search for: numericderivative

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

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,

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

...he 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 estimated values of *numerical* derivatives according to ecdf output. And that's it. Now look into output of numericDerivative in my example: [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 attr(,"gradient") [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0 0 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 0 0 [3,] 0 0 0 0 0 0 0...

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

Thanks; definitely a bug. I've submitted it to the bug tracker at https://bugs.r-project.org/bugzilla/show_bug.cgi?id=17831 Best, luke On Mon, 15 Jun 2020, Raimundo Neto wrote: > 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,

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 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

...rm(10000) ) dx <- diff(x) ed <- 1/10000/dx plot(x[-1], ed, log="y") # my "empirical density" lines(x,dnorm(x),col=2) Now I could have estimation for differential entropy like this: -sum(ed*log(ed)*dx) That's it. > What is this function `numericDerivative': do you mean `numericDeriv'? yes. Sorry, there is no auto-completion function in my non-emacs email client as in emacs' ESS environment ;-) kind regards, Valery A.Khamenya --------------------------------------------------------------------------- Bioinformatics Department BioVisioN...

Hello, I don't understand the description / help-text for the numericDeriv() function. Why is there a new environment used? And what is meant with an environment here? Is it similar or the same as a local workspace, like an environment in functional languages? And why is it needed here? numericDeriv could just calculate the difference bewtween two values and divide this difference by the

Hello All: I have defined the following function (fitterma as a sum of exponentials) that best fits my cumulative distribution. I am also attaching the "xtime" values that I have. I want to try two things as indicated below and am experiencing problems. Any help will be greatly appreciated. Best, Parmee ----------------------- *fitterma <- function(xtime) { * *a <-

Dear All, I am trying to solve a Generalized Method of Moments problem which necessitate the gradient of moments computation to get the standard errors of estimates. I know optim does not output the gradient, but I can use numericDeriv to get that. My question is: is this the best function to do this? Thank you Jean,

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

UseRs, I used the optim function valor.optim <- optim(c(1,1,1),logexp1,method ="BFGS",control=list(fnscale=-1),hessian=T); and I want to calculate the derivates, psi1<-valor.optim$par[1] psi2<-valor.optim$par[2] psi3<-valor.optim$par[3] a0=exp(psi1); a1=exp(psi2)/(20+exp(psi2)+exp(psi3)); a2=exp(psi3)/(20+exp(psi2)+exp(psi3))

Moved from r-help: On Thu, 14 Aug 2003 09:08:26 -0700, Spencer Graves <spencer.graves@pdf.com> wrote : >p.s. The following command in S-Plus 6.1 seems to work fine but >produces an error in R 1.7.1: > >nls(y~a, data=tstDf, start=list(a=1)) >Error in nlsModel(formula, mf, start) : singular gradient matrix at >initial parameter estimates This looks like a bug in

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

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.

Ah... I searched for half an hour for this function... you know, the help function in R could really be a lot better... But wait a minute... looking at this, it appears you have to pass in an expression. What if it is an unknown function, where you only have a handle to the function, but you cannot see it's implementation ? Will this work then ? -----Original Message----- From: Berton Gunter

Hi Everyone, I am trying to use NLS to fit a dataset using a Kappa function, but I am having problems. Depending on the start values that I provide, I get either: Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model Or Error in nls(FldFatRate ~ funct3(MeanDepth_m, h, k, z, a), data = data1, : singular gradient I think these

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