similar to: How to get the result of eval()

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))

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 attempting to extract the derivative/ gradient from this expression df1p <- deriv(f1, "P") > df1p expression({ .value <- s - c - a * P .grad <- array(0, c(length(.value), 1L), list(NULL, c("P"))) .grad[, "P"] <- -a attr(.value, "gradient") <- .grad .value }) So in this case I want to extract the "-a".

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: >

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

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, 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 All, I would like to compute the simple finite-difference approximation to the gradient of a scalar function of a large number of variables (on the order of 1000). Although a one-time computation using the following function grad() is fast and simple enough, the overhead for repeated evaluation of gradient in iterative schemes is quite significant. I was wondering whether there are

Hello R community, I have been using R's inbuilt maximum likelihood functions, for the different methods (NR, BFGS, etc). I have figured out how to use all of them except the maxBHHH function. This one is different from the others as it requires an observation level gradient. I am using the following syntax: maxBHHH(logLik,grad=nuGradient,finalHessian="BHHH",start=prm,iterlim=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,

Dear R People: When the D function is used for a symbolic derivative, an expression is returned, which is fine. How do you change that expression to a function, please? I've been experimenting with substitute and deparse, but no success yet. This is R 1.8.0 for Windows. thanks in advance for the help! Sincerely, Erin Hodgess mailto: hodgess at gator.uhd.edu

Dear R-helpers I have a problem related to the use of NLME I think is simply a matter of getting the nlme coding correct, but i cannot get my brain around it I am analysing some 24 growth curves of some cells , and i wanted to say that there are significant differences between the curves in two parameters that describe the pattern of growth. these parameters are from a logistic (r & k)

Hi R-users, Recently, I use maxNR function to find maximizer. I have error appears as follows Error in maxNRCompute(fn = fn, grad = grad, hess = hess, start = start, : NA in the initial gradient My code is mu=2 s=1 n=300 library(maxLik) set.seed(1004) x<-rcauchy(n,mu,s) loglik<-function(mu) { log(prod(dcauchy(x,mu,s))) } maxNR(loglik,start=median(x))$estimate Does anyone know how

Dear Rhelpers I have the problem with initial values, could you please tell me how to solve it? Thank you June > p = summary(maxLik(fr,start=c(0,0,0,1,0,-25,-0.2))) Error in maxRoutine(fn = logLik, grad = grad, hess = hess, start = start, : NA in the initial gradient > p = summary(maxLik(fr,start=c(0,0,0,1,0,-25,-0.2),method="BFGS")) Error in optim(start, func, gr =

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)))

Merry Xmas to all, I am writing a function and curiously this runs sometimes on one data set and fails on another and i cannot figure out why. Any help much appreciated. If i run the code below with data <- iris[ ,1:4] The code runs fine, but if i run on a large dataset i get the following error (showing data structures as matrix is large) > str(cluster.data) num [1:9985, 1:811] 0 0 0 0

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

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:

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)),

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.