SAS程序运行慢，求高手改进

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下面的程序运行的超慢 ，各位高手能否帮我看看，是不是有可以改进的地方，在此多谢了。
libname DATA 'c:\SASDATA';
data data.ab;
a=-0.618;
b=1.618;
v=1+a*a;
u=1+(1-a)**2;
t=1+b*b;
s=1+(b-1)**2;
la=2*log(v/u);
lb=2*log(t/s);
c=la/log(1+a*a);
d=lb/log(1+b*b);
keep c d;
run;
options cmplib=sasuser.funcs;
proc fcmp outlib=sasuser.funcs.math  ;
  function lstar(u,c,d);
    if (u le  -0.618) then return (c*log(1+u*u));
    if (u ge  1.618) then  return (d*log(1+u*u));
else
      return(2*log((1+u*u)/(1+(u-1)*(u-1))));
endsub;run;
proc fcmp outlib=sasuser.funcs.statistics;
  function g(u);
    if (u le  -0.25) then return (2*log((1+0.25*0.25)/(1+1.25*1.25)));
    if (u ge  1.25) then  return (2*log((1+1.25*1.25)/(1+0.25*0.25)));
else
      return(2*log((1+u*u)/(1+(u-1)*(u-1))));
endsub;run;
proc printto log=_null_;
run;
options cmplib=sasuser.funcs;
%macro arim(r);
%do m=1 %to &r;
data data.lstar&m;
set ab;
do i=1 to 2000;
x=rand('T',3);
y=x/sqrt(3);
lstar=lstar(y,c,d);
l=g(y);
output;
end;
run;
data data.ci&m;
set data.lstar&m;
retain sum1 0;
sum1=sum1+lstar;
if (sum1 ge 0) then sum1=sum1;
else sum1=0;
output;keep i sum1;
run;
data data.si&m;
set data.lstar&m;
retain sum2 0;
sum2=sum2+l;
if (sum2 ge 0) then sum2=sum2;
else sum2=0;
output;keep i sum2;
run;
%end;
%mend arim;
%arim(10000);
proc printto log=_null_;
run;
libname data 'c:\SASDATA';
%macro ar(u);
%do n=1 %to &u;
%macro arim(r);
%do m=1 %to &r;
data shifts&m&n;
set data.ci&m;
if (sum1 ge (2.01+0.01*(&n-1))) then output;
run;
data shift&m&n;
set shifts&m&n;
if _n_=1 then output;keep i;
run;
data drifts&m&n;
set data.si&m;
if (sum2 ge (2.01+0.01*(&n-1))) then output;
run;
data drift&m&n;
set drifts&m&n;
if _n_=1 then output;keep i;
run;
%end;
%mend arim;
%arim(10000);
%macro test;
data sumshift&n;
set %do m=1 %to 10000;
shift&m&n
%end;
;
data sumdrift&n;
set %do m=1 %to 10000;
drift&m&n
%end;
;
%mend test;
%test;
%end;
%mend ar;
%ar(10);

%macro ar(u);
%do n=1 %to &u;
proc means data=sumshift&n;
var i;
output out=ci_shift&n(drop =_freq_ _type_)mean=ARL;
run;
proc means data=sumdrift&n;
var i;
output out=si_drift&n(drop =_freq_ _type_)mean=sARL;
run;
%end;
%mend ar;
%ar(10);
data ci_mean;
set ci_shift1-ci_shift10;
run;
data si_mean;
set si_drift1-si_drift10;
run;
data h;
do h=2.01 to 3 by 0.01;
output;end;
run;
libname SASDATA 'c:\SASDATA1';
data SASDATA.RLCUSUM6;
merge ci_mean si_mean h;
run;
proc gplot data=SASDATA.RLCUSUM6;
plot ARL*h=1 sARL*h=2/overlay;
symbol1 i=none v=dot;
symbol2 i=none v=dot c=black;
run;
proc gplot data=SASDATA.RLCUSUM6;
plot ARL*h=1 sARL*h=2;
symbol1 i=join v=none;
symbol2 i=join v=none;
run;
proc gplot data=SASDATA.RLCUSUM6;
plot ARL*h=1 sARL*h=2/overlay;
symbol1 i=join v=none;
symbol2 i=join v=none;
run;

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关键词：sas程序 求高手 Statistics statistic function function return

1. libname DATA 'c:\SASDATA';
   
2. proc printto log=_null_; run;
   
3. 
   
