similar to: Selecting from matrix and then stacking in array.

Hi there, I'm trying to solve a ML problem where the likelihood function is a function of two numerical procedures and I'm having some problems figuring out how to do this. The log-likelihood function is of the form L(c,psi) = 1/T sum [log (f(c, psi)) - log(g(c,psi))], where c is a 2xT matrix of data and psi is the parameter vector. f(c, psi) is the transition density which can be

Dear R users, When running the program below I receive the following error message: fit <- optim(parm, objective, yt = tyield, hessian = TRUE) Error in as.vector(data) : no method for coercing this S4 class to a vector I can't figure out what the problem is exactly. I imagine that it has something to do with "tyield" being a matrix. Any help on explaining what's going on

Hi, I'm trying to estimate the parameters of a state space model of the following form measurement eq: z_t = a + b*y_t + eps_t transition eq y_t+h = (I -exp(-hL))theta + exp(-hL)y_t+ eta_{t+h}. The problem is that the distribution of the innovations of the transition equation depend on the previous value of the state variable. To be exact: y_t|y_{t-1} ~N(mu, Q_t) where Q is a diagonal

Dear R-users, I'm writing a program that constructs a dataset. I wish to save the dataset to a file. Here's a very simple example of what I'm trying to do function(x=peter){ y <- x/2 write.csv(y, file = "...\x") } The problem is that I want to name the dataset as whatever the name of the input is. In this case peter. How do I do this? Thank you in advance. Kristian

Dear R-users I'm running a maximum likelihood procedure using the spg package. I'd like to save some output produced in each iteration to a file, but if I put the capture.output() within the function I get the following message; Error in spg(par = startval, fn = loglik, gr = NULL, method = 3, lower = lo, : Failure in initial function evaluation!Error in -fn(par, ...) : invalid argument

Dear R-users, I'd like to plot a three-dimensional surface and at the meantime I'm using an array. I would like to have the values of my first matrix in the heights of the plot and the colors of the single facet taking into account the second matrix. I hope that the next code will help all of you to understand better my issue, Thanks in advance, Giancarlo ############################ ##

Dear R-users, I'm estimating a two- dimensional state-space model using the FKF package. The resulting log likelihood function is maximized using auglag from the Alabama package. The procedure works well for a subset of my data, but if I try to use the entire data set I get the following error message. Error in eigen(a$hessian) : infinite or missing values in 'x' What's even

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1]

Hi list, I am unsuccessfully trying to produce a serious of trellis barcharts from within a for-loop. The barcharts work outside the loop. What am I missing? Example attached. Thanks Herry #XXXXXXXXXXXXXXXXXXXXXX trellis.device(bg="white") trellis.par.get("fontsize")->fontsize fontsize$default<-16 trellis.par.set("fontsize",fontsize)

Hello again, after obtaining the keytab file I tried to use kinit keytab.file followed by the spn $ samba-tool spn list z1 z1 User CN=z1,CN=Users,DC=home,DC=lan has the following servicePrincipalName: zookeeper/ap42.home.lan $ samba-tool domain exportkeytab z1.ktab --principal=z1 $ samba-tool domain exportkeytab z1.ktab --principal=zookeeper/ap42.home.lan $ kinit -V -k -t z1.ktab

Hi Team, I am trying to very simple plot with command plot. Question : I am trying to plot (x,y) based on the value of Column e. If column e value is greater than 0 then plot(x,y) otherwise do not plot it. Data : structure(list(x = c(1, 1, 1, 2, 2, 2, 3, 3, 3), y = c(1, 2, 3, 1, 2, 3, 1, 2, 3), e = c(0, -1, -2, 1, 0, -1, 2, 1, 0)), row.names = c(NA, -9L), .Names = c("x",

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should

Dear R users, I have posted this question several days ago and received not a single suggestion. I believe I have provided sufficient information for at least some help. Here I repost the question with several modifications. I want to run nested fixed-factor Anova in R on different experiments. I have 48 levels of the main factor x1 and 242 levels of the nested factor z1, and continuous response

hello, this is my script: #1) read in data: daten<-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt', header=TRUE, sep="\t") daten<-as.matrix(daten) #2) create empty matrix: indxind<-matrix(nrow=617, ncol=617) indxind[1:20,1:19] #3) compare cells to each other, score: for (s in 3:34) { #walks though the matrix colum by colum, starting at

Hi,   I use fCopulae package to draw different graphs of univariate and bivariate skew t.  But the plots titles overlap.  I tried using cex.main, font.main to adjust the size but they still overlaps.  Here is my code: par(mfrow = c(3, 1)) mu = 0 Omega = 1 alpha1 = 0 alpha2 = 1.5 alpha3 = 2 alpha4 = 0.5 Z1 = matrix(dmvst(x, 1, mu, Omega, alpha1, df = Inf), length(x)) Z2 = matrix(dmvst(x, 1, mu,

Dear R-group members, I use: platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32 status major 1 minor 4.1 year 2002 month 01 day 30 language R I try to fit a 2 compartment model. The compartments are open, connected to each other and

Dear Rusers, I would like some comments about the following results (under R-2.2.0) > m = matrix(1:6 , 2 , 3) > m [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 > z1 = m[(m[,1]==2),] > z1 [1] 2 4 6 > is.matrix(z1) [1] FALSE > z2 = m[(m[,1]==0),] > z2 [,1] [,2] [,3] > is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about

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 would like to perform a multivariate regression analysis to model the relationship between m responses Y1, ... Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. Based on my readings of the R help archives and R documentation, the function lm() should be able to

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im(z2) [1] -2