如何从SAS运行得到的众多结果中提取想要的结果并进行分析？

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工作中遇到一个难题，跪求各位高手大侠的帮助，将不胜感激！数据如下；
             x1                    x2

0	0
0	0
0	0.194524181
0	0.193628445
0	0.324881475
0	0.62184472
0	0.620721755
0	0.951667341
0	0.950444341
0	1.125868305
0	0.947999226
0.463629947	0.94677711
0.662146589	1.122072013
1.337229367	1.887472204
1.334961048	3.191768201
1.33269345	4.152501155
1.833266353	5.722478385
1.830842011	8.047885729
3.234993863	12.75743694
4.828810466	16.64131955
6.230338604	19.63580102
8.540719884	22.39520766
10.65538505	25.34358262
12.94672109	27.95276074
15.42105955	30.14475545
16.46129195	30.70560581
17.53262754	32.43165634
19.20164089	34.82412985
19.77129671	35.43639784
20.34921228	37.32067156
22.14289494	38.61075375
21.53021689	39.92961874
22.13351014	36.6747002
22.74547163	37.95048576
22.74073388	39.25470498
23.99583954	41.26761358
23.99100983	40.58462724
23.98618031	41.26070298
23.34707735	41.25724776
24.61984585	40.57433489
23.97169283	41.93735249
23.33272929	41.24688242
23.96203542	41.93039321

步骤是先对x1和x2做相关，得到R值（相关系数）；然后将X2向下错一位得到新的一列n1，再做x1和n1相关，得到R1；比较R和R1，如果R1>R则将n1再错位和x1做相关，如果R1<=R则停止，最后并输出错位次数。

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关键词：不胜感激 相关系数 如何 不胜感激

1. data a;
   
2. input x1  x2;
   
3. cards;
   
4. 0        0
   
5. 0        0
   
6. 0        0.194524181
   
7. 0        0.193628445
   
8. 0        0.324881475
   
9. 0        0.62184472
   
10. 0        0.620721755
    
11. 0        0.951667341
    
12. 0        0.950444341
    
13. 0        1.125868305
    
14. 0        0.947999226
    
15. 0.463629947        0.94677711
    
16. 0.662146589        1.122072013
    
17. 1.337229367        1.887472204
    
18. 1.334961048        3.191768201
    
19. 1.33269345        4.152501155
    
20. 1.833266353        5.722478385
    
21. 1.830842011        8.047885729
    
22. 3.234993863        12.75743694
    
23. 4.828810466        16.64131955
    
24. 6.230338604        19.63580102
    
25. 8.540719884        22.39520766
    
26. 10.65538505        25.34358262
    
27. 12.94672109        27.95276074
    
28. 15.42105955        30.14475545
    
29. 16.46129195        30.70560581
    
30. 17.53262754        32.43165634
    
31. 19.20164089        34.82412985
    
32. 19.77129671        35.43639784
    
33. 20.34921228        37.32067156
    
34. 22.14289494        38.61075375
    
35. 21.53021689        39.92961874
    
36. 22.13351014        36.6747002
    
37. 22.74547163        37.95048576
    
38. 22.74073388        39.25470498
    
39. 23.99583954        41.26761358
    
40. 23.99100983        40.58462724
    
41. 23.98618031        41.26070298
    
42. 23.34707735        41.25724776
    
43. 24.61984585        40.57433489
    
44. 23.97169283        41.93735249
    
45. 23.33272929        41.24688242
    
46. 23.96203542        41.93039321
    
47. ;
    
48. 
    
49. proc corr data=a out=R(where=(_type_='CORR')) noprint;
    
50. var x1 x2;
    
51. run;
    
52. 
    
53. data R(keep=x1 _NAME_);
    
54.   set R end=final;
    
55.   if final then do;
    
56.     call symput('R',put(x1,best.));
    
57.     output;
    
58.   end;
    
59. run;
    
60. 
    
61. %macro prg;
    
62. %let num=1;
    
63. %let lastnum=0;
    
64. %let R1=1;
    
65. 
    
66. %do %while(&R<&R1);
    
67. 
    
68. data a;
    
69.   set a;
    
70.   if &num=1 then n&num=lag(x2);
    
71.   else n&num=lag(n&lastnum);
    
72. run;
    
73. 
    
74. proc corr data=a out=R1(where=(_type_='CORR')) noprint;
    
75. var x1 n&num;
    
76. run;
    
77. 
    
78. data R1(keep=x1 _NAME_);
    
79.   set R1 end=final;
    
80.   if final then do;
    
81.     call symput('R1',put(x1,best.));
    
82.     output;
    
83.   end;
    
84. run;
    
85. 
    
86. data R;
    
87.   set R R1;
    
88. run;
    
89. 
    
90. %let num=%eval(&num+1);
    
91. %let lastnum=%eval(&lastnum+1);
    
92. %end;
    
93. 
    
94. %put "The number of iteration is" %eval(&num-1);
    
95. 
    
96. %mend;
    
97. 
    
98. %prg

复制代码

在数据集R可以找到x1和X2，n1,n2的相关系数

这位高人真是大好人，太感谢了！再次谢谢您！

data crackman;
input x1 x2@;
%num(&num.-1);
cards;
0 0
0 0
0 0.194524181
0 0.193628445
0 0.324881475
0 0.62184472
0 0.620721755
0 0.951667341
0 0.950444341
0 1.125868305
0 0.947999226
0.463629947 0.94677711
0.662146589 1.122072013
1.337229367 1.887472204
1.334961048 3.191768201
1.33269345 4.152501155
1.833266353 5.722478385
1.830842011 8.047885729
3.234993863 12.75743694
4.828810466 16.64131955
6.230338604 19.63580102
8.540719884 22.39520766
10.65538505 25.34358262
12.94672109 27.95276074
15.42105955 30.14475545
16.46129195 30.70560581
17.53262754 32.43165634
19.20164089 34.82412985
19.77129671 35.43639784
20.34921228 37.32067156
22.14289494 38.61075375
21.53021689 39.92961874
22.13351014 36.6747002
22.74547163 37.95048576
22.74073388 39.25470498
23.99583954 41.26761358
23.99100983 40.58462724
23.98618031 41.26070298
23.34707735 41.25724776
24.61984585 40.57433489
23.97169283 41.93735249
23.33272929 41.24688242
23.96203542 41.93039321
;
run;
proc sql;
select count(*) as num into:num from crackman;
quit;
proc corr data=crackman out=a(keep=x1 where=(x1<=1));
var _all_;
run;
proc sql ;
select x1 into:nu from a where _name_='x2';
quit;
data b;
set a;
n=&nu;
m=n-x1;
if m<0 then group+0 ;
else group+1;
obs=_n_-2;
where group=1;
run;
proc print data=b;
run;
%macro num(num);
%do i=1 %to &num.;
%let j=%eval(&i.+2);
x&j.=lag&i.(x2);
%end;
output;
%mend num;

结果里面的X1为R值
OBS为错位次数
是5

5# crackman

方法虽然不尽相同但总是大同小异.

二楼兄弟写的程序比我的好
没有额外的数据集