4. data data.ab;
   
5.   a=-0.618;
   
6.   b=1.618;
   
7.   v=1+a*a;
   
8.   u=1+(1-a)**2;
   
9.   t=1+b*b;
   
10.   s=1+(b-1)**2;
    
11.   la=2*log(v/u);
    
12.   lb=2*log(t/s);
    
13.   c=la/log(1+a*a);
    
14.   d=lb/log(1+b*b);
    
15.   keep c d;
    
16. run;
    
17. 
    
18. proc fcmp outlib=sasuser.funcs.math_stat;
    
19.   # not clear to me how it works
    
20.   function lstar(u,c,d);
    
21.     if (u le  -0.618) then return (c*log(1+u*u));
    
22.     if (u ge   1.618) then return (d*log(1+u*u));
    
23.       else  return(2*log((1+u*u)/(1+(u-1)*(u-1))));
    
24.   endsub;
    
25.   function g(u);
    
26.     if (u le  -0.25) then return (2*log((1+0.25*0.25)/(1+1.25*1.25)));
    
27.     if (u ge   1.25) then return (2*log((1+1.25*1.25)/(1+0.25*0.25)));
    
28.       else return(2*log((1+u*u)/(1+(u-1)*(u-1))));
    
29.   endsub;
    
30. run;
    
31. 
    
32. options cmplib =sasuser.funcs;
    
33. 
    
34. %let r =10000;
    
35. data data.lstar;
    
36.   call streaminit(1234); # for reproduce results
    
37.   set ab;
    
38.   do sample =1 to &r;
    
39.     call missing(sum1, sum2);
    
40.     do i =1 to 2000;
    
41.       x     =rand('T',3)  ;
    
42.       y     =x/sqrt(3)    ;
    
43.       lstar =lstar(y,c,d) ;
    
44.       l     =g(y)         ;
    
45.       sum1 ++lstar; sum2 ++l;
    
46.       if (sum1 lt 0) then sum1=0;
    
47.       if (sum2 lt 0) then sum2=0;
    
48.       output;
    
49.     end;
    
50.   end;
    
51.   keep sample i sum1 sum2;
    
52. run;
    
53. 
    
54. data shifts_drifts;
    
55.   set data.lstar;
    
56.   by sample;
    
57.   array _n[10] _temporary_ (1 2 3 4 5 6 7 8 9 10);
    
58.   array _s[10] _temporary_;  array _d[10] _temporary_;
    
59.   if first.sample then call missing(of _s[*] _d[*]);
    
60.   do k =1 to dim(_n);
    
61.     if _s[k] <1 then if (sum1 ge (2.01+0.01*(_n[k]-1))) then do;
    
62.       _s[k] =1; n =_n[k]; cat ='shifts';
    
63.       output shifts_drifts(keep =sample n i cat);
    
64.     end;
    
65.     if _d[k] <1 then if (sum2 ge (2.01+0.01*(_n[k]-1))) then do;
    
66.       _d[k] =1; n =_n[k]; cat ='drifts';
    
67.       output shifts_drifts(keep =sample n i cat);
    
68.     end;
    
69.   end;
    
70. run;
    
71.   
    
72. proc means data=shifts_drifts;
    
73.   class n cat;
    
74.   var i;
    
75.   output out=ci_shifts_drifts(drop =_freq_ _type_) mean=ARL_sARL;
    
76. run;

复制代码

没有SAS无法测试。也可能没有完全理解你的程序（程序实在太长了）。我想小错误肯定有，自己修改。我的理解是，在模拟数据的时候，使用macro的效率几乎总是最差的。请注意：你的fcmp的逻辑不知道是否合理（是你想要的？）。
京剧

That is what I am doing:
1. Sampling and calculating lstar and l 2000 times;
2. Accumulating lstar and l to get sum1 and sum2 while adjusting them to 0 when their values are negative;
3. Using a threshold f(n), n =1, 2, 3, ... to fetch the first observation number where y >= f(n) was satisfied; when y =sum1, then called shifts, sum2 called drifts. Repeat n from 1 to 10, i.e. f(1), f(2), ...
4. Almost sure that each sample can return an qualified observation number since the sample size (=2000) is fairly large;
5. repeating above 1-4 10000 times (&r.), you may get 10000*10(=n) shifts and drifts each;
6. finding the shifts and drifts means for each n;
7. you will get 10 different (maybe) means for  each drifts and shifts;
8. plot the means for mean vs. h (from 2.00 to 2.09 by 0.01).
By the way, the FCMP may be right for you. I miss the return function at first time.
some typo in the code, such as to drop the keep= in separate output statement, etc.
And that what i got. Then run-tme is about 50s. Jingju

对于每一个n，我是想将ci1，... ...，ci10000中的那些大于或者等于2.01+0.01*（n-1）的那些sum1以及对应的i
存放到数据集shifts1n，... ...，shifts10000n，然后再分别输出输出shifts1n，... ... , shifts10000n中的第一个观测对应得i，将这些i 存放到shift1n，... ... ，shift10000n，再将这10000个数据集合并，并求i 得均值。我感觉我在这里搞得复杂了，程序应该就是慢在这里。


另外proc fcmp 应该没错

jingju11 发表于 2013-4-3 10:52
没有SAS无法测试。也可能没有完全理解你的程序（程序实在太长了）。我想小错误肯定有，自己修改。我的理解是 ...

54.data shifts_drifts;
55.  set data.lstar;

56.  by sample;

57.  array _n[10] _temporary_ (1 2 3 4 5 6 7 8 9 10);

58.  array _s[10] _temporary_;  array _d[10] _temporary_;

59.  if first.sample then call missing(of _s

_d

);

60.  do k =1 to dim(_n);

61.    if _s[k] <1 then if (sum1 ge (2.01+0.01*(_n[k]-1))) then do;
62.      _s[k] =1; n =_n[k]; cat ='shifts';
63.      output shifts_drifts(keep =sample n i cat);
64.    end;

65.    if _d[k] <1 then if (sum2 ge (2.01+0.01*(_n[k]-1))) then do;
66.      _d[k] =1; n =_n[k]; cat ='drifts';

67.      output shifts_drifts(keep =sample n i cat);
68.    end;

69.  end;
70.run;




这一段程序我看不懂，能否逐字逐句句帮忙解释下，的多谢您了，我是个菜鸟。另外call streaminit(1234); 括号里面的1234是随便选的吗，
call missing(sum1, sum2);是什么意思

And that what i got. Then run-tme is about 50s. Jingju

call missing(sum1, sum2);是把 sum1,sum2置为缺失值。

call missing(sum1, sum2);是将sum1,sum2置为缺失。

jingju11 发表于 2013-4-3 10:52
没有SAS无法测试。也可能没有完全理解你的程序（程序实在太长了）。我想小错误肯定有，自己修改。我的理解是 ...

太感谢您了！真的不知道还能再说出什么样的感谢的话来，再次感谢！

webgu 发表于 2013-4-3 22:33
call missing(sum1, sum2);是把 sum1,sum2置为缺失值。

309    do k =1 to dim(_n);
310
311      if _s[k] <1 then if (sum1 ge (2.01+0.01*(_n[k]-1))) then do;
312        _s[k] =1; n =_n[k]; cat ='shifts';
313        output shifts_drifts(keep =sample n i cat);
                               -
                               22
                               76
ERROR 22-322: 语法错误，期望下列之一: 名称, 带引号的字符串, ;, RC, _DATA_, _LAST_, _NULL_.

ERROR 76-322: 语法错误，语句将被忽略。

314      end;
上面的程序在SAS中运行后，出现以下结果，您能否帮忙查下，多谢您了！

jingju11 发表于 2013-4-3 10:52
没有SAS无法测试。也可能没有完全理解你的程序（程序实在太长了）。我想小错误肯定有，自己修改。我的理解是 ...

这个图我用proc gplot没有画出来，求您指教。