# [R] Is there a way to not use an explicit loop?

Dan Davison davison at stats.ox.ac.uk
Wed Sep 17 21:44:49 CEST 2008

```Both shape parameters of rbeta can be vectors; for

x <- rbeta(n, shape1, shape2)

x[i] ~ Beta(shape1[i], shape2[i])

so

bbsim <- function(m=1000, num.post.draws=1e4, size.a=100, prob.a=.27, prior.count=1) {
data.count <- rbinom(m, size.a, prob.a)
shape1 <- rep(prior.count + data.count, each=num.post.draws)
shape2 <- rep(prior.count + size.a - data.count, each=num.post.draws)
matrix(rbeta(m * num.post.draws, shape1, shape2), num.post.draws, m)
}

Then you can do

beta.draws <- bbsim()
means <- apply(beta.draws, 2, mean)
medians <- apply(beta.draws, 2, median)
etc

Dan

On Wed, Sep 17, 2008 at 11:56:36AM -0700, Juancarlos Laguardia wrote:
> I have a problem in where i generate m independent draws from a binomial
> distribution,
> say
>
> draw1 = rbinom( m , size.a, prob.a )
>
>
> then I need to use each draw to generate a beta distribution.  So, like
> using a beta prior, binomial likelihood, and obtain beta posterior, m
> many times.  I have not found out a way to vectorize draws from a beta
> distribution, so I have an explicit for loop within my code
>
>
>
> for( i in 1: m ) {
>
> beta.post = rbeta( 10000, draw1[i] + prior.constant  ,  prior.constant +
> size.a  - draw1[i] )
>
> beta.post.mean[i] = mean(beta.post)
> beta.post.median[i] = median(beta.post)
>
> etc.. for other info
>
> }
>
> Is there a way to vectorize draws from an beta distribution?
>
> UC Slug
>
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