search for: namevec

...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))) try #1 deriv(nls.fn) #get an Error in .Internal(deriv.default(expr, namevec, function.arg, tag, hessian)) : 'namevec' is missing try #2 deriv(nls.fn, namevec=c("asym", "xmid", "scal")) #this doesn't seem to give me the expression, and the gradients are zero. I've tried to do this with Ryacas as well, but I'm lost. Can an...

Hi experts, I am trying to write a very flexible method that allows me to add a new column to an existing data frame. This is what I have so far: add.column <- function(df, new.col, name) { n.row <- dim(df)[1] length(new.col) <- n.row names(new.col) <- name return(cbind(df, new.col)) } df <- NULL df <- data.frame(a=c(1,2,3)) df # corect: added NA to new collumn df <-

...a"), function(phi, theta) { : Funci?n '`[`' no est? en la tabla de derivadas dfg<-for(n in 1:2) as.formula(sprintf("y~(1 - phi * theta) * (phi - theta)*phi^ %d",n)) dfg y ~ (1 - phi * theta) * (phi - theta) * phi^2 deriv(dfg[[1]],phi) Erro en deriv.default(expr, namevec, function.arg, tag, hessian) : nombres de variable inv?lidos Give me a help, please. Bernardo.

...ress expression(y ~ "A"/(1 + exp((4 * "mu"/"A") * ("lambda" - "time")) + 2)) > > d1<-deriv(express) Error in deriv.default(express) : element 2 is empty; the part of the args list of '.Internal' being evaluated was: (expr, namevec, function.arg, tag, hessian) #### Why is this not working and how do I get the second derivative? Thank you for reading my post, all help is appreciated, Kitty [[alternative HTML version deleted]]

...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, ...) This doesn't seem to include an "order of derivative" argument, in fact, in the examples section, it is outlined, how one can build a higher deriv. function oneself... Any hints are much appreciated! alex -- View t...

...ients to build the polynomial. coef<-as.matrix(model1$coefficients) In the next step I need to define the polynomial to derive it. If I write the coefficients manually (writing the numbers by hand) the deriv command works fine! bb<-deriv(~2847.22015 -463.06063*x+ 25.43829*x^2 -0.17896*x^3, namevec="x", function.arg=40) I would like to automate this step by being able to extract the coefficients from the linear model and adding them into the polynomial (and not write them by hand)! But if I build the polynomial with the function(x) command calling the * coef* values, the numeric...

Hello, and please excuse this off-topic question, but I have not been able to find an answer elsewhere. Consider a value Z that is calculated using the product (or ratio) of two means X_mean and Y_mean: Z=X_mean*Y_mean. More generally, Z=f(X_mean, Y_mean). The standard error of Z will be a function of the standard errors of the means of X and Y. I want to calculate this se of Z. Can someone

...^2 likelihood <- const*exp(exponent) #defining the merit function -sum(log(likelihood)) } deriv2 <- deriv( expr = ~ -log(1/(sqrt(2*pi)*S)*exp((-1/(2*(S^2)))* (log.conc-(log(dose*Ke*Ka/(Cl*(Ka-Ke))) +log(exp(-Ke*time)-exp(-Ka*time))))^2)), namevec = c("Ke","Ka","Cl","S"), function.arg = function(Ke, Ka, Cl, S, dose, time, log.conc) NULL ) gradient2.1compart <- function(parms, dose, time, log.conc) { Ke <- parms[1]; Ka <- parms[2]; Cl <- parms[3]; S <- parms[4] colSums(attr(deriv2.1...

...* exp( a + b*t ) graphmodeldpdt <- coef( graphmodeld )[ 2 ] * graphmodelp graphmodeldpdt #> 1 2 3 #> -0.31464113 -0.11310757 -0.08042364 # Ellison suggestion - fancy, only works for simple functions dc <- deriv( expression( exp( a + b * t ) ) , namevec = "t" ) dcf <- function( t ) { cgm <- coef( graphmodeld ) a <- cgm[ 1 ] b <- cgm[ 2 ] eval(dc) } result <- dcf( nd ) result #> [1] 46.13085 16.58317 11.79125 #> attr(,"gradient") #> t #> [1,] -0.31464113 #> [2,...

...0.65.1, and later someone added to help(deriv) 's description of function.arg \item{function.arg}{If specified, a character vector of arguments for a function return, or a function (with empty body) or \code{TRUE}, the latter indicating that a function with argument names \code{namevec} should be used. the comment >>>> \bold{Note:} this is incompatible with S.} Today, I've been looking a bit at the C code and I can't understand why giving the function.arg as a function doesn't work anymore. (it just returns the 'empty' fun...

> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > >> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote: >> >>> Sent: Friday, April 06, 2018 at 5:55 AM >>> From: "David Winsemius" <dwinsemius at comcast.net> >>> >>> >>> Not correct. You already have

Dear R-users, I would like to calculate elasticities and sensitivities of each parameters involved in the following transition matrix: A <- matrix(c( sigma*s0*f1, sigma*s0*f2, s, v ), nrow=2, byrow=TRUE,dimnames=list(stage,stage)) The command "eigen.analysis" avaliable in package "popbio" provides sensibility matrix and elasticity matrix

Hi, I need to write a function that would look something like this: S <- function(b=betas){ expression(b[1] * f(b[2] * x * f(b[3] * x * f(...b[n-1] * x * f(b[n] * x)))...) } Where n is the number of element in b. Further I need to be able to evaluate S at some x numerically of course and I need to use "deriv" and produce dS/dx such that I can evaluate it also at some x. I