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...xpression of this derivative. For example: Suppose that I have a function of this form: f(x,y)=x^3+y^3+(x^2)*(y^2). With "deriv3" I can evaluate the first derivative and the hessian matrix, as follows: > d<-deriv3(~x^3+y^3+(x^2)*(y^2),c("x","y")) > d[[1]] { .expr4 <- x^2 .expr5 <- y^2 .expr9 <- 2 * x .expr15 <- 2 * y .value <- x^3 + y^3 + .expr4 * .expr5 .grad <- array(0, c(length(.value), 2), list(NULL, c("x", "y"))) .hessian <- array(0, c(length(.value), 2, 2), list(NULL, c("x", "y"), c("x&q...

...nte, sin default 1.Construction to objectiv functin with n=1 data >fllwform<-formula(~log(a)-log(b)-(a-1)*(log(x)-log(b))- (x^a)/(b^a)) > fllwfuncH <-deriv(fllwform,c("a","b"),function(a,b,x){}) > fllwfuncH function (a, b, x) { .expr2 <- log(b) .expr4 <- a - 1 .expr5 <- log(x) .expr6 <- .expr5 - .expr2 .expr9 <- x^a .expr10 <- b^a .expr19 <- .expr10^2 .expr23 <- 1/b .value <- log(a) - .expr2 - .expr4 * .expr6 - .expr9/.expr10 .grad <- array(0, c(length(.value), 2), list(NULL, c(&...

...on({ .value <- dnorm(z, 0, s) .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(z * dnorm(z)) attr(.value, "gradient") <- .grad .value }) deriv(~exp(-z^2/(2*s^2))/s/sqrt(2*pi),"z") expression({ .expr4 <- 2 * s^2 .expr6 <- exp(-z^2/.expr4) .expr9 <- sqrt(2 * pi) .value <- .expr6/s/.expr9 .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(.expr6 * (2 * z/.expr4)/s/.expr9) attr(.value, "gradient") &lt...

...iv(*, *, 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) { .expr1 <- cos(x) .expr2 <- .expr1 * y .expr4 <- cos(.expr2) .value <- sin(.expr2) .grad <- array(0, c(length(.value), 2), list(NULL, c("x", "y"))) .grad[, "x"] <- -.expr4 * (sin(x) * y) .grad[, "y"] <- .expr4 * .expr1 attr(.value, "gradient") <- .grad...

...z, SSmicman, SSweibull, etc). my question is, how to derive the function of parameters. and also which model to use for get the initials values. In the following example's the red color coding requires the explanation?? thanks SSgompertz function (x, Asym, b2, b3) { .expr2 <- b3^x .expr4 <- exp(-b2 * .expr2) .value <- Asym * .expr4 .actualArgs <- as.list(match.call()[c("Asym", "b2", "b3")]) if (all(unlist(lapply(.actualArgs, is.name)))) { .grad <- array(0, c(length(.value), 3L), list(NULL, c("Asym",...

...rg = ) 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) { .expr1 <- cos(x) .expr2 <- .expr1 * y .expr4 <- cos(.expr2) .value <- sin(.expr2) .grad <- array(0, c(length(.value), 2), list(NULL, c("x", "y"))) .grad[, "x"] <- -.expr4 * (sin(x) * y) .grad[, "y"] <- .expr4 * .expr1 attr(.v...

...e with the dataset provided hereunder (see end of this message). As a workaround, I propose a modified self-start function for SSfpl that remedy to this: --------------------------------------- fpl <- function (input, A, B, xmid, scal) { .expr1 <- B - A .expr2 <- xmid - input .expr4 <- exp((.expr2/scal)) .expr5 <- 1 + .expr4 .expr8 <- 1/.expr5 .expr13 <- .expr5^2 .value <- A + (.expr1/.expr5) .actualArgs <- as.list(match.call()[c("A", "B", "xmid", "scal")]) if (all(unlist(lapply(.actualArgs, is.n...

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

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)))

...platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32 status major 1 minor 3.0 year 2001 month 06 day 22 language R ----------------------------------------------------------------- > deriv(~offv+Asym/(1+exp(-(x-xmid)/scal)),"x") expression({ .expr4 <- exp(-x - xmid/scal) #### Wrong !!!! > deriv(~offv+Asym/(1+exp(-x/scal+xmid/scal)),"x") expression({ .expr5 <- exp(-x/scal + xmid/scal) ## Correct > deriv(~offv+Asym/(1+exp((xmid-x)/scal)),"x") expression({ .expr3 <- exp((xmid - x)/scal) ## Coorrect...

...not try-inside-try: > > > > foo() { > > try { > > expr1; > > } finally { > > try { > > expr2; > > } finally { > > try { > > expr3; > > } finally { > > try { > > expr4; > > } finally > > … > > } > > } > > } > > } > > } > > > > With duplication, the non-EH path through foo will include each of expr1 > through exprn. The outermost funclet will include (in its non-EH pat...

I think adding transitions to cleanupblocks on the normal execution path would be an optimization barrier. Most passes would see the cleanupblock instruction and run the other way. It's definitely appealing from the perspective of getting the smallest possible code, but I'm OK with having no more than two copies of everything in the finally block. I think with the addition of the