;

data ffile6;
set ffile5;
retain value;
if x^=. then value=x;
if x=. then x=value;
drop value;
run;

proc print ;run;

问题解决了，楼主再简单调下就行了

Dear my friend,I will try, and thank you for your reply.

今天用第二种方法跑了一遍，计算量确实很大，LOG平均2分钟满一次屏

这不就是前一个帖子的答案吗