Is there a way to not use an explicit loop?

Hi,

you might try this:

set.seed(100)

m         <- 10
size.a    <- 10
prob.a    <- 0.3
prior.constant = 0

draw1 = rbinom( m , size.a, prob.a )

beta.draws <- function(draw, size.a, prior.constant, n) {
	rbeta(n, prior.constant + draw, prior.constant + size.a - draw)
}

bdraws <- sapply(draw1, beta.draws, size.a = size.a, prior.constant =
prior.constant, n = 10000)
beta.post <- apply(bdraws, 2, function(x) c(post.mean = mean(x),
post.median = median(x)) )
beta.post
                 [,1]      [,2]      [,3]       [,4]      [,5]
[,6]      [,7]      [,8]
post.mean   0.2017118 0.1996809 0.2991173 0.10069613 0.3001924
0.2991149 0.4033310 0.2003104
post.median 0.1804893 0.1791630 0.2845427 0.07505278 0.2858155
0.2844503 0.3961419 0.1790511
                 [,9]     [,10]
post.mean   0.3013020 0.1990232
post.median 0.2886199 0.1786447



best


V?ctor H Cervantes




2008/9/17 Juancarlos Laguardia <brassman785 at gmail.com>:

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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