similar to: Extracting formulae from expression() / deriv()

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,

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))

The deriv() function takes an 'expression' as its first argument). I was wondering if the this function can take an array or a vector of expressions as its first argument. Aside, I saw how to give a vector argument to the second argument. like to have something like: deriv(c(~x^2+y^3, ~x^5+y^6), c("x","y")) the documentation for this function talks about being able to

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, 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,

Hi, Suppose I have an arbitrary function: arbfun<-function(x) {...} Is there a robust implementation of a numerical derivative routine in R which I can use to take it's derivative ? Something a bit more than simple division by delta of the difference of evaluating the function at x and x+delta... Perhaps there is a way to do this using D or deriv but I could not figure it out.

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)),

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)))

------- 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)

I'm trying to obtain numerical derivative of a probability computed with mvtnorm with respect to its parameters using grad() and jacobian() from NumDeriv. To simplify the matter, here is an example: PP1 <- function(p){ thetac <- p thetae <- 0.323340333 thetab <- -0.280970036 thetao <- 0.770768082 ssigma <- diag(4) ssigma[1,2] <- 0.229502120

I just discover the deriv function but I have a minor problem at the end when using its result: For example: dx2x <- deriv(~ A*x^2, "x") ; dx2x # it works fine: # expression({ # .value <- A * x^2 # .grad <- array(0, c(length(.value), 1L), list(NULL, c("x"))) # .grad[, "x"] <- A * (2 * x) # attr(.value, "gradient") <- .grad # .value # }) A

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

I fit my CDF to sum of exponentials and now I want to take the numerical derivative of this function to obtain probability density.I will really appreciate your help reagrding the error messages I am getting which I don't understand. * * > fitterma <- function(xtime) { a <- -0.09144115 b <- -0.01335756 c <- -2.368057 d <- -0.00600052

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Hi, I have been trying to do numerical differentiation using R. I found some old S code using Richardson Extrapolation which I managed to get to work. I am posting it here in case anyone needs it. ######################################################################## richardson.grad <- function(func, x, d=0.01, eps=1e-4, r=6, show=F){ # This function calculates a numerical approximation

Dear group, Here is my data.frame : df <- structure(list(DESCRIPTION = c("PRM HGH GD ALU", "PRM HGH GD ALU", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ",

Found this solution. It is maybe not the most elegant way, but it does the job. > a=as.data.frame(substr(lme$DESCRIPTION,1,14)) > colnames(a)=c("DESCRIPTION") > lme=as.data.frame(c(a,lme[,2:3])) > lme DESCRIPTION CLOSING.PRICE POSITION 1 PRIMARY NICKEL 25,755.7100 0 2 PRIMARY NICKEL 25,760.8600 0 3 PRM HGH GD ALU 2,415.9000 0

Hi there, I have a function which has a variable called show as an input: richardson.grad <- function(func, x, d=0.01, eps=1e-4, r=6, show=F){ # do some things if(show) { cat("\n","first order approximations", "\n") print(a.mtr, 12) } #do more things and return } The show variable is being used as a flag to show intermediate