similar to: numericDeriv

I have to compute some standard errors using the delta method and so have to use the command "numericDeriv" to get the desired gradient. Befor using it on my complicated function, I've done a try with a simple exemple : x <- 1:5 numericDeriv(quote(x^2),"x") and i get : [1] 1 8 27 64 125 216 attr(,"gradient") [,1] [,2] [,3] [,4] [,5] [,6] [1,] Inf

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

Hi all. I have a numerical function f(x), with x being a vector of generic size (say k=4), and I wanna take the numerically computed gradient, using deriv or numericDeriv (or something else). My difficulties here are that in deriv and numericDeric the function is passed as an expression, and one have to pass the list of variables involved as a char vector... So, it's a pure R programming

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.

> On only ten points, what did you expect ? Even with 1000 > observations, estimating a density is difficult, and has > been the subject of a century of research. Kernel density > estimates are among the most successful. For your immediate > application, try plot(density(rnorm(10)), type="l"), etc. wait, you misunderstood me! I'd like to see 10 or 9 points with

Hi, I am stuck on something for a couple days, I am almost about to give up. This looks simple, but I can't figure out. I hope I can get some help here. I am trying to do some symbolic and numerical derivations. Let me explain the problem. Let's say, I have a matrix as follows: > load <- matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE) > > load [,1] [,2] [1,] 3 0

Hi, I am a non-expert user of R. I am essaying the fit of two different functions to my data, but I receive two different error messages. I suppose I have two different problems here... But, of which nature? In the first instance I did try with some different starting values for the parameters, but without success. If anyone could suggest a sensible way to proceed to solve these I would be

Dear R developers, I've run into a weird behavior of the numericDeriv function (from the stats package) which I also posted on StackOverflow (question has same title as this email, except for the version of R). Running the code bellow we can see that the numericDeriv function gives an error as the derivative of x^a wrt a is x^a * log(x) and log is not defined for negative numbers. However,

Dear all As far as I could trace, looking at the function C function numeric_deriv, this unwanted behavior comes from the inner most loop in, at the very end of the function, for(i = 0, start = 0; i < LENGTH(theta); i++) { for(j = 0; j < LENGTH(VECTOR_ELT(pars, i)); j++, start += LENGTH(ans)) { SEXP ans_del; double origPar, xx, delta; origPar = REAL(VECTOR_ELT(pars, i))[j];

Hi, df <- read.table("data.txt", header=T); library(nls); fm <- nls(y ~ a*(x+d)^(-b), df, start=list(a=max(df->y,na.rm=T)/2,b=1,d=0)); I was using the following routine which was giving Singular Gradient, Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an Infinity produced when evaluating the model errors. I also tried the

Dear All, I am running a simulation to obtain coverage probability of Wald type confidence intervals for my parameter d in a function of two parameters (mu,d). I am optimizing it using "optim" method "L-BFGS-B" to obtain MLE. As, I want to invert the Hessian matrix to get Standard errors of the two parameter estimates. However, my Hessian matrix at times becomes

R Help List -- I have defined two time-series-vector-valued-functions, let them be f and g, and want to find the numeric derivative of f with respect to the variable x where f depends on x through g: (d/dx)(f (g(x) ) Moreover, x is a vector I tried this out the long way (naming every element of the x vector and then making the 'theta' argument in numericDeriv() the character vector of

Dear R users, I am running a maximum likelihood model with optim. I chose the simulated annealing method (method="SANN"). SANN is not performing bad, but I guess it would be much more effecive if I could set the `parscale' parameter. The help sais: `parscale' A vector of scaling values for the parameters. Optimization is performed on `par/parscale' and these

Hi All, following expression: x <- sort(rnorm(10)); e <- ecdf(x); d <- numericDeriv(e(x),"x"); makes d far from approximation of one dimensional pdf. What's wrong then here? Kind regards. --------------------------------------------------------------------------- Valery A.Khamenya Bioinformatics Department BioVisioN AG, Hannover

GENERAL REFERENCE ON NONLINEAR OPTIMIZATION? What are your favorite references on nonlinear optimization? I like Bates and Watts (1988) Nonlinear Regression Analysis and Its Applications (Wiley), especially for its key insights regarding parameter effects vs. intrinsic curvature. Before I spent time and money on several of the refences cited on the help pages for "optim",

Dear All, I am trying to find MLE by using "OPTIM" function. Difficult in differentiating some parameter in my objective function, I would like to use the returned hessian matrix to yield an estimate of Fisher's Information matrix. My question: Since the hessian is calculated by numerical differentiate, is it a reliable estimate? Otherwise I would have to do a lot of work to

I've been working on a new package and I have a few questions regarding the behaviour of the nlm function. I've been (for better or worse) using the nlm function to fit a linear model without suppling the hessian or gradient attributes in the objective function. I'm curious as to why the nlm requires 31 iterations (for the linear model), and then it doesn't work when I try to add

Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial

Dear R-users Coming from a proc mixed (SAS) background I am trying to get into the use of (n)lme. In this connection, I have some (presumably stupid) questions which I am sure someone out there can answer: 1) With proc mixed it is easy to get a hold on the estimated variance parameters as they can be put out into a SAS data set. How do I do the same with lme-objects? For example, I can see the

Hello: Below is a toy logistic regression problem. When I wrote my own code, Newton-Raphson converged in three iterations using both the gradient and the Hessian and the starting values given below. But I can't get nlm() to work! I would much appreciate any help. > x [1] 10.2 7.7 5.1 3.8 2.6 > y [1] 9 8 3 2 1 > n [1] 10 9 6 8 10 derfs4=function(b,x,y,n) {