similar to: mle2(): logarithm of negative pdfs

Dear all, I need to compute tensor product of B-spline defined over equi-spaced break-points. I wrote my own program (it works in a 2-dimensional setting) library(splines) # set the break-points Knots = seq(-1,1,length=10) # number of splines M = (length(Knots)-4)^2 # short cut to splineDesign function bspline = function(x) splineDesign(Knots,x,outer.ok = T) # bivariate tensor product of

Dear R-users, I learned today that there exists an interesting topic in numerical analysis names "best polynomial approximation" (BSA). Given a function f the BSA of degree k, say pk, is the polynomial such that pk=arginf sup(|f-pk|) Although given some regularity condition of f, pk is unique, pk IS NOT calculated with least square. A quick google tour show a rich field of research

Hi all, I am running a bivariate regression with the following: p1=c(184,155,676,67,922,22,76,24,39) p2=c(1845,1483,2287,367,1693,488,435,1782,745) I1=c(1530,1505,2505,204,2285,269,1271,298,2023) I2=c(8238,6247,6150,2748,4361,5549,2657,3533,5415) R1=I1-p1 R2=I2-p2 x1=cbind(p1,R1) y1=cbind(p2,R2) fit1=lm(y1~-1+x1) summary(fit1) Response 2: Coefficients: Estimate Std. Error t value

Hi everybody, I am using the L-BFGS-B method of the mle2() function to estimate the values of 6 parameters. mle2 uses the methods implemented in optim. As I got it from the descriptions available online, one can use the parscale parameter to tell R somehow what the values of the estimated parameters should be . . . Could somebody please help me understand what one has to do actually with the

Dear R-helpers, I'm having a problem in using plot.design in Design Library. Tho following example code produce the error: > n <- 1000 # define sample size > set.seed(17) # so can reproduce the results > age <- rnorm(n, 50, 10) > blood.pressure <- rnorm(n, 120, 15) > cholesterol <- rnorm(n, 200, 25) > sex <-

Dear all, I'm stuck in a seemingly trivial task that I need to perform for many datasets. Basically, I want to replace NA with 0 in a specified range of columns in a dataframe. I know the first and last column to be recoded only by its name. I can select the columns starting like this a[match('first',names(a)): match('last',names(a))] The question is how can replace all NA

Hi R-users, I have this code to uniformise the data using gamma: > length(dp1) [1] 696 > dim(dp1) [1] 58 12 > dim(ahall) [1]  1 12 > dim(bhall) [1]  1 12 > trans_dt <- function(dt,a,b) + { n1 <- ncol(dt) +   n2 <- length(dt) +   trans  <- vector(mode='numeric', length=n2) +   dim(trans) <- dim(dt) +   for (i in 1:n1) +   {  dt[,i] <- as.vector(dt[,i])

Dear R Users, I am executing the following command to produce a line graph: matplot(aggregate_1986[,1], aggregate_1986[,2:3], type="l", col=2:3) On the x-axis I have values of Latitude (in column 1) ranging from -60 to +80 (left to right on the x-axis). However, I wish to have these values shown in reverse on the x-axis, going from +80 to -60 (ie. North to South in terms of Latitude).

Hi, I would like to generate two correlated variables. I found that funktion for doing that: a <- rmvnorm(n=10000,mean=c(20,20),sigma=matrix(c(5,0.8*sqrt(50), 0.8*sqrt(50),10),2,2)) (using library(mvtnorm)) Now I also want to generate two correlated variables where the error variance vary over the variable-correlation. And I want to plot this for showing heteroscedasticity. Like shown

Hi, At each iteration in my program,I need to generate tree vectors,X1,X2,X3, from exponential distribution with parameters a1,a2,a3. Can you help me please how can I do it such that it take a little time? thank you khazaei

Dear group, Here are my objects in my environment: > ls() [1] "Pos100415" "Pos100416" "posA" "pose15" "pose16" "pose16t" "position" "trade" "x" I need to pass the object "Pos100415" to a function. This element is a data.frame, obtained through a function: Pos(x)<-myfun(x)

hi all, I have a data frame such as: 1 blue 0.3 1 NA 0.4 1 red NA 2 blue NA 2 green NA 2 blue NA 3 red 0.5 3 blue NA 3 NA 1.1 I wish to find the last non-missing value in every 3ple: ie I want a 3 by 3 data.frame such as: 1 red 0.4 2 blue NA 3 blue 1.1 I have written a little script data = structure(list(V1 = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L ), V2 = structure(c(1L, NA,

Hi again! Suppose I have the following: > xy <- round(rexp(20),1) > xy [1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9 2.5 0.1 1.5 0.4 > table(xy) xy 0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4 1 4 2 1 2 1 2 1 1 2 1 1 1 > Is there a way to set things up to have 0 - 0.4 0.5 - 0.9 etc. please? I know there is the cut

Hello, I am trying to call a FORTRAN subroutine within R and something really strange happens: I have a dll-library, that I load with dyn.load('mpbvv.dll'). I have checked the [Ordinal/Name Pointer] Table for the function within the library that I want to call - it is there (objdump - p mpbvv.dll). Then, I have written an R-wrapper to call the FORTRAN subroutine, which works fine. SInce

Hi R, Let, x=1:80 I want to sum up first 8 elements of x, then again next 8 elements of x, then again another 8 elements..... So, my new vector should look like: c(36,100,164,228,292,356,420,484,548,612) I used: aggregate(x,list(rep(1:10,each=8)),sum)[-1] or rowsum(x,group=rep(1:10,each=8)) But without grouping, can I achieve the required? Any other ways of doing

For a function that takes an argument as a list of lists of parameters, I'd like to be able to convert that to a data.frame and vice versa, but can't quite figure out how. pats <- list(structure(list(shape = 0, shape.col = "black", shape.lty = 1, cell.fill = "white", back.fill = "white", label = 1, label.size = 1, ref.col = "gray80",

Hello, I am trying to plot a few correlograms on the same figure, with the function corrgram() from the package corrgram. However, the function does not seem to use the base graphic system, as setting out the multiple figure layout with, e.g., par(mfrow=c(2, 2,)) does not work. Does anybody know a workaround for this? Many thanks in advance for any advice best giuseppe -- Giuseppe Pagnoni,

Dear all, I am estimating a mixed-model in Ubuntu Raring (13.04¸ amd64), with the code: fm0 <- lme(rt ~ run + group * stim * cond, random=list( subj=pdSymm(~ 1 + run), subj=pdSymm(~ 0 + stim)), data=mydat1) When I check the approximate variance-covariance matrix, I get: > fm0$apVar [1] "Non-positive definite

Dear all, I am having quite a hard time in trying to figure out how to correctly spell out a model in R (a repeated-measures anova with a within-subject covariate, I guess). Even though I have read in the posting guide that statistical advice may or may not get an answer on this list, I decided to try it anyway, hoping not to incur in somebody's ire for misusing the tool. For the sake of

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) ,