You can cut execution time by a factor 2 simply using the fact that the
double summation is symmetric in the indices j and k:
2 * sum(sapply(1:(m-1), function(k){sum(sapply((k-1):m,
function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))}))
+ sum(x^2*dnorm((2*mu)/sqrt(2*sig))/ sqrt(2*sig))
Best,
Giovanni Petris
> Date: Thu, 28 Aug 2008 21:47:38 -0700 (PDT)
> From: kathie <kathryn.lord2000 at gmail.com>
> Sender: r-help-bounces at r-project.org
> Precedence: list
>
>
> Dear R users...
>
> I made the R-code for this double summation computation
>
>
> http://www.nabble.com/file/p19213599/doublesum.jpg
>
> -------------------------------------------------
> Here is my code..
>
>
> sum(sapply(1:m, function(k){sum(sapply(1:m,
>
function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))}))
>
>
> -------------------------------------------------
>
> In fact, this is a part of optimization. I think if it is changed more
> efficiently, then the running time could be shortened.
>
> How could I change this to more efficiently? Any suggestion will be greatly
> appreciated.
>
> Kathryn Lord
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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>
--
Giovanni Petris <GPetris at uark.edu>
Associate Professor
Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)
http://definetti.uark.edu/~gpetris/