search for: rexps

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)

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

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

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

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

Hi Simon, Thanks so much for your help. Your advice has been taken to heart. I now pass in blocks of 100,000 records, and it does 100,000 predictions in seconds and returns a logical vector with the predictions to an int array. It works like a charm! I want to reference what we were talking about earlier. Let?s say we evaluate an R expression. Is there a way to just print, verbatim, what R

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 =

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 userRs, "playing around" with combinations of replicate() and random number generating functions inside a self-defined "wrapper" function I encounterd a puzzling behaviour. The following are intentionally simple (and rather nonsense-) examples to isolate the relevant aspects. Please, note the seemingly "inconsistent" behaviour for the second call of

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 ##

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

Hello, I want to import R-output via Rserve to Java, especially for the function ctree from the package party. Rserve is working properly. Yet, I only get the predictions with the Java code try{ RConnection c = new RConnection(); ... c.voidEval("modell <- ctree(...)"); REXP y = c.eval("nodes(modell,1)[[1]]$prediction"); ...

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, 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.

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

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))

Hi, I am new to R. My question is: how to get the 'hazard ratio' using the 'coxph' function in 'survival' package? thanks, karena -- View this message in context: http://r.789695.n4.nabble.com/About-hazard-ratio-urgent-tp3595527p3595527.html Sent from the R help mailing list archive at Nabble.com.

...= ((REXPVector) ((RList) tableRead).get(3)).asDoubles(); String[] d5 = ((REXPVector) ((RList) tableRead).get(4)).asStrings(); // create data frame with data.REXP myDf = REXP.createDataFrame(new RList( new REXP[] { new REXPDouble(d1), new REXPDouble(d2), new REXPDouble(d3), new REXPDouble(d4), new REXPString(d5) })); // assign the data to a variable as was suggested.rConnection.assign("boxMVariable", myDf); // create a string command with that variable name.String boxVariable = "boxM(boxMVariable [,-5], boxMVariable[,5]"; // try to execute the command... // FAILS with org.ro...

i would like to add a variable to an existing matrix by manipulating 2 previous variables eg for the data m treat strata censti survTime [1,] 1 2 284.684074 690.4961005 [2,] 1 1 172.764515 32.3990335 [3,] 1 1 2393.195400 24.6145279 [4,] 2 1 30.364771 8.0272267 [5,] 1 1 523.182282 554.7659501 l

l want to create a column of 1 and 2 randomly what command should l use eg treatment strata 1 1 2 0 1 1 2 1 2 0 2 1 2 0 1 0 these should be created randomly secondly if l have