similar to: ?subexpressions, D, deriv

Also, this should have gone in R-bugs quite a while ago : ------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)),

------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)), c("x","y"), function(x,y){}) function (x, y)

Dear all, for demonstration purposes I want to display the basis functions used by a thin plate regression spline in a gamm model. I've been searching the help files, but I can't really figure out how to get the plots of the basis functions. Anybody an idea? Some toy code : require(mgcv) require(nlme) x1 <- 1:1000 x2 <- runif(1000,10,500) fx1 <- -4*sin(x1/50) fx2 <-

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

Dear list # I have a function ff <- function(a,b=2,c=4){a+b+c} # which I programmatically want to modify to a more specialized function in which a is replaced by 1 ff1 <- function(b=2,c=4){1+b+c} # I do as follows: vals <- list(a=1) (expr1 <- as.expression(body(ff))) expression({ a + b + c }) (expr2 <- do.call("substitute", list(expr1[[1]], vals))) { 1 +

I've been trying to get comfy with arima() and associated functions in the ts() package. I'm thinking seriously about using this package, and R generally, in a 4th year intro time series course that I'm teaching this autumn. I have a couple of questions about arima: (1) The help file says that residuals component of the value returned by arima() consists of the

It is again perhaps my recent spate of bad sleeping that has prevented my brain from wrapping around this explanation, or it is perhaps my inherent hatred of regular expression syntax. However, I have been unable to put this into a working form after staring at it for a while and trying different recipes. If anyone wants to take a stab at this, I'd appreciate it. (from

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

Hello, I have a function for reading a data-frame from a file, which contains E = read.table(file = filename, header = T, colClasses = c(rep("integer",6),"numeric","integer",rep("numeric",8)), ...) Now a small variation arose, where colClasses =

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()

Hello, nls library provides 6 self-starting models, among them: SSfp, a four parameters logistic function. Its self-starting procedure involves several steps. One of these steps is: pars <- as.vector(coef(nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xydata, start = list(lscal = 0), algorithm = "plinear"))) which assumes an initial value of lscal equal to 0. If lscal

Hello, I'm running mixed models in GAMM4 with 2 (non-nested) random intercepts and I want to include a spline term for one of my exposure variables. However, when I include a spline term, I always get reported degrees of freedom of less than 1, even when I know that my spline is using more than 1 degree of freedom. For example, here is the code for my model: >

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: >

Today I ran into another aspect of the poison problem... Basically, SimplifyCFG wants to take expr1 && expr2 and flatten it into x = expr1 y = expr2 x&y This isn't safe when expr2 might execute UB. The consequence is that no LLVM shift instruction is safe to speculatively execute, nor is any nsw/nuw/exact variant, unless the operands can be proven to be in

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

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

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

Hi >i have aproblem withe execution of my function >first, i wrote my function in the script of R >nom_fonction <- function(arg1[=expr1], arg2[=expr2], ...){ bloc d'instructions } > when i want to have the result i mean the laste instruction in the bloc of > instruction , i try to >wrote the name of function >source(aj.fun) Error in readLines(file, warn =