search for: expr8

...t 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 value of F...

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 <-

...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...

...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(0, c(length(.value), 4), lis...

...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...

...ated. 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 va...

...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 (of course the return v...

...he 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 ent...

...he 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...

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

...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...

...uation. > >> > >> 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...

...ated. 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 co...

...uation. > >> > >> 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...

...ated. 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...

...ated. 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 co...

...uation. > >> > >> 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...

...uation. > >> > >> 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...

...;> > >> 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 enta...