similar to: Help in using unique count by match function

Dear list, I've recently came across a problem that I think I've solved and that I wanted to share with you for two reasons: - Maybe others come across the same problem. - Maybe someone has a much simpler solution that wants to share with me ;-) The problem is as follows: expand.grid() allows you to generate a data.frame with all combinations of a set of values, e.g.: >

Hi, I wonder how to pass several functions and their arguments as arguments to a function. For example, the main function is f = function(X ) { process1(X) ... process2(X) } I have a few functions that operate on X, e.g. g1(X, par1), g2(X, par2), g3(X, par3). par1, par2 and par3 are parameters and of different types. I would like to pass g1, g2, g3 and their arguments to f and g1,

Hi, I have a data frame DATA, which (simplified of course) looks like this: know1 = c("Y","N","N","Y","N","N","Y","Y","N") par1=c(1,4,5,3,3,2,3,3,5) know2 = c("Y","Y","N","Y","N","N","N","Y","Y")

Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =

Hi R users: I'm trying to make a function where two of the parameters are functions, but I don't know how to put each set of parameters for each function. What am I missing? I try this code: f2<-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){ force(fun1) force(fun2) force(n) p1<-unlist(par1) p2<-unlist(par2) force(p1) force(p2)

Dear R users: I have a problem about catch the value from function. I have following two functions (part): sbolus1 <- function() { ....... for( i in 1:Subject) { kel<-par1 Vd<-par2 PKindex<-sbolus1.out(PKtime,kel,Vd,defun,par1,par2,Dose,i) } savefile(PKindex) } sbolus1.out<-function(PKtime,kel,Vd,defun,par1,par2,Dose,i) { time<-PKtime$time

Hi R-devel, I am facing a situation where the number of options I would like to propose to the user is somewhat big (and could easily increase more and more as I will code up a little more - even coming to a point where an user should be able to implement his own options). What we have to handle options is the couple: options(par=value) and getOption("par") I was aking myselft what

Hi all, I'm trying to use lapply on a list with the following command: out<-lapply(mylist,myfun,par1=p,par2=d) (1) where myfun<-function(x,par1,par1) {.....} (2) now this function is in fact a wrapper for some Fortran code I have written so I think this might be the problem. When I call lapply() as in (1) I get the following message: Error in get(x,

Hi there again, I have a problem with the "parse"-command. First of all, I show you in a simplified way, what I am trying to do and what "R" answers: > test [1] "u.g$par1, u.g$par2" > mode(test) [1] "character" > ausdruck <- parse(text = test) Error in parse(file, n, text, prompt) : syntax error in "u.g$par1,"

Hi, I wonder how to pass several functions and their arguments as arguments to a function. For example, the main function is f = function(X ) { process(X) ... process(X) } I have a few functions that operate on X, e.g. g1(X, par1), g2(X, par2), g3(X, par3). par1, par2 and par3 are parameters and of different types.

After a few days of work, I think I nearly have it. Unfortunately, theta is unchanged after I run this (as a script from a file). I thought that theta would contain the fitted parameters. The goal here is to find the least squares fit according to the function defined as "rss" subject to the constraints defined as ui and ci. I defined ui and ci to (hopefully) force par2 and par3

Hi, I am currently working with the sem package in R, to create pathway diagrams. Id like to use the standardized path coeffcients. To get these, I use std.coef. However, using this yields only the standardized coefficients, but does not give me the standard error. Does someone know how to get std.coef to show the standard error of the standardized path coefficients as well? Thanks, Bastiaan

Dear all, Using: library(vegan) data(BCI) mod <- radfit(BCI[1,]) mod RAD models, family poisson No. of species 93, total abundance 448 par1 par2 par3 Deviance AIC BIC Null 39.5261 315.4362 315.4362 Preemption 0.042797 21.8939 299.8041 302.3367 Lognormal 1.0687 1.0186 25.1528 305.0629 310.1281

Hello all, I am trying to do a Latin Hypercube Sampling (LHS) to a 5-parameter design matrix. I start as follows: library(lhs) p1<-randomLHS(1000, 5) If I check the distribution of each parameter (column), they are perfectly uniformly distributed (as expected).For example, hist(p1[,1]) Now the hard (maybe strange) question. I want the combination of the first three parameters to sum up to

I also have several packages in my RW0633 in Windoze and I would like to keep them in one place when I update the R system. Brian say to modify Rprofile to ..lib.loc <- c("c:/MyR/library",.Library) now Rprofile has a line like .lib.loc <- unique(c(unlist(strsplit(getenv("RLIBS"),":")),.Library) Sorry to be dense but where should I put the

Petr, there was a thinko in your response. tmp <- data.frame(m=factor(letters[1:4]), n=1:4) tmp tmp$m <- factor(tmp$m, levels=c("c","b","a","d")) ## right tmp[order(tmp$m),] tmp <- data.frame(m=factor(letters[1:4]), n=1:4) levels(tmp$m) <- c("c","b","a","d") ## wrong tmp[order(tmp$m),] changing levels

Hi After melt you can change levels of your factor variable. Again with the toy example. > levels(temp$variable) [1] "y1" "y2" "y3" "y4" > levels(temp$variable) <- levels(temp$variable)[c(2,4,1,3)] > levels(temp$variable) [1] "y2" "y4" "y1" "y3" > And you will get graphs with this new levels ordering.

I''m looking through the api for something like find(:all, :distinct => true) so that multiple instances of a row won''t be returned - is this possible? Thanks!

Dear R useRs, I have a problem with nls.lm function of minpackl.lm package. I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic conductivity with nls.lm function of minpack.lm package. For the first fit, the parameter estimates keep changing even after 1000 iterations (Th) and I have a following error message for fit of hydraulic conductivity (k); Reason for

Hi Petr and Richard; Thanks for your responses and supports. I just faced a different problem. I have the following R codes and work well. p <- ggplot(a, aes(x=Phenotypes, y=Metabolites, size=abs(Beta), colour=factor(sign(Beta)))) + theme(axis.text=element_text(size = 5)) p1<-p+geom_point() p2<-p1+theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(),