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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, g2, g3 may appear to be in different orders. So that final effect of the passing is f = function(X ) { process1(X) g1(X, par1) g2(X, pa...

Hi My code looks like this I have two parameters x and par1. X contains values and par1 contains the function which i required to use if par1 is max then output should be max(x). FUN <- match.fun(par1) result=FUN(x) Is it possible to incorporate the unique count of x within this code eg x=("a","b","a","...

...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.: > expand.grid(par1=-1:1,par2=c('a','b')) par1 par2 1 -1 a 2 0 a 3 1 a 4 -1 b 5 0 b 6 1 b There is nothing wrong with this nice function except when you have too many combinations to fit in your computer memory, and that was my problem: I wanted to do something fo...

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 = 25)) df <- data.frame(par1, par2, fac1, fac2) Now,...

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") par2=c(3,4,4,3,5,2,4,3,2) DATA=data.frame(know1,par1,know2,par2) it represents answers in a questionnaire, where respondents evaluate two t...

width <- 7 height <- 5 # create whatever device type # and have cex=1 before setting par(mai=) windows(width=width, height=height, rescale="R") oldmai <- par("mai") par(cex=1) par(mai=oldmai) plot.new() par(usr=c(0,2,0,2)) par(cex=2) par1 <- par() text(1,1,"there is hardly anything i know for sure\nneither of the future nor of the past\nnor should i know what to ask\nif the person answering was you\n\n(Ernst Jandl, stanzen)") dev.off() # now set cex != 1 before setting par(mai=) # shouldn't change anything but doe...

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) localfun1 <- function() fun1(n, p1) localfun2 <- function() fun2(n, p2) vp<-replicate(nsim,t.test(localfun1(), localfun2())$p...

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 parms<-c(kel=kel,Vd=Vd) C1.lsoda<-data.frame(lsoda(Dose/Vd,c(0,time),defun,parms)...

...itmap(file="tmp") > densityplot(~y,col="black",main="x") > dev.off() Basically, they are the same lines inside the script, I got the expected output. I appreciate you advice, The script code: totalgraphics <- function(server,status) { par2 = '0' par1 = '0' filename <- paste(server,status,sep='.') filename <- paste('./pre-processed/',filename,sep='') v <- read.csv(file=filename,sep=',') x <- v[[status]] par1 <- as.numeric(par1) par2 <- as.numeric(par2) x <- as.ts(x) library(lattice) fi...

...ons(par=value) and getOption("par") I was aking myselft what would be the "better" strategy to handle a bunch of options for a package. I ended up with the idea of storing a list, as my options would also be classified, with something like: -- MyPkgOptions = list(set1=list(par1=1,par2=2),set2=list(subset1=list(par1=11,par2=22),subset2=list(par1=111,par2=222))) options(PkgName=MyPkgOptions) -- Then, to make easier the access to an element, I tweaked a little bit getOption, with the following version: -- getOption <- function(x,...) { op = options(x)[[1]] if (le...

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, envir, m...

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," That is what I did and I can't find the mistake. I just want to have "test" without quotes (to do eva...

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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.

...m it gives an error which is due to the outer function. I have tested the program with another simpler function and it works. So, I know that the problem is in the function that I have written.  Below I copy the error and the part of the program (the function) which causes the error:   Error: outer(par1, par2, y1, y2, FUN = function.subplot, KERNEL = KERNEL,: dims [product 25921] [1] does not match the length of the object.     par1 <- seq(-3,3,0.1) par2<- seq(-3,3,0.1) x<-cbind(par1,par2)       # par1:parameter1, par2: parameter2, sigma is the third parameter that I fix densityfunction &...

...1 <- c(0,0.978723404255319,0.087378640776699,0.549295774647887,0.0596026490066225 ,0.61578947368421); x2 <- c(1,3,4,5,6,7); x3 <- c(3600,169200,185400,255600,271800,342000); observs <- data.frame(y, x1, x2, x3); # function definition rss <- function(par, y, x1, x2, x3) { par1 <- par[1] par2 <- par[2] par3 <- par[3] ressq <- (y - par1 * (x1 + 1) * x2^(-par2) * x3^par3)^2 sum(ressq) } #call to optimizer opti <- constrOptim(c(0.5, 0.5, 0.1), rss, NULL, ui = rbind(c(0, 0), c(1, 0), c(0, 1)), ci = c(0, 1, 1)); [[alternative HTML version de...

...ients, 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 PS: When I use std.coef, all I get is this: std.coef(path.model.SSI4) Std. Estimate par1 par1 0.39499 com_veg <--- tempm par2 par2 0.35231 SNutBili <--- tempm par3 par3 -0.68170 S_SSI4 <--- tempm par4 par4 -0.39145 com_veg <--- Wdeficit par5 par5 -0.60025 SNutBili <--- Wdeficit par6 par6 -0.20562 S_SSI4 <--- Wdeficit par7...

...... but it doesn't happen all the time... description is below and you can download the video files form the link below if you like... (1MB rar file with 5 video files) https://download.yousendit.com/UmNKckhWUnIwZ21Ga1E9PQ video files and info: 01_original_avi_captured_with_camera.avi 320x240 PAR1:1 progressive 16.716fps mjpg avi this is the original footage captured by the camera 02_mpeg_converted_from_original_avi.mpg 320x240 par1:1 progressive 25fps mpeg2 this is the file i have converted to mpeg form original file with mainconcept mpeg pro.. the video is OK 03_ogv_converted_from_origi...

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 Zipf 0.11033 -0.74705 61.0465 340.9567 346....

...he combination of the first three parameters to sum up to 1 (which obviously do not) s<-p1[,1]+p1[,2]+p1[,3] s==1 It occurred to me to divide each of these parameters with the sum (vector "s" above). However the uniform distribution is lost (example for parameter 1 - first column): par1.transf<-p1[,1]/s hist(par1.transf) So, is there a way to maintain the random LHS (with uniformly distributed parameters) so that the refered condition is fulfilled? Any suggestions would be much welcome. Thanks, Duarte