similar to: Reversing axis label order

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

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 <-

Hi, In order to use the mle2-function, one has to define the likelihood function itself. As we know, the likelihood function is a sum of the logarithm of probability density functions (pdf). I have implemented myself the pdfs that I am using. My problem is, that the pdfs values are negative and I cann't take the logarithm of them in the log-likelihood function. So how can one take the

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])

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)

Dear R Users, I have a data frame of the nature: > head(aggregate_1986) Latitude Mean Annual Simulated Runoff per 1? Latitudinal Band 1 -55 574.09287 2 -54 247.23078 3 -53 103.40756 4 -52 86.99991 5

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

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, Could anybody offer any advice about implementing a Kohonen self organizing map for categorical data? Specifically I am wondering if there are any pre-existent packages that can deal with categorical data and/or how one would compare the input vector of categoricals with the self organizing map nodes. Thanks in advance. George Chen ----- Original Message ----- From: r-help-request at

Dear all, I use at least two pc to perform my data analysis. Both are powered with R updated at the latest release (currently 2.9.0 under windows xp pro). I bring .Rdata on my portable drive and use them on any of my pc (this worked also under (k)ubuntu equipped machines). Please notice that not all my pc have the same libraries installed. If a not installed library was called in .First() I got an

Dear all, I am currently facing a problem related to the spatial autocorrelation of a sample of stations; these stations supply weekly data for a fixed time-window during the year (namely, 4-6 months per year). For this reason I'm trying to use the R package 'spdep' (specifically Moran's I) in order to get rid of it. Does anyone know how is it possible (if it is...) to

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,

Hello, All: ????? Roughly a decade ago, I added "matplot.Date" and "matplot.POSIXct" to the "fda" package, so we could get reasonable labeling of the horizontal axis when "x" was class "Date" or "POSIXct".? I also added a local version of "matplot.default" that just changes the defaults for "xlab" and