search for: expr7

...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 return val...

...n 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 <- array(0, c(length(.value), 2), list(NULL...

...ed. 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 (...

...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...

...ed. 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 (...

...on 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 r...

...ity 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 >> hel...

...is 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...

...on. >>> >>> 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? Ple...

...ibility 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...

...ast 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...

...on 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 thi...

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 #

...ast 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...

...on 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 (...

...on 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 thi...

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 <-

...ast 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...

...ast 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...

...> >> > >> 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...