Hello,
I'm glad it helped.
Inline.
Em 27-08-2012 19:00, Zuofeng Shang escreveu:
Hi Rui,
Many thanks for your excellent code! That works very well. But I still
have a kind of naive question about the set.seed(1)
Consider shape1=c(1,3) and shape2=c(2,4)
Using R I got that
> set.seed(1)
> rbeta(2,c(1,3),c(2,4))
[1] 0.5803922 0.4679171
> set.seed(1)
> c(rbeta(1,1,2),rbeta(1,3,4))
[1] 0.5803922 0.4679171
They have the same outputs. So rbeta(2,shape1,shape2) is the correct
solution which is equivalent to operating rbeta() componentwise to
shape1 and shape2. This is exactly the solution I wanted. Thanks a lot!
But I also found the following:
set.seed(1)
> rbeta(1,1,2)
[1] 0.5803922
> set.seed(1)
> rbeta(1,3,4)
[1] 0.3016368
This is pretty strange. Why is the second one 0.3016368 instead of
being 0.4679171? Probably I was wrong somewhere here.
No, you are not wrong, and no, it is not strange. You are restarting the
random number generator so your _first_ random beta number cannot be
eqaul to the _second_ one. Just look:
set.seed(1)
rbeta(1,3,4) # 0.3016368
set.seed(1)
rbeta(2,3,4) # 0.3016368 0.4679171
The second is, as expected, what you had. When yo change the beta
parameters, the numbers must also change (obvious!)
Rui Barradas
The binomial code works very well. Thanks for this very much!
Have a nice day.
Jeff
δΊ 8/27/2012 1:25 PM, Rui Barradas ει:
Hello,
Try the following.
set.seed(1)
X <- rbeta(2*length(shape1), rep(shape1, each = 2), rep(shape2, each
= 2))
X <- matrix(X, ncol = 2, byrow = TRUE)
#-- binomial
n <- 10
p <- runif(10000)
Y1 <- matrix(nrow = 10000, ncol = 10)
set.seed(6292)
for(i in 1:10000){
Y1[i, ] <- rbinom(n, 2, p[i])
}
set.seed(6292)
Y2 <- rbinom(n*length(p), 2, rep(p, each = 10))
Y2 <- matrix(Y2, ncol = n, byrow = TRUE)
identical(Y1, Y2) #TRUE
The trick is not to use argument recycling.
Hope this helps,
Rui Barradas
Em 27-08-2012 15:49, JeffND escreveu:
Hi folks,
I have a question about how to efficiently produce random numbers
from Beta
and Binomial distributions.
For Beta distribution, suppose we have two shape vectors shape1 and
shape2.
I hope to generate a 10000 x 2 matrix X whose i th rwo is a sample from
reta(2,shape1[i]mshape2[i]). Of course this can be done via loops:
for(i in 1:10000)
{
X[i,]=rbeta(2,shape1[i],shape2[i])
}
However, the above code is time consuming. It would be better to
directly
generate X without any loops. I tried the following code
X<- rbeta(2,shape1,shape2)
but it looks not right. So I was wondering if there is an R code
doing this.
For Binomial distribution, I have a similar question. I hope to
generate a
10000 x n matrix Y whose i th row is a sample from rbinom(n,2,p[i]),
where p
is a vector of binomial probabilities. We can also do it by loops
for(i in 1:10000)
{
Y[i,]=rbinom(n,2,p[i])
}
But it would be nice to do it without using any loops. Is this
possible in
R?
Many thanks for great help and suggestions!
Jeff
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