similar to: Generate a random bistochastic matrix

Displaying 20 results from an estimated 500 matches similar to: "Generate a random bistochastic matrix"

2006 Oct 16
5
Re : Generate a random bistochastic matrix
Thanks, I tried someting like this, but computation takes times for large matrices btransf <- function(y,X=length(y)^4) { N <- length(y) bm <- matrix(rep(1/N,N^2),N,N) for(j in 1:X){ coord <- sample(1:N,4,replace=T) d <- runif(1,0,min(bm[coord[1],coord[2]],bm[coord[3],coord[4]]))
2006 Oct 16
2
Re : Re : Generate a random bistochastic matrix
Yes, you're right. In fact, it's just an adaptation of a matlab command and the author advises using N^4 replications that's 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
2007 Aug 01
1
Re : Custom axis
Maybe I do not explain well what I would like to do. I do not want to change the labels of the axis, but the scale. What I want is a general procedure for changing the scale. Its like using a logarithmic 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 :
2006 Oct 12
1
OT: jobs for asterisk lovers
Sorry for being out of topic... My client, a leading international VoIP carrier, is looking to hire two Asterisk "freaks" for a permanent role based in Brussels (Belgium). One role would be for technical customer support, and the other role would be for developments. If any interest, please respond to my email mcarrette8@yahoo.fr with your resume and I will forward more details.
2005 Nov 28
3
optimization with inequalities
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,
2007 Aug 24
2
Applying a function to an array
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
2005 Dec 05
3
The gamma function and infinity
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
2005 Oct 31
3
Applying a function to a vector
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
2006 Jan 12
2
tapply and weighted means
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$w) I just get NA. What's the problem ? What can I do ?
2007 Jan 15
2
Kernel density output
Hi, I'm using the density() command for a given vector x and I would like to know how to get the estimated value of the density for each element of the vector x instead of values corresponding to points from a grid. Thanks Florent Bresson
2005 Oct 19
1
Problem with na in nls
I'm trying to run a nls on a subset of a data.frame. In the subset, one observation is NA. So I drop the observation but when I ask for : >sm <- nls(machin$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.
2009 Jan 07
1
Importing data from SPSS with Arabic encoding
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
2006 Jul 19
3
Progress in a loop
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 ?
2005 Nov 16
2
numericDeriv
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
2008 Mar 15
0
Naissance de fr.centos.org / Birth of fr.centos.org
(For non native french speakers : that will be my only announce here in another language than english ;-) ) Le projet CentOS est heureux de vous annoncer la naissance du site http://fr.centos.org . En r?ponse ? la demande croissante de la communaut? des utilisateurs francophones de CentOS, le forum fr.centos.org a vu le jour. Nous profitons de cette annonce pour relancer l'appel aux
2003 Apr 22
4
fisher exact vs. simulated chi-square
Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with
2006 Oct 19
1
Re : CI
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2005 Dec 11
1
Quantile function for the generalized beta distribution of the 2nd kind
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)
2006 Nov 06
1
Subset and levels
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 <-
2005 Dec 29
1
search in matrix
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 ?