similar to: deriv() to take vector of expressions as 1st arg?

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".

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))

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,

Does anybody know of a function that implements the derivative (gradient) of the multivariate normal density with respect to the *parameters*? It?s easy enough to implement myself, but I?d like to avoid reinventing the wheel (with some bugs) if possible. Here?s a simple example of the result I?d like, using numerical differentiation: library(mvtnorm) library(numDeriv) f=function(pars, xx, yy)

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: >

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,

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

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

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

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)))

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)),

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

Dear R useRs, I have a problem with nls.lm function of minpackl.lm package. I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic conductivity with nls.lm function of minpack.lm package. For the first fit, the parameter estimates keep changing even after 1000 iterations (Th) and I have a following error message for fit of hydraulic conductivity (k); Reason for

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

Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide

Hi, I am interested in using the numeric output from the "gradient" attribute of deriv's output in subsequent analyses. But, I have so far been unable to determine how to do so. I will use the example from the deriv help to illustrate. > ## function with defaulted arguments: > (fx <- deriv(y ~ b0 + b1 * 2^(-x/th), c("b0", "b1", "th"),

A non-text attachment was scrubbed... Name: not available Type: text Size: 1856 bytes Desc: not available Url : https://stat.ethz.ch/pipermail/r-help/attachments/19970430/db1498ee/attachment.pl

------- 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)