similar to: bugs in deriv(*, *, function.arg = ) (PR#953)

------- 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)

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

Full_Name: bernardo moises lagos alvarez Version: 2.4.0 OS: Windows XP professional Submission from: (NULL) (152.74.219.16) I need obtained the MLE of weibull parameters using the nlm with exact gradient an hessian. I am doing. bug report :Erro en log(b) : el argumento "b" est? ausente, sin default 1.Construction to objectiv functin with n=1 data

Dear all i want to fit the self start model in nls. i have two question. i have a function, (asfr ~ I(((a*b)/c))+ ((c/age)^3/2)+ exp((-b^2)*(c/age)+(age/c)-2) i am wondering how to build the selfstart model. there is lost of example, (i.e. SSgompertz, SSmicman, SSweibull, etc). my question is, how to derive the function of parameters. and also which model to use for get the initials values. In the

Full_Name: Joerg Polzehl Version: 2.3.1 OS: x86_64, linux-gnu Submission from: (NULL) (62.141.176.22) I observed an incorrect behavior of function deriv when evaluating arguments of dnorm deriv(~dnorm(z,0,s),"z") expression({ .value <- dnorm(z, 0, s) .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(z * dnorm(z))

Hi all, I am not used to using the computer to do calculus and have up to now done my differentiation "by hand" , calling on skills I learned many years ago and some standard cheat sheets. My interest at present is in getting the second derivative of a gaussian, which I did by hand and results in a somewhat messy result involving terms in sigma^5 .. I have done some spot checks

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 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, 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 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 #

Hello, I am wondering if someone can help me. I have the following function that I derived using nls() SSlogis. I would like to find its derivative. I thought I had done this using deriv(), but for some reason this isn't working out for me. Here is the function: asym <- 84.951 xmid <- 66.90742 scal <- -6.3 x.seq <- seq(1, 153,, 153) nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))

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

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"

Dear participants to the list, this is my problem: I want to obtain an expression that represents the second derivative of one function. With "deriv3" (package "stats") it is possible to evaluate the second derivative, but I do not know how I can get the (analytical) expression of this derivative. For example: Suppose that I have a function of this form:

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 +

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

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

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,

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 =