为什么rantbl()对cards里的分数概率不认识？

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程序如下：data e;       
input p1 p2;
do i=1 to 16;
a=rantbl(100,p1,p2);
end;
cards;
1/2 1/2
;
run;

日志里显示数据行“有对“p1”无效的数据”，不读。
但是直接用 a=rantbl(100,1/2,1/2);就可以。
因为我的概率有40多项，所以想用cards输入，请问高手这个怎么解决呢？

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关键词：rantbl cards card RAN Ant 程序

问题不在rantbl,而是datalines/cards不认识这种分数表示。。那该怎么表示啊？

墨岚 发表于 2013-5-2 16:21
问题不在rantbl,而是datalines/cards不认识这种分数表示。。那该怎么表示啊？

One more reason for the problem, is the function won't take characters as its arguments although you can read the fractions from cards. In fact, a more elegant way to do that is to load/write the values to a macro variable. That will give you a lot more flexibility on programing. Jingju

1. data have;
   
2.         input (frac1-frac4) (4* $ :12.);
   
3.         array p[4] _temporary_; array frac[4];
   
4.         do i =1 to 4;
   
5.                 if index(frac[i], '/') then p[i] =scan(frac[i],1, '/')/scan(frac[i],2, '/');
   
6.                 else p[i] =frac[i] +0;
   
7.         end;
   
8.         do i =1 to 100;
   
9.                 a =rantbl(123,of p[ * ]); output;
   
10.         end;
    
11. cards;
    
12. .1 1/2 1/3 1/15
    
13. ;

复制代码

jingju11 发表于 2013-5-3 08:11
One more reason for the problem, is the function won't take characters as its arguments although y ...

Thanks! The idea that input fractions as characters solves my problem indeed('_temorary_' is very useful!). But I recognize code like "a =rantbl(123,.1,1/2,1/3,1/15);" can be read as well .
By the way, can rantbl() generate several radom numbers, instead of one number at a time? I need different numbers in one repeat...

Yeah, it takes fractions (1/2)but not the character ('1/2').
I don't think think so. as you know, sas function returns one value at a time. If you want to have multiple columns, just insert one more loop, like:

1. array t[3] a  b c;
   
2. do i =1 to 100;
   
3.    do j =1 to 3;
   
4.       t[i] =rand('tabled', of p{*});
   
5.    end;
   
6.    output;
   
7. end;

复制代码

jingju11 发表于 2013-5-3 10:31
Yeah, it takes fractions (1/2)but not the character ('1/2').
I don't think think so. as you know,  ...

That's what I did.. Unfortunately, these three numbers are in fact three samples but not three numbers in one sample which is preferred. Still thank u for the confirmation!

random numbers not distinguished by samples. i tend to understand your samples as different datasets. using output in different positions will give you different forms of data...

jingju11 发表于 2013-5-3 10:56
random numbers not distinguished by samples. i tend to understand your samples as different datasets ...

比如现在有一个20项的多项分布，按照这个分布取得一批随机数不是我的最终目的，这批随机数实际上是我想做出的另一个分布的一个样本，因为这个样本是符合前面的多项分布的。每次抽样就是按照这个分布抽取固定数目的随机数比如10个，因为模型的原因这10个数不能重复，类似于一次不放回的抽样，所以我想问可否一次抽取多个不重复的随机数..rantbl的sas帮助里也没有看到这方面的说明.

There is a tricky way to avoid your input fraction problem. It may also give you more options to programm with array or loop.

Or just type a few more letters and numbers, it could save your time working out how the macro or array, or loops could make the program looks pretty.
data e;
input n1 d1 n2 d2 n3 d3 n4 d4 n5 d5;
     p1 = n1/d1;
         p2 = n2/d2;
         p3 = n3/d3;
         p4 = n4/d4;
         p5 = n5/d5;
         do i=1 to 16;
    a=rantbl(100,p1,p2,p3,p4,p5);
        output;
        end;
datalines;
1 2 1 3 1 4 1 5 1 6
;

yongyitian 发表于 2013-5-3 12:09
There is a tricky way to avoid your input fraction problem. It may also give you more options to pro ...

Tricky indeed.