关于字符串的排列组合

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假设宏变量  %let str=v1 v2 v3;       /*v的个数可能超过3，v1 v2 v3 v4.....vn*/
我要生成几个新的宏变量，每一个新宏变量的值是 v1 v2 v3组合的一种情况，如：
str1=v1
str2=v2
str3=v3
str4=v1 v2
str5=v1 v3
str6=v2 v3

请教下，哪位能有好的，快捷实现的方法，谢谢！！

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关键词：字符串 let chuan 字符串

let me try:

data test;
  input serial var $;
  datalines;
  1 v1
  2 v2
  3 v3
  ;
  run;

proc sql;
    create table tmp as(
    select var as newvar from test union all
    select cat(a.var,b.var) as newvar from test a,test b where  a.serial<b.serial union all
    select cat(a.var,b.var,c.var) as newvar from test a,test b,test c where  a.serial<b.serial and b.serial<c.serial);
quit;

data fstr;
     set tmp;
     serial +1;
run;

proc sql;
   select newvar into :str1 from fstr where serial=1;
   select newvar into :str2 from fstr where serial=2;
   select newvar into :str3 from fstr where serial=3;
   select newvar into :str4 from fstr where serial=4;
   select newvar into :str5 from fstr where serial=5;
   select newvar into :str6 from fstr where serial=6;
quit;

%put &str1;
%put &str2;
%put &str3;
%put &str4;
%put &str5;
%put &str6;

1. Functions allcomb, allperm
http://support.sas.com/documenta ... .htm#a003112302.htm
2. PROC PLAN

PhoenixHuang 发表于 2013-9-22 22:49
let me try:

data test;

嗯，我后来用allcomb, 好像实现起来还比较快捷，你的方法也不错：

data _null_;
   array x[5] $3 ('ant' 'bee' 'cat' 'dog' 'ewe');
   n=dim(x);
   k=3;
   ncomb=comb(n,k);
   do j=1 to ncomb+1;
      call allcomb(j, k, of x

);
      put j 5. +3 x1-x3;
   end;
run;
SAS writes the following output to the log:
    1   ant bee cat
    2   ant bee ewe
    3   ant bee dog
    4   ant cat dog
    5   ant cat ewe
    6   ant dog ewe
    7   bee dog ewe
    8   bee dog cat
    9   bee ewe cat
   10   dog ewe cat
   11   dog ewe cat

邓贵大 发表于 2013-9-23 00:41
1. Functions allcomb, allperm
http://support.sas.com/documentation/cdl/en/lrdict/64316/HTML/default ...

嗯，谢谢，我后来自己找到了这种方法，挺不错的

使用allcomb()方程是非常美妙的程序.在以前我也有此需要,但是没有静下来考虑是否可以用宏来编写.今天决心试一试.

我的程序此处可见
但是,在运行超过计13个字符串的时候,有BUG.也许大家有什么意见.
京剧

jingju11 发表于 2013-9-24 05:14
使用allcomb()方程是非常美妙的程序.在以前我也有此需要,但是没有静下来考虑是否可以用宏来编写.今天决心试 ...

why bother with slower recursions?
There's an iterative approach
http://support.sas.com/techsup/technote/ts498.html

indeed. thanks

邓贵大 发表于 2013-9-24 12:50
why bother with slower recursions?
There's an iterative approach
http://support.sas.com/techsup/ ...

的确如此.
感觉楼主真有一览众山小的意味.无论提出问题还是解决问题,都非常不凡.
值得学习.
京剧

jingju11 发表于 2013-9-25 00:41
的确如此.
感觉楼主真有一览众山小的意味.无论提出问题还是解决问题,都非常不凡.
值得学习.

京大师过誉了，看不出你程序的毛病
但有个笨方法

1. %macro comb(
   
2.         k,
   
3.         words,
   
4.         out=
   
5. );
   
6. %local i;
   
7. %if &k=0 %then %do;
   
8.         data &out;
   
9.                 length x $ 200;
   
10.                 x=' ';
    
11.         run;
    
12. %end;
    
13. %else %if &k=%sysfunc(countw(&words)) %then %do;
    
14.         data &out;
    
15.                 length x $ 200;
    
16.                 x = "&words";
    
17.                 output;
    
18.         run;
    
19. %end;
    
20. %else %do;
    
21.         %local tmpdsn1 tmpdsn2;
    
22.         data;stop;run;        %let tmpdsn1=%substr(&syslast, 6);
    
23.         data;stop;run;        %let tmpdsn2=%substr(&syslast, 6);
    
24.         %comb(%eval(&k-1), %substr(&words, %index(&words, %str( ))), out=&tmpdsn1)
    
25.         %comb(&k, %substr(&words, %index(&words, %str( ))), out=&tmpdsn2)
    
26.         data &out;
    
27.                 set &tmpdsn1(in=in1) &tmpdsn2;
    
28.                 if in1 then x = catx(' ', scan("&words",1), x);
    
29.         run;
    
30.         proc datasets nolist;
    
31.                 delete &tmpdsn1 &tmpdsn2;
    
32.         quit;
    
33. %end;
    
34. %exit:
    
35. %mend;
    
36. 
    
37. options nonotes;
    
38. %comb(0,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
39. proc append base=hell1 data=hell;
    
40. %comb(1,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
41. proc append base=hell1 data=hell;
    
42. %comb(2,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
43. proc append base=hell1 data=hell;
    
44. %comb(3,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
45. proc append base=hell1 data=hell;
    
46. %comb(4,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
47. proc append base=hell1 data=hell;
    
48. %comb(5,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
49. proc append base=hell1 data=hell;
    
50. %comb(6,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
51. proc append base=hell1 data=hell;
    
52. %comb(7,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
53. proc append base=hell1 data=hell;
    
54. %comb(8,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
55. proc append base=hell1 data=hell;
    
56. %comb(9,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
57. proc append base=hell1 data=hell;
    
58. %comb(10,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
59. proc append base=hell1 data=hell;
    
60. %comb(11,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
61. proc append base=hell1 data=hell;
    
62. %comb(12,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
63. proc append base=hell1 data=hell;
    
64. %comb(13,one two three four five six seven eight nine ten 11 12 13,out=hell)
    
65. proc append base=hell1 data=hell;
    
66. proc print data=hell1;
    
67. options notes;

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