similar to: R-beta: Small problems with R0.49

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 <-

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

Hey, everybody--- Ignorance CAN be bliss, at least for a while, but, .... Just so you know... A week or two ago, some upgrades to the expression parser (you know, the expressions you put in $[ ... ] in your extensions.conf file) that I submitted, have been merged into the CVS HEAD of the source. Hopefully, for around 99.9% of you, it won't make any difference to you. The Makefile has also

Dear list, Hi, I am trying to get the second derivative of a logistic formula, in R summary the model is given as : ### >$nls >Nonlinear regression model >model: data ~ logistic(time, A, mu, lambda, addpar) >data: parent.frame() > A mu lambda >0.53243 0.03741 6.94296 ### but I know the formula used is #

Quick question: Which function do you use to calculate partial derivatives from a model equation? I've looked at deriv(), but think it gives derivatives, not partial derivatives. Of course my equation isn't this simple, but as an example, I'm looking for something that let's you control whether it's a partial or not, such as: somefunction(y~a+bx, with respect to x,

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

Hi, I was hoping to get some advice on how to derive estimates of slopes from four parameter logistic models fit with SSfpl. I fit the model using: model<-nls(temp~SSfpl(time,a,b,c,d)) summary(model) I am interested in the values of the lower and upper asymptotes (parameters a and b), but also in the gradient of the line at the inflection point (c) which I assume tells me my rate of

Hello! I have the following problem: I have a function to construct three surfaceplots with a marker for an optimum, each of the plots has as title paste("Estimated ",pred.var.lab," for ",var.lab[1]," vs. ",var.lab[2],sep="") with different var.lab[1,2] each time. My problem is now that I need to allow for plotmath expressions in the variables pred.var.lab

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?

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: >

Dear everyone, the following is obviously used to compute the nth derivative, which seems to work (deriv(sqrt(1 - x^2),x,n)) However, before using this, I wanted to make sure it does what I think it does but can't figure it out when reading the ?deriv info or any other documentation on deriv for that matter: deriv(expr, namevec, function.arg = NULL, tag = ".expr", hessian = FALSE,

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

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

Hi, I am using R a lot to make plots relating to radioactivity, I am often using expression() to label the plots with nuclide names written with superscripts, e.g. expression(paste("Releases of ", { }^{99},Tc," (TBq/year)"))->ywtext But, is there any simple way to change the number and name of the nuclide through a variable? I tried nuccode=expression({ }^{99},Tc)

First, thanks to all who helped me with my question about rescaling axes on the fly. Using unlist() and range() to set the axis ranges in advance worked well. I've since plotted about 300 datasets with relative ease. Now I'm trying to fit a lossy oscillator resonance to (the square root of) a lorentzian (testframe$y is oscillator amplitude, testframe$x is drive frequency): lorentz

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

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()

I have a situation where this is fine: > if (length(x)>15) { clever <- rr.ATM(x, maxtrim=7) } else { clever <- rr.ATM(x) } > clever $ATM [1] 1848.929 $sigma [1] 1.613415 $trim [1] 0 $lo [1] 1845.714 $hi [1] 1852.143 But this variant, using ifelse(), breaks: > clever <- ifelse(length(x)>15, rr.ATM(x, maxtrim=7), rr.ATM(x))