similar to: nls

Is there anyway to change any y[i] value (i=2,...6) to make following NLS workable? x <- c(0,5,10,15,20,25,30) y <- c(1.00000,0.82000,0.68000,0.64000,0.66667,0.68667,0.64000) lm(1/y ~~ x) nls(1/y ~~ a+b*x^c, start=list(a=1.16122,b=0.01565,c=1), trace=TRUE) #0.0920573 : 1.16122 0.01565 1.00000 #Error in numericDeriv(form[[3]], names(ind), env) : # Missing value or

how to realize the following SQL command in R? select distinct A, B, count(C) from TABLE group by A, B ; quit; Best Regards --------------------------------- Sponsored Link Get a free Motorola Razr! Today Only! Choose Cingular, Sprint, Verizon, Alltel, or T-Mobile. [[alternative HTML version deleted]]

How to obtain the FUNCTION for the following smooth curve? x 0 100 250 500 1000 4000 y 1.8 1.2 1.02 0.99 0.97 0.85 Thanks, SJ --------------------------------- [[alternative HTML version deleted]]

I typed the following commands but it still use maxiter=50 after the 2nd command: nls.control(maxiter = 1000) nls(......) Thanks! --------------------------------- [[alternative HTML version deleted]]

x<-c(-1.873....,-0.121) # 23 numerics; x.arma12 <- armaFit(x ~ arma(1,2)) #estimates y[t]= -0.11465 - 0.23767 y[t-1] - 0.14230 e[t-1] -0.85770 e[t-2] + e[t]; # ? how to predict 46 steps ahead based on 23 data points? # the following doesn't work since n is in armaSim rather than armaFit; predict(x.arima12, n.ahead=46) # Thanks ---------------------------------

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

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

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

> 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

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

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,

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, 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, 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

I'm running into problems trying to use the nls function to fit the some data. I'm invoking nls using nls(s~k/(a+r)^b, start=list(k=1, a=13, b=0.59)) but I get errors indicating that the step has been reduced below the minimum step size or an inifinity is generated in numericDeriv. I've tried to use a variety of starting values for a, b, k but get similar errors. Is there

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

Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown