[R] Simulation help

ONKELINX, Thierry Thierry.ONKELINX at inbo.be
Wed Sep 20 16:07:31 CEST 2006


This should work. I try to avoid for loops because large (or nested) for
loops tend to consume a lot of memory. 

one.dataset <- function(dummy = 0, slope = 0.01){
  return(c(yr1= mean(rbinom(50,1,.5)), yr2 =mean(rbinom(50,1,.5 +
slope)), yr3 =mean(rbinom(50,1,.5 + 2 * slope)), yr4
=mean(rbinom(50,1,.5 + 3 * slope)), yr5 =mean(rbinom(50,1,.5 + 4 *
slope))))
}

datasets.slope01 <- t(sapply(1:100, one.dataset, slope = 0.01))
datasets.slope05 <- t(sapply(1:100, one.dataset, slope = 0.05))

Cheers,

Thierry

------------------------------------------------------------------------
----

ir. Thierry Onkelinx

Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature
and Forest

Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance

Gaverstraat 4

9500 Geraardsbergen

Belgium

tel. + 32 54/436 185

Thierry.Onkelinx op inbo.be

www.inbo.be 

-----Oorspronkelijk bericht-----
Van: r-help-bounces op stat.math.ethz.ch
[mailto:r-help-bounces op stat.math.ethz.ch] Namens Zodet, Marc W. (AHRQ)
Verzonden: woensdag 20 september 2006 15:29
Aan: r-help op stat.math.ethz.ch
Onderwerp: [R] Simulation help

I'm trying to simulate trend data over a five year period.  I want
different trend profiles...the simplest being a linear trend.  I've been
using the following code:

 

patBdta1 <- NULL

for(i in 1:100)

patBdta1 <- rbind(patBdta1,c(yr1= mean(rbinom(50,1,.50)),

                             yr2 =mean(rbinom(50,1,.51)),

                             yr3 =mean(rbinom(50,1,.52)),

                             yr4 =mean(rbinom(50,1,.53)),

                             yr5 =mean(rbinom(50,1,.54))))

 

This code creates 100 data sets each with a 5 yr binomial trend profile
with a slope of approximately .01.

 

Now, what I want to do is pass this code (or some code) in such a way
that I can simulate various trend slopes (i.e., pass in a loop or
vectorize vs. copying/repeating code for each slope).

 

Any guidance is much appreciated.  Thanks!

 

Marc

 

Marc W. Zodet, MS

Senior Health Statistician

Agency for Healthcare Research and Quality


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