Sorry about that; s <- 1
-----Original Message-----
From: Gabriela Cendoya [mailto:gabrielacendoya.rl...@gmail.com]
Sent: Monday, October 04, 2010 1:44 PM
To: Doran, Harold
Cc: R-help
Subject: Re: [R] Help with apply
You are missing "s" in your definitions so I can't reproduce your code.
> tmp <- data.frame(var1 = sample(c(0:10), 3, replace = TRUE), var2 =
> sample(c(0:10), 3, replace = TRUE), var3 = sample(c(0:10), 3, replace = TRUE))
>
> str(tmp)
'data.frame': 3 obs. of 3 variables:
$ var1: int 9 3 9
$ var2: int 4 6 2
$ var3: int 2 9 3
>
> #I can run the following double loop and yield what I want in the end (rr1)
> as:
>
> library(statmod)
> Q <- 2
> b <- runif(3)
> qq <- gauss.quad.prob(Q, dist = 'normal', mu = 0, sigma=1)
> rr1 <- matrix(0, nrow = Q, ncol = nrow(tmp))
> L <- nrow(tmp)
> for(j in 1:Q){
+ for(i in 1:L){
+ rr1[j,i] <- exp(sum(log((exp(tmp[i,]*(qq$nodes[j]-b))) /
(factorial(tmp[i,]) *
+ exp(exp(qq$nodes[j]-b)))))) * ((1/(s*sqrt(2*pi))) *
exp(-((qq$nodes[j]-0)^2/(2*s^2))))/dnorm(qq$nodes[j]) * qq$weights[j]
+ }
+ }
Error: objeto 's' no encontrado
> rr1
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
>
Gabriela
2010/10/4, Doran, Harold <hdo...@air.org>:
> Suppose I have the following data:
>
> tmp <- data.frame(var1 = sample(c(0:10), 3, replace = TRUE), var2 =
> sample(c(0:10), 3, replace = TRUE), var3 = sample(c(0:10), 3, replace =
> TRUE))
>
> I can run the following double loop and yield what I want in the end (rr1)
> as:
>
> library(statmod)
> Q <- 2
> b <- runif(3)
> qq <- gauss.quad.prob(Q, dist = 'normal', mu = 0, sigma=1)
> rr1 <- matrix(0, nrow = Q, ncol = nrow(tmp))
> L <- nrow(tmp)
> for(j in 1:Q){
> for(i in 1:L){
> rr1[j,i] <-
> exp(sum(log((exp(tmp[i,]*(qq$nodes[j]-b))) / (factorial(tmp[i,]) *
> exp(exp(qq$nodes[j]-b)))))) *
>
> ((1/(s*sqrt(2*pi))) * exp(-((qq$nodes[j]-0)^2/(2*s^2))))/dnorm(qq$nodes[j])
> * qq$weights[j]
> }
> }
> rr1
>
> But, I think this is so inefficient for large Q and nrow(tmp). The function
> I am looping over is:
>
> fn <- function(x, nodes, weights, s, b) {
> exp(sum(log((exp(x*(nodes-b))) / (factorial(x) *
> exp(exp(nodes-b)))))) *
> ((1/(s*sqrt(2*pi))) *
> exp(-((nodes-0)^2/(2*s^2))))/dnorm(nodes) * weights
> }
>
> I've tried using apply in a few ways, but I can't replicate rr1 from the
> double loop. I can go through each value of Q one step at a time and get
> matching values in the rr1 table, but this would still require a loop and
> that I think can be avoided.
>
> apply(tmp, 1, fn, nodes = qq$nodes[1], weights = qq$weights[1], s = 1, b =
> b)
>
> Does anyone see an efficient way to compute rr1 without a loop?
>
> Harold
>
>> sessionInfo()
> R version 2.10.1 (2009-12-14)
> i386-pc-mingw32
>
> locale:
> [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
> States.1252 LC_MONETARY=English_United States.1252
> [4] LC_NUMERIC=C LC_TIME=English_United
> States.1252
>
> attached base packages:
> [1] stats graphics grDevices utils datasets methods base
>
> other attached packages:
> [1] minqa_1.1.9 Rcpp_0.8.6 MiscPsycho_1.6 statmod_1.4.6
> lattice_0.17-26 gdata_2.8.0
>
> loaded via a namespace (and not attached):
> [1] grid_2.10.1 gtools_2.6.2 tools_2.10.1
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
_________________________
Lic. María Gabriela Cendoya
Magíster en Biometría
Profesor Adjunto
Facultad de Ciencias Agrarias
UNMdP - Argentina
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