similar to: Can't invert matrix

Hi, I sometimes play around with extreme parameters for distributions and found that qbeta is not always monotone as the following example shows. I don't know whether this is serious enough to submit a bug report (as this example is near to the limitations of floating point arithmetic). Josef > x <- qbeta((0:100)/100,0.01,5) > x [1] 0.000000e+00 1.253990e-201 1.589622e-171

Should one expect minor numerical differences between 64bit and 32bit R on Windows? Hunting around the lists I've not been able to find a definitive answer yet. Seems plausible using different precision arithmetic, but waned to confirm from those who might know for sure. BACKGROUND A colleague was trying to replicate some modelling results (from a soon to be published book) using rpart, ada,

Dear R users, I'm using the function coxme of the package coxme in order to build Cox models with complex random effects. Unfortunately, I sometimes get surprising estimations of the variances of the random effects. I ran models with different fixed covariates but always with the same 3 random effects defined by the argument varlist=coxmeMlist(list(mat1,mat2,mat3), rescale = F, pdcheck = F,

Same behavior was noted on R 2.9.0 OS X and WIndows. > t(sapply(seq(0.01, .99, length=20), function(x) c(x, length(agrep("Staatssekretar im Bundeskanzleramt","Bundeskanzler",max.distance=x))))) [1,] 0.01000000 0 [2,] 0.06157895 0 [3,] 0.11315789 0 [4,] 0.16473684 0 [5,] 0.21631579 0 [6,] 0.26789474 0 [7,] 0.31947368 0 [8,] 0.37105263 0 [9,]

Hi, I am using rpart as a part of my masters' project. I am trying to print out the resulting model using plot() function along with text() function. I am having difficulties with labels being cut-off. In text() function, I am using use.n=T option to get the number of people in each nodes but the on the lower and left part of the plot, the numbers get cut off. Thanks! Linus [[alternative

Hello all, I'm was going through the help for ?rpart.object And noticed some inconsistencies, Some might be a mistake in the help file and some might be my misunderstanding. The help in the section: value -> frame (first paragraph), states that: > yval, the fitted value of the response at each node, *and splits, a two > column matrix of left and right split labels for each node. *

> I'm using rpart function for creating regression trees. > now how to measure the fitness of regression tree??? > > thanks n Regards, > Vibha I read R-help as a digest so often come late to a discussion. Let me start by being the first to directly answer the question: > fit <- rpart(time ~ age +ph.ecog,lung) > summary(fit) Call: rpart(formula = time ~ age +

R-help, I'm trying to estimate some parameters using the Maximum Likehood method. The model describes fish growth using a sigmoidal-type of curve: fn_w <- function(params) { Winf <- params[1] k <- params[2] t0 <- params[3] b <- params[4] sigma <- params[5] what <- Winf * (1-exp(- k *(tt - t0)))^b