search for: f_bresson

...ic scale on a plot. Labels are the same, but the increases of x along the x-axis are defined by a known monotone and continuous function. Florent Bresson ----- Message d'origine ---- De : "joris.dewolf at cropdesign.com" <joris.dewolf at cropdesign.com> ? : Florent Bresson <f_bresson at yahoo.fr> Cc : r-help at stat.math.ethz.ch; r-help-bounces at stat.math.ethz.ch Envoy? le : Mercredi, 1 Ao?t 2007, 12h01mn 58s Objet : Re: [R] Custom axis x <- 1:10 y <- rnorm(10,10,1) x2 <- 3*x + 2 plot(y ~ x, xaxt = "n") axis(side=1,at = x, labels = x2) Joris...

...y.btransf <- y.btransf+(mean(y)-mean(y.btransf)) as.vector(y.btransf) } the fonction is designed to perform a mean-preserving transformation of a vector. ----- Message d'origine ---- De : Richard M. Heiberger <rmh at temple.edu> ? : Florent Bresson <f_bresson at yahoo.fr>; r-help at stat.math.ethz.ch Envoy? le : Lundi, 16 Octobre 2006, 14h58mn 13s Objet : Re: [R] Generate a random bistochastic matrix bistochastic.3x3 <- function() { B <- matrix(0, 3, 3) ## 2 df tmp.1 <- runif(3) B[1,] <- tmp.1/sum(tmp.1) ## 1 df tmp.2 &...

...why it's the default in the function. The bistochastic matrix is not my subject of interest, but I need it to perform some random tranformation of a vector of incomes. Florent Bresson ----- Message d'origine ---- De : Richard M. Heiberger <rmh at temple.edu> ? : Florent Bresson <f_bresson at yahoo.fr>; r-help at stat.math.ethz.ch Envoy? le : Lundi, 16 Octobre 2006, 16h23mn 39s Objet : Re: Re : [R] Generate a random bistochastic matrix I am sorry, I can't figure out what your function is doing. Why do you have N^4 in the argument? A matrix should have N rows and N columns, t...

Dear R-users, I would like to apply a function (more precisely sd()) over the third dimension of a three-dimension array. The function apply would be interesting but the chosen function can only be applied on the rows and columns of the array according to the help file. I can use a loop to cut the array in matrices and then use apply for each replication, but it's not very nice. A small

I have to estimate the following model for several group of observations : y(1-y) = p[1]*(x^2-y) + p[2]*y*(x-1) + p[3]*(x-y) with constraints : p[1]+p[3] >= 1 p[1]+p[2]+p[3]+1 >= 0 p[3] >= 0 I use the following code : func <- sum((y(1-y) - p[1]*(x^2-y) + p[2]*y*(x-1) + p[3]*(x-y))^2) estim <- optim( c(1,0,0),func, method="L-BFGS-B" , lower=c(1-p[3], -p[1]-p[3]-1,

I have to calculate some formula like: gamma(x)/(gamma(x+y) and I observed that for relatively big values of x, R returns infinity and so cannot compute the formula. Is it possible to force R to give the real value of gamma(x) instead of Inf ? thanks

I'm trying to run a nls on a subset of a data.frame. In the subset, one observation is NA. So I drop the observation but when I ask for : >sm <- nls(machin$revcum ~ Lc.singh(machin$popcum,p), start=list(p=c(2,3))) I get : Erreur dans parse(file, n, text, prompt) : syntax error in "~ " If I put some value for the non available observation instead of droping it, it works.

Please, I would like to generate a random bistochastic matrix, that is a squared matrix of non-negative numbers with each row and each column sum to 1, for example : .2 .3 .5 .6 .3 .1 .2 .4 .4 I don't know of to code this. Do you have any idea ? Thanks Florent Bresson ___________________________________________________________________________ Demandez ?

I' m trying to compute weighted mean on different groups but it only returns NA. If I use the following data.frame truc: x y w 1 1 1 1 2 2 1 3 1 1 4 2 0 2 1 0 3 2 0 4 1 0 5 1 where x is a factor, and then use the command : tapply(truc$y,list(truc$x),wtd.mean, weights=truc$w) I just get NA. What's the problem ? What can I do ?

I have defined a function to compute the value of a beta distribution of the second kind (the existing beta distribution of th stats package is the beta distribution of the first kind). It works perfectly for a single value, but I want to apply it to a vector of 22 000 values. I can use a loop for the calculation of each value but it runs very very slowly. So, what can I change ? Hers's the

I have succeded in defining the cdf of the generalized beta of the second kind, eg. pgbeta2 <- function(quint,b,a,p1,p2) { integrate(function(x) {exp(log(a)+(a*p1-1)*log(x)-(a*p1)*log(b)-log(beta(p1,p2))-(p1+p2)*log(1+(x/b)^a))},0,quint)$value } but I'm facing problems with the quantile function. I tried something like qgbeta2 <- function(proba,b,a,p1,p2) { optimize(function(z)

I'm dealing with a matrix like : "x" "y" "z" [1,] 2 4 1 [2,] 6 1 2 ... [n,] 7 3 1 For each row I would like to know the header of the column which corresponds to the minimum value. In the case of my matrix, I would like to obtain the following vector : z y ... z Any idea ?

Hi, I've got a very simple problem but cannot find the solution. I'm using two data frames (say X and Y) and I want to get a subset of one according to the different levels of a variable "code" of the other data frame. I tried something like Z <- subset(X, code==levels(Y$code)) (1) but it does not work. I do not want to do something like Z <-

Hi, I'm using the density() command for a given vector x and I would like to know how to get the estimated value of the density for each element of the vector x instead of values corresponding to points from a grid. Thanks Florent Bresson

Dear R-users, I'm facing a problem with the import of data in R. I have a sav file that, I presume, uses some Arabic encoding (but I don't know which one) and I would like to read it with R. When I use the function read.spss (I also tried spss.get(Hmisc)), I get the following message: > read.spss("Hhld.sav") Erreur dans read.spss("Hhld.sav") : erreur ? la lecture

I have to compute some standard errors using the delta method and so have to use the command "numericDeriv" to get the desired gradient. Befor using it on my complicated function, I've done a try with a simple exemple : x <- 1:5 numericDeriv(quote(x^2),"x") and i get : [1] 1 8 27 64 125 216 attr(,"gradient") [,1] [,2] [,3] [,4] [,5] [,6] [1,] Inf

Hi, I have to use a loop to perform a quite computer intensive estimation and I would like to know the progress of the loop during the process. I tried to include something like print(paste(k,date(),sep=" : ")) where k is the number of the iteration, but the result appears only at the end of the loop. Can someone help me please ?