similar to: Avoiding a loop

Given the following, one of the things I am trying to see is what % of draws are below a certain number: lambda <- 3 rate <- 5 n <- 5 set.seed(123) v <- replicate(n, rexp(rpois(1,lambda), rate)) vv <- unlist(v) cat("% of draws below 0.1:", round(length(subset(vv, vv < 0.1))/length(vv)*100,0), "%\n") In actuality, my lambda, rate, and n are 26, 10, 1000000,

Having not much success with my previous question I try to reformulate it: I'm simulating a Markow chain in Fortran interfaced with R. Each loop of my Fortran calls various functions of the R RNG through the wrapper given below. In a run of 100 iterations of the Markov kernel, after 20 iterations, the RNG seems like frozen. For example, the first call to the RNG in my loop is:

Hello! I am running a very simple mini Monte-Carlo below using the function tstatistic (right below this sentence): tstatistic = function(x,y){ m=length(x) n=length(y) sp=sqrt( ((m-1)*sd(x)^2 + (n-1)*sd(y)^2)/(m+n-2) ) t=(mean(x)-mean(y))/(sp*sqrt(1/m+1/n)) return(t) } alpha=.1; m=10; n=10 # sets alpha, m, n - for run 1 N=10000 # sets the number of simulations n.reject=0 # counter of num.

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

In splus rexp() and rweibull() are related: > set.seed(153) > rexp(1) [1] 0.0493267 > set.seed(153) > rweibull(1, shape=1) [1] 0.0493267 (you can also try shape =2, then rweibull = sqrt(rexp) ) However, in rw0.64.1 (on Win NT) they are different > .Random.seed <- 1:4 > rexp(1) [1] 1.412030 > .Random.seed <- 1:4 > rweibull(1, shape=1) [1] 2.054032 May be rweibull

Hello you all from the R Help mailing list! I am working on a PowerBook with Mac Os X and use R 2.5.0. I used the distance function from the analogue package to perform a similarity analysis using the Gowers Index and weighted Variables. My variables are bivariate data and measurements as well as interval data transformed into minimum and maximum variables. I used this Code:

I am using rexp to generate several exponential distributions. I am passing rexp a vector of rates , r. I am wanting to simulate a sample of size 200 for each rate so the code looks like: rexp(n=200*length(r),rate=r) this gives me a vector of the random exponential variables, but they are all disjointed b/c rexp goes through and simulates an exponential variable for each rate and it does that 200

Trying to get past a frustrating error to do a "simple" Box's M test using Java. The Box's M test says it will work with a data.frame. Here's the setup code: REXP myDf = REXP.createDataFrame(new RList( new REXP[] { new REXPDouble(d1), new REXPDouble(d2), new REXPDouble(d3), new REXPDouble(d4), new REXPInteger(d5) })); Here's the call: REXP boxMResult =

Hello, I call a function via .Call passing to it a raw vector(D) and an integer(I) The vector is a series K1,KData1, V1,VData1, K2, KData2, ... where the integer K1 is the length of Data1 and similarly for Ki (wrt Datai)(similarly for V*) There 2*I such pairs( (Ki,KDatai), (Vi,VDatai)) The numbers Ki(and Vi) are written in network order. I am returning a list of I elements each element a

Hi Everyone, I did a quick search of the list and it looks like this may not have been asked before... I'm trying to generate a matrix of random numbers between 0 and 1, with 6 columns, 10000 rows. About all I know is that runif(1) gives me the random number I'm looking for. Any help would be great! thanks, -Max

Hi Everyone, I did a quick search of the list and it looks like this may not have been asked before... I'm trying to generate a matrix of random numbers between 0 and 1, with 6 columns, 10000 rows. About all I know is that runif(1) gives me the random number I'm looking for. Any help would be great! thanks, -Max

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Dear all, while looking for some inspiration of how to organise some code, I studied the code of random.c and noticed that for distributions with 2 or 3 parameters the user is not warned if NAs are created while such a warning is issued for distributions with 1 parameter. E.g: R version 2.7.0 Under development (unstable) (2008-02-29 r44639) [...] > rexp(2, rate=Inf) [1] NaN NaN Warning

Hi all, I just run into this today. Apparently rexp() sometimes gives different slightly results for the same seed on 32 bit and 64 bit machines. runif() is the same for both, so the problem seems to be in rexp(). 64 bit Linux is the same as 64 bit OSX, and R-devel gives the same results as R-3.0.2. Best, Gabor # --------------------------------------------- > options(digits=22) ;

Dear all, First, let's create some data to play around: set.seed(1) (df <- data.frame(Group=rep(c("Group1","Group2","Group3"), each=10), Value=c(rexp(10, 1), rexp(10, 4), rexp(10, 10)))[sample(1:30,30),]) ## Now we need the empirical distribution function: edf <- function(x) ecdf(x)(x) # empirical distribution function evaluated at x ##

This might be a very beginner question, but I'm looking for an example for something that I have never done. Suppose that I wanted to express actions with respect to lifted semantics of CPU instructions to an intermediate representation, BAP IL or LLVM IR. How might I go about providing a Backus Naur Form specification and then dynamically interpreting those lifted instructions by also

Hi R, I have a matrix, > x1=matrix(NA,6,6,dimnames=list(letters[1:6],LETTERS[1:6])) > x1 A B C D E F a NA NA NA NA NA NA b NA NA NA NA NA NA c NA NA NA NA NA NA d NA NA NA NA NA NA e NA NA NA NA NA NA f NA NA NA NA NA NA > x2=matrix(rpois(9,1),3,3,dimnames=list(c("b","a","f"),c("D","E","F"))) > x2

Hi, Wondering if anyone could help me out with this error.Im trying to fill a matrix with random numbers taken from an exponential distribution using a loop: x.3<-matrix(rep(0,3000),nrow=1000,byrow=T)for(i in 1:1000){x[i,]<-rexp(3,rate=2/3)} I get the error message: Error in x[i, ] <- rexp(3, rate = 2/3) : incorrect number of subscripts on matrix Any ideas??? Appreciate any thoughts.

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

hie R users l have the following matrix n=20 m<-matrix(nrow=n,ncol=4) colnames(m)=c("treatmentgrp","strata","survivalTime") for(i in 1:n) m[i,]<-c(sample(c(1,2),1,replace=TRUE),sample(c(1:2),1,replace=TRUE),rexp(1,0.07),rexp(1,0.02))