SAS变量迭代计算如何实现？

比如：一个变量xt,
xt=lag3(xt)+m;
赋予x初始值，x0=1,x1=2,x3=3,x4=x0+m,以此类推
请问这个如何解决？

1. %let m = 10;
   
2. 
   
3. data test;
   
4.   retain x0 x1 x2 (1 2 3);
   
5.   do i = 1 to 20;
   
6.     if i in (1,2,3) then xt = i;
   
7.         else do;
   
8.       if mod(i,3) = 1 then do;
   
9.             xt = x0 + &m;
   
10.         x0 = xt;
    
11.           end;
    
12.           else if mod(i,3) = 2 then do;
    
13.             xt = x1 + &m;
    
14.             x1 = xt;
    
15.           end;
    
16.           else if mod(i,3) = 0 then do;
    
17.             xt = x2 + &m;
    
18.             x2 = xt;
    
19.           end;
    
20.         end;
    
21.     output;
    
22.   end;
    
23.   drop i x0 x1 x2;
    
24. run;

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这很容易！有点像dynamic array！

1. %let m=25;
   
2. data tmp;
   
3.    array n(*) x1-x&m;
   
4.    do i=1 to &m;n[i]=i*1;end;
   
5.    keep x1-x&m;
   
6. proc print;run;

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what do you want to talk about?I am confused by your answer.
what i want to say is not what you try to solve.
please check it again.it seems like autoregression,AR process.
who can use LAGn to solve it?

3# 郑小知 what u want to say?i don't know.

lag3(xt)＝log(xt)/log3

用循环语句以及retain应该就能实现迭代计算，由于不是很理解你的问题，没法给你程序

7# guoluo
谢谢高手！最佳答案非你莫属了。你明白我的意思。就是一个AR过程，引用变量滞后n期的数值，请问LAGn函数是否可以用得上？