similar to: nls() fit to a lorentzian - can I specify partials?

I am more than a little puzzled by your question. In the construct {expr1; expr2; expr3} all of the expressions expr1, expr2, and expr3 are evaluated, in that order. That's what curly braces are FOR. When you want some expressions evaluated in a specific order, that's why and when you use curly braces. If that's not what you want, don't use them. Complaining about it is like

R's { expr1; expr2; expr3} acts much like C's ( expr1, expr2, expr3) E.g., $ cat a.c #include <stdio.h> int main(int argc, char* argv[]) { double y = 10 ; double x = (printf("Starting... "), y = y + 100, y * 20); printf("Done: x=%g, y=%g\n", x, y); return 0; } $ gcc -Wall a.c $ ./a.out Starting... Done: x=2200, y=110 I don't like that

Dear members, I have the following code: > TB <- {x <- 3;y <- 5} > TB [1] 5 It is consistent with the documentation: For {, the result of the last expression evaluated. This has the visibility of the last evaluation. But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems?

I would like to be able, inside a function, to create a new function, and use it as part of a formula as an argument to, say, gnls or nlme. for example: MyTop <- function(data=dta) { Cexp <- function(dose,A,B,m){...} Model <- as.formula(paste("y","~ Cexp(",paste(formals(Cexp),collapse =", "),")")) MyCall <-

Hello Akshay, R is quite inspired by LISP, where this is a common thing. It is not in fact that {...} returned something, rather any expression evalulates to some value, and for a compound statement that is the last evaluated expression. {...} might be seen as similar to LISPs (begin ...). Now this is a very different thing compared to {...} in something like C, even if it looks or behaves

With rw2010dev I get a strange protect(): protection stack overflow error with a small data frame which otherwise is usable: If anybody wants to have a look I can provide an RData file with the problematic data frame. Doesn't seem to be necessary, the following simulated example generates the error: > testmat <- matrix(1:80, 20,4) > dim(testmat) [1] 20 4 > str(testmat) int

Hello all R gurus, I have a following problem which I hope someone will help me to solve. I have a data.frame in form similar to below. > testframe<-data.frame("Name"=c("aa","aa","aa","aa","aa","bb","bb","bb","bb","bb"),"Value"=c(1,100,1,1,1,100,100,100,100,1))

Is it possible to use R as a data-mining tool? Here's the problem I've got. I have a couple of data sets consisting of results from a cDNA microarray experiment - the details about the biology don't really matter here, the same theory applies for any other data-mining task (that's why I thought it'd be more appropriate to post this on r-user). Each of these datasets consists

Apologees the novice question. Currently climbing up the learning curve of R. Suppose I have the following function and the data.frame: testfun<-function(x=1,y=2) x+y testframe=data.frame(col1=c(1,2),col2=c(3,4)) When evaluating testfun, I want to use the default value for y (which is 2) and for x, I want to feed (one by one) the values in col2 of testframe. How can I achieve this please?

R-Help community, I understand that data.frames can hold elements of type double, string etc but NOT objects (such as a matrix etc). This is not convenient for me in the following situation. I have a function that takes 2 inputs and returns a vector: testfun <- function (x,y) seq(x,y,1) I have a data.frame defined as follows: testframe<-data.frame(xvalues=c(2,3),yvalues=c(4,5)) I would

Dear R list, I have been trying to do a linear model, extracting the effect of a covariate.... and the results do not match, when I do it with other programs (e.g. minitab).... so it is obvious that I was doing something wrong. Whan I do it with minitab, I have this results: (sector is a factor and depth is the covariate): Source DF Seq SS Adj SS Adj MS F P

Hmm. Chuck: I don't see how this example represents incomplete/incommensurate recycling. Doesn't TRUE replicate from length-1 to length-3 in this case (mat[c(TRUE,FALSE),2] would be an example of incomplete recycling)? William: clever, but maybe too clever unless you really need the speed? (The clever way is 8 times faster in the following case ...) x <- rep(1,1e6)

Dear all, I am using the nlsLM function to fit a Lorentzian function to my experimental data. The LM algorithm should allow to specify limits, but the upper limit appears not to work as expected in my code. The parameter 'w', which is peak width at half maximuim always hits the upper limit if the limit is specified. I would expect the value to be in-between the upper and lower limit with

Hi, 2 questions: Question 1: example of what I currently do: for(i in 1:6){sink("temp.txt",append=TRUE) dput(i+0) sink()} x=scan(file="temp.txt") print(prod(x)) file.remove("C:/R-2.5.0/temp.txt") But how to convert the output of the loop to a vector that I can manipulate (by prod or sum etc), without having to write and append to a file? Question 2: >

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

Hello! I work with: R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2 I think there is a bug in the conditional execution if (expr1) {expr2} else {expr3} If I try: "if (expr1) expr2 else expr3" it works well but when I put the expression expr2 and expr3 between {} I receive an error message

Unless you do something special within a function, only the value(s) returned are available to the caller. That is the essence of functional-type programming languages. You need to read up on (function) environments in R . You can search on this. ?function and its links also contain useful information, but it may too terse to be explicable to you. There are of course many available references on

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

?s 14:47 de 09/01/2023, akshay kulkarni escreveu: > Dear members, > I have the following code: > >> TB <- {x <- 3;y <- 5} >> TB > [1] 5 > > It is consistent with the documentation: For {, the result of the last expression evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the