search for: expr5

...sult of the last expression evaluated. This has the visibility of the last evaluation. But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} Both expr5 and expr7 are created, and are accessible by the code outside of the nested braces right? But the "return value" of the nested braces is expr7. So doesn't this mean that only expr7 should be accessible? Please help me entangle this (of course the ret...

...ression evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code outside of the nested braces right? But the "return value" of the nested braces is expr7. So doesn't this mean that only expr7 should be accessible? Please help me entangle this (of cour...

...gnificant differences between the curves in two parameters that describe the pattern of growth. these parameters are from a logistic (r & k) . i have attempted to construct a self starting routine for nlme ie: SSGrowth_function(x, r, k) { .expr2 <- (k - 100000)/100000 .expr5 <- exp(((r * -1) * x)) .expr7 <- 1 + (.expr2 * .expr5) .expr13 <- .expr7^2 .value <- k/.expr7 .actualArgs <- match.call()[c("r", "k")] if(all(unlist(lapply(as.list(.actualArgs), is.name)))) { .grad <...

...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", "y&quo...

...evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be > advantageous for solving practical problems? Specifically, consider the > following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code outside > of the nested braces right? But the "return value" of the nested braces is > expr7. So doesn't this mean that only expr7 should be accessible? Please > help me entangle...

...truction 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("a",...

...evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be > advantageous for solving practical problems? Specifically, consider the > following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code outside > of the nested braces right? But the "return value" of the nested braces is > expr7. So doesn't this mean that only expr7 should be accessible? Please > help me entangle...

...xpression evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code outside of the nested braces right? But the "return value" of the nested braces is expr7. So doesn't this mean that only expr7 should be accessible? Please help me entangle this (of cours...

...visibility of the last evaluation. >> >> But both x AND y are created, but the "return value" is y. How can this >> be advantageous for solving practical problems? Specifically, consider the >> following code: >> >> F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} >> >> Both expr5 and expr7 are created, and are accessible by the code outside >> of the nested braces right? But the "return value" of the nested braces is >> expr7. So doesn't this mean that only expr7 should be accessible? Please >&...

...ted. This has the visibility of the last evaluation. >> >> But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: >> >> F <- function(X) {? expr; expr2; { expr5; expr7}; expr8;expr10} >> >> Both expr5 and expr7 are created, and are accessible by the code outside of the nested braces right? But the "return value" of the nested braces is expr7. So doesn't this mean that only expr7 should be accessible? Please help me entangle this (...

...valuation. >>> >>> But both x AND y are created, but the "return value" is y. How can this >>> be advantageous for solving practical problems? Specifically, consider the >>> following code: >>> >>> F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} >>> >>> Both expr5 and expr7 are created, and are accessible by the code outside >>> of the nested braces right? But the "return value" of the nested braces is >>> expr7. So doesn't this mean that only expr7 should be accessib...

...the visibility of the last evaluation. > > > > But both x AND y are created, but the "return value" is y. How can this > be advantageous for solving practical problems? Specifically, consider the > following code: > > > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > > > Both expr5 and expr7 are created, and are accessible by the code outside > of the nested braces right? But the "return value" of the nested braces is > expr7. So doesn't this mean that only expr7 should be accessible? Please > help m...

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 <-

...f the last evaluation. > >> > >> But both x AND y are created, but the "return value" is y. How can this > be advantageous for solving practical problems? Specifically, consider the > following code: > >> > >> F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > >> > >> Both expr5 and expr7 are created, and are accessible by the code > outside of the nested braces right? But the "return value" of the nested > braces is expr7. So doesn't this mean that only expr7 should be accessible? > Pleas...

...xpression evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code > outside of the nested braces right? But the "return value" of the > nested braces is expr7. So doesn't this mean that only expr7 should be > accessible? Please help me entan...

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 #

...f the last evaluation. > >> > >> But both x AND y are created, but the "return value" is y. How can this > be advantageous for solving practical problems? Specifically, consider the > following code: > >> > >> F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > >> > >> Both expr5 and expr7 are created, and are accessible by the code > outside of the nested braces right? But the "return value" of the nested > braces is expr7. So doesn't this mean that only expr7 should be accessible? > Pleas...

...xpression evaluated. This has the visibility of the last evaluation. > > But both x AND y are created, but the "return value" is y. How can this be advantageous for solving practical problems? Specifically, consider the following code: > > F <- function(X) { expr; expr2; { expr5; expr7}; expr8;expr10} > > Both expr5 and expr7 are created, and are accessible by the code > outside of the nested braces right? But the "return value" of the > nested braces is expr7. So doesn't this mean that only expr7 should be > accessible? Please help me entangle...

...r (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.name)))) { .grad <- array(...

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