在相同的种子基础上，复制在R中产生的随机的正常生成(rancor) ?

Given the same seed, is there a way to produce the exact same random normal numbers generated in SAS, using the rannor function, in R?

给定相同的种子，是否有办法在R中使用rannor函数生成在SAS中生成的完全相同的随机数?

3 个解决方案

#1

To make these match, you need two things:

要使这些匹配，你需要两样东西:

The seed used to generate the random number
用来产生随机数的种子
The formula used to generate the random number
用来生成随机数的公式

SAS uses for rannor (and I think also for rand, but I haven't seen confirmation of this), the following algorithm (found in Psuedo-Random Numbers: Out of Uniform, by Robert Johnson and Hui Liu):

SAS用于rannor(我认为也适用于rand，但我没有看到对此的证实)，下面的算法(在Psuedo-Random Numbers: Out of Uniform, Robert Johnson and Hui Liu著):

rannor_result = (−2* log(U1))**.5*cos(2*constant('pi')*U2)

where U1 and U2 are two numbers from the uniform number stream. (A second number could also be obtained, but that number is discarded as far as I can tell.)

U1和U2是来自均匀数流的两个数。(也可以得到第二个数字，但据我所知，这个数字被丢弃了。)

See the following SAS datastep:

参见下面的SAS datastep:

data test1;
  U1 = ranuni(7);
  U2 = ranuni(7);
  X1 = (−2* log(U1))**.5*cos(2*constant('pi')*U2);
  U1 = ranuni(7);
  U2 = ranuni(7);
  X2 = (−2* log(U1))**.5*cos(2*constant('pi')*U2);
run;
data test2;
  x1 = rannor(7);
  x2 = rannor(7);
run;

test1 and test2 have identical values for the rannorm numbers.

test1和test2对于rannorm数有相同的值。

I doubt that R and SAS share common PRNG algorithms, particularly as the one rannor uses is not very good (you should use rand which is much, much better, using a Mersenne Twister). However, you can easily ask SAS to output the seeds it used - as long as you output the RANUNI results.

我怀疑R和SAS是否共享常见的PRNG算法，特别是rannor使用的那个算法不是很好(您应该使用rand，使用Mersenne Twister会更好)。但是，您可以很容易地要求SAS输出它使用的种子——只要您输出RANUNI结果。

You can ask SAS to do that in this manner:

你可以让SAS这样做:

data rands_uni;
  seed=7;
  do _i_=1 to 10;
    seed1 = seed;
    call ranuni(seed,x1);
    seed2=seed;
    call ranuni(seed,x2);
    output;
  end;
run;

From that you could then in R calculate the rannor result (ie, from x1 and x2). I include the seeds there for reference - perhaps R does have the ability to use them. The above paper does reference the algorithm SAS used for ranuni, but it also states that you cannot replicate it perfectly due to some corrections used (or, perhaps, due to floating point precision issues?).

然后你可以用R计算rannor结果(即，从x1和x2)。我把种子包括在那里作为参考——也许R确实有能力使用它们。上面的文章确实引用了用于ranuni的算法SAS，但是它也指出，由于使用了一些修正(或者，可能是由于浮点精度问题)，您无法完美地复制它。

#2

I test both, it seem it did not.

我两项都考了，好像没有。

data _null_;
CALL STREAMINIT(1245);
do i=1 to 10;
x2 = rand('NORMAL',0,1);
put x2;
end;
run;
/*
-1.295643934
0.0651085355
-0.391011151
1.1024205822
2.0712099724
-0.296163515
0.898941256
-0.584950901
0.2027164546
-0.986522152*/

set.seed(1245)
df<-rnorm(10,0,1)
df
 [1]  0.41914007 -1.45583361 -1.45588008 -1.01564637 -0.53309914 -0.68360608  0.32590880 
 [8]  0.92256137 -0.05085288  0.29722361

#3

Proc IML;
Call RandSeed(6);
U1 = J(6,1);
U2 = J(6,1);
/* Weibull Distribution */
Call RandGen(U1,"WEIBULL",1.7737152,13.164695);
Print "Weibull Distribution:" U1;
/* Unifrom Distribution */
Call RandGen(U2,"UNIFORM",0,1);
Print "Uniform Distribution:" U2;

Submit/R;
    set.seed(6);
    ## Weibull Distribution;
    U1 <- rweibull(6,1.7737152,13.164695);
    cat("Weibull Distribution:", U1, "\n");
    ## Uniform Distribution;
    U2 <- runif(6,0,1);
    cat("Unifrom Distribution:", U2, "\n");
EndSubmit;

Quit;

#1