Shawn Way

2018-May-30 18:14 UTC

### [R] Evaluation failure of IAPWS95 functions in a rowwise manner (tidyverse style)

I'm trying to use the IAPWS95 package with the tidyverse packages. For some reason, the function is not outputting the correct rho. A minimal example with results is below. I've also included the definition of the DTp function from the IAPWS95 library. ===================================library(IAPWS95) library(tidyverse) initial <- data.frame(T=c(279,294),p=c(0.46,0.46)) initial2 <- initial %>% mutate(rho=DTp(T,p))> DTpfunction (T, p) { y <- 0 icode <- 0 res <- .Fortran("DTp", as.double(T), as.double(p), as.double(y), as.integer(icode)) options(digits = 9) if (res[[4]] != 0) { error <- as.character(errorCodes[which(errorCodes[, 1] == res[[4]]), 2]) print(error) } print(res[[3]]) } <bytecode: 0x0000000006f520e0> <environment: namespace:IAPWS95> Results:> initial2T p rho 1 279 0.46 1000.12283 2 294 0.46 1000.12283 What the results should be:> initial2T p rho 1 279 0.46 1000.12283 2 294 0.46 998.19167 I think something is evaluating incorrectly, but I don't know enough about NSE vs SE to figure this out. Any thoughts to what could be the reason why the second row has the same rho as the first? When I run the function DTp individually, I get the right results. Shawn Way ???

Ista Zahn

2018-May-30 19:28 UTC

### [R] Evaluation failure of IAPWS95 functions in a rowwise manner (tidyverse style)

Hi Shawn, I don't think it has anything to do with the tidyverse. If you keep simplifying your example you'll get all the way down to> DTp(T=c(279,294),p=c(0.46,0.46))[1] 1000.12283 --Ista On Wed, May 30, 2018 at 2:14 PM, Shawn Way <SWay at meco.com> wrote:> I'm trying to use the IAPWS95 package with the tidyverse packages. For some reason, the function is not outputting the correct rho. > > A minimal example with results is below. I've also included the definition of the DTp function from the IAPWS95 library. > ===================================> library(IAPWS95) > library(tidyverse) > > initial <- data.frame(T=c(279,294),p=c(0.46,0.46)) > initial2 <- initial %>% > mutate(rho=DTp(T,p)) > >> DTp > function (T, p) > { > y <- 0 > icode <- 0 > res <- .Fortran("DTp", as.double(T), as.double(p), as.double(y), > as.integer(icode)) > options(digits = 9) > if (res[[4]] != 0) { > error <- as.character(errorCodes[which(errorCodes[, 1] => res[[4]]), 2]) > print(error) > } > print(res[[3]]) > } > <bytecode: 0x0000000006f520e0> > <environment: namespace:IAPWS95> > > Results: > >> initial2 > T p rho > 1 279 0.46 1000.12283 > 2 294 0.46 1000.12283 > > > What the results should be: > >> initial2 > T p rho > 1 279 0.46 1000.12283 > 2 294 0.46 998.19167 > > I think something is evaluating incorrectly, but I don't know enough about NSE vs SE to figure this out. > > Any thoughts to what could be the reason why the second row has the same rho as the first? When I run the function DTp individually, I get the right results. > > > