similar to: Code find exact distribution for runs test?

Dear R users, I am doing Cox regression using coxph(Survival) or cph(Design). I have time varying effects (diagnosed with schoenfeld residuals Chi2 test and graph) so I first want to split time into 2 separate intervals : t<6months and t>=6months, to estimate one hazard ratio (hr) for each interval. I am analysing Overall survival according to 3 prognostic factors (age,deep,ldh).

Dear all, I just noticed that the 0.9 update for FrF2 did not work out for Mac OS due to an error in an example that ran without error on all other platforms. I do not find any reason for this. In the past, umlauts or tab characters have sometimes been an issue, but I didn't find any of these. The function definition is FrF2(nruns = NULL, nfactors = NULL, factor.names = if

I am having problem loading packages in the newest version of R (2.13.0) on my Mac. I have tried to install various packages (e.g. lawstat, Rcmdr, car) and load them using the Package Installer and Package Manager menu options but I get the follow error: > library(lawstat) Loading required package: mvtnorm Error: package 'mvtnorm' could not be loaded In addition: Warning

I'm having issues when trying to do a filled State plot using the maps() package. It seems that the color values aren't being plotted correctly, and I'm not sure why... I've started debugging it, but I'm stuck. I've attached the code that I've got so far. The for loop plots states 1:i. What I'm noticing is that the colors are correct until after Michigan gets

hi all i know that one should try and limit the amount of looping in R programs. i have supplied some code below. i am interested in seeing how the code cold be rewritten if we dont use the loops. a brief overview of what is done in the code. ============================================== ============================================== ============================================== 1. the input

Hi, I'm interested in incorporating some interaction terms between my explanatory variables (distance matrices in this case) when I'm using the function MRM of the package ecodist. The function is doing a multiple regression on distance matrices. I can get the function to work perfectly on my explanatory matrices by adding a (+) sign as known. However, when I try to use the (*) sign or

I would like to perform partial mantel tests on only within group values, with "between group" values assigned to NA. This is possible in package ncf partial.mantel.test, however this sues a different permutation to that used in ecodist.ecodist will not accept data with NA values, returning a "matrix is not square error. is it possible to perform this test in ecodist? many thanks

Dear R Programmers, I am trying to run a Poisson regression on all pairs of variables in a data set and obtain the permutation distribution. The number of pairs is around 100000. It seems my code will take weeks to run, unless I try something else. Could you give me any suggestions on how to improve the speed of the code below, or any general suggestions on how I may accomplish this task. Thanks

Hello everyone, I am looking at the joint spatial distribution of 2 kinds of organisms (estimated on a grid of points) and want to test for significant association or dissociation. My first question is: do you know a nice technique to do that, considering that I have a limited number of points (36) but that they are repeated (4 times)? I did GLMs to test for correlations between the

Dear R-users, While making a prediction using the randomForest function (package randomForest) I'm getting the following error message: "Error in predict.randomForest(model, newdata = CV) : No forest component in the object" Here's my complete code. For reproducing this task, please find my 2 data sets attached ( http://www.nabble.com/file/p17855119/data.rar data.rar ).

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

Folks, I'm a bit puzzled by the fact that if I generate 100,000 standard normal variates using rnorm() and perform the Jarque-Bera on the resulting vector, I get p-values that vary drastically from run to run. Is this expected? Surely the p-val should be close to 1 for each test? Are 100,000 variates sufficient for this test? Or is it that rnorm() is not a robust random number generator?

Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-

Hello! I have a problem with integrate() in my function nctspa(). Integrate produces an error message "evaluation of function gave a result of wrong length". I don't know what that means. Could anyone suggest me what is wrong with my function? These are the examples of function calls that work OK: nctspa(a=1:10,n=5) nctspa(a=1:10, n=5, mu=2, theta=3, renorm=0) This does not work:

My machine has 8GB memory. I had quit all other programs that might take a lot of memory when I try the script (before I post the first message in this thread). The cdf file is of only 741 MB. It is strange to me to see the error. On Mon, Dec 28, 2009 at 2:38 AM, Wolfgang Huber <whuber at embl.de> wrote: > Dear Peng Yu > > how big is the RAM of your computer? You could try with

Hello R-Users, I'm new to R so I apologize in advance for any big mistake I might be doing. I'm trying to fit a set of samples with some probabilistic curve, and I have an important question to ask; in particular I have some data, from which I calculate manually the CDF, and then I import them into R and try to fit: I have the x values (my original samples) and the y values

Please give me some help, many thanks. I graphed a expected CDF line of a binomial distribution on a graph, And I have some observed points (observed CDF) from 4 groups fall on the smooth CDF line. I cannot really get the legend I want legend ('topleft', c('a, 'b', 'c', 'd', 'expected CDF'), col=c(1,2,3,4), pch=c(0,1,2, 3, '' ),

Hi group, I would like to draw multiple Lorenz curves in a single plot using data already prepared. Here is a simple example: require("lawstat") lorenz.curve(c(1,2,3),c(4,5,4)) lorenz.curve(c(1,2,3),c(4,2,1)) This example draws two separate graphs. How can I combine them in a distinguishable way? I tried ?polygon without success... Ralf

Hi, What package(s) must I install so that I can apply the Levene' test in my data? I tried 'lawstat' but dependency ?VGAM? is not available for this package. I am using Rstudio Version 1.1.383 Tks [[alternative HTML version deleted]]

Hi, I need to write two inequalities depend on cumulative distribution (CDF) of beta binomial distribution where alpha and beta are unknown and need to find them. CDF of betabinomial(2,10,alpha,beta) <0.3<=CDF of betabinomial(3,10,alpha,beta) and CDF of betabinomial(5,10,alpha,beta) <0.8<=CDF of betabinomial(6,10,alpha,beta) How I can do that using r studio package? I tried to do