similar to: higher derivatives using deriv

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,

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 #

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)))

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)),

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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))

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

------- 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)

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: >

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, 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 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)

Dear ExpeRts, I have trouble implementing a function which computes the k-th derivative of a specified function f and returns it as a function. I tried to adapt what I found under ?deriv but could not get it to work. Here is how it should look like: ## specify the function f <- function (x,alpha) x^alpha ## higher derivatives DD <- function(expr, variable, order = 1) { if(order <

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.

Hi all, I need to automate a process in order to prepare a a big loop in the future but I have a problem with the *command function* First I fit a model with lm > model1<-lm(data2[,2]~data2[,1]+I(data2[,1]^2)+I(data2[,1]^3)+I(data2[,1]^4)) I extract the coefficients to build the polynomial. coef<-as.matrix(model1$coefficients) In the next step I need to define the polynomial to

I would like to be able, inside a function, to create a new function, and use it as part of a formula as an argument to, say, gnls or nlme. for example: MyTop <- function(data=dta) { Cexp <- function(dose,A,B,m){...} Model <- as.formula(paste("y","~ Cexp(",paste(formals(Cexp),collapse =", "),")")) MyCall <-

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

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

Hello expeRts, I need generate symbolize the autocovariances matrix of a Gaussian ARMA(1,1), for derivate it and evaluate. I try this codes, but whitout sucess vacv<-NULL vacv[1]<-1-2*phi*theta-theta^2 vacv[2]<-(1-phi*theta)*(phi-theta) vacv[3:n]<-acv[2]*(phi^(1:(n-2))) facv<-list() for(i in 1:2)