similar to: Deparsing part of a list argument

I don't understand how this function can subset by i when i is missing.... ## My function: myfun = function(vec, i){ ret = vec[i] ret } ## My data: i = 10 vec = 1:100 ## Expected input and behavior: myfun(vec, i) ## Missing an argument, but error is not caught! ## How is subsetting even possible here??? myfun(vec) Is there a way to check for missing function arguments, *and*

Suppose I have f1 <- function(x) x f2 <- function(x) x^2 funlist <- list(f1,f2) Then I would like to evaluate funlist such that when x is 10 I should get a list with 10 and 100. A naive way of doint this is myf <- funlist[[1]] do.call(paste(quote(myf)), list(x=10)) myf <- funlist[[2]] do.call(paste(quote(myf)), list(x=10)) - but there has to be much

Full_Name: Lixin Han Version: 2.4.0 OS: Windows 2000 Submission from: (NULL) (155.94.110.222) A character vector c('a','b') is supplied to rm(). As a result, 'c' is deleted unintentionally. > a <- 1:5 > b <- 'abc' > c <- letters > ls() [1] "a" "b" "c" > rm(c('a','b')) > ls() character(0)

When trying out some variations with `[.data.frame` I noticed some (to me) odd behaviour, which I found out has nothing to do with `[.data.frame`, but rather with the way arguments are matched, when mixing named/unnamed and missing/non-missing arguments. Consider the following example: myfun <- function(x,y,z) { ? print(match.call()) ? cat('x=',if(missing(x)) 'missing'

In both R and JSON (and many other languages), unicode characters can be escaped using a backslash followed by a lowercase "u" and a 4 digit hex code. However when deparsing a character vector in R on Windows, the non-latin characters get escaped as "<U+" followed by their 4 digit hex code and ">": > x <- "I like \u5BFF\u53F8" > cat(x) I like

I am using nlsLM {minpack.lm} to find the values of parameters a and b of function myfun which give the best fit for the data set, mydata. mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

Hi, The following function works, but is there a neater way to write it? f = function(x,...) { # return a character vector of the arguments passed in after 'x' gsub(" ","",unlist(strsplit(deparse(substitute(list(...))),"[(,)]")))[-1] } > f(x,a,b,c*d) [1] "a" "b" "c*d" > Thanks. [[alternative HTML

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

Dear All, In the title of a plot, I would like to mix greek letters with numbers, where the numbers are obtained from a particular function to a vector (e.g., max(x)); in each call, the value of this vector can change. Without greek symbols I use something like: title(sub=paste("x1=", deparse(x[1]),"beta = ",deparse(max(x)), "rho = ", deparse(min(x)))) but I'd

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

R-Developers, I'm looking for some help computing on the R language. I'm hoping to write a function that parses a language or expression object and returns another expression with all instances of certain argument of a given function altered. For instance, say I would like my function, myFun to take an expression and whenever the argument 'x' appears within the function FUN inside

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

Full_Name: Douglas Bates Version: Version 0.64.0 Unstable (April 3, 1999) OS: Debian GNU/Linux 2.1 Submission from: (NULL) (128.105.5.97) R> fm2 <- lm( 1/Surv ~ Poison + Antidote + Poison:Antidote, data = BC ) R> anova( fm2 ) Analysis of Variance Table Response: / Response: 1 Response: Surv Df Sum Sq Mean Sq F value Pr(>F) Poison 2 34.9 17.4

Constant propagation: %sum = add <N x float> %a, %b @llvm.reduce(ext <N x double> %sum) if %a and %b are vector of constants, the %sum also becomes a vector of constants. At this point you have @llvm.reduce(ext <N x double> %sum) and don't know what kind of reduction do you need. - Elena -----Original Message----- From: Renato Golin [mailto:renato.golin at linaro.org]

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

It surprised me that prod(numeric(0)) is 1. I guess if you say (operation(nothing) == identity element) this makes sense, but ?? Looking in the code, this makes sense: basically (s=1; for i=0 to length(x), multiply s by x[i]) -- which comes out to 1. What *should* prod(numeric(0)) produce? I couldn't find the answer documented anywhere. (And how about sum(numeric(0))==0, which for

Hi interested readers, I have a function that creates several functions within a loop and I would like them to be returned for further use as follows: Main.Function(df,...){ # df is a multivariate data funcList<-list(NULL) for (i in 1:ncol(df)){ temp<-logspline(df[,i],...) # logspline density estimate funcList[[i]]<-function(x){expression(temp,x)} } return(funcList) } I have tried

Dear R users, I am currently writing a R package and to do so I am following the guidelines in manual 'Writing R extensions'. In Section 3.1, it is suggested to tidy up the code using a file containing the following: options(keep.source = FALSE) source("myfuns..R") dump(ls(all = TRUE), file = new.myfuns.R") I have done this for my own packages and although it runs, I get

Hi. 1)I have two variables: call a<-c(e.g.0,3,6,7...) b<-c(e.g.6,8,3,4...) I want to create a third vector z wich contain the pairs values z<-c(0,6,3,8,6,3,7,4....and so on for each pairs (a,b)). There is a specific function? How can i write my own function? 2)When i try to write a function and then i save it like "function.R" file, i try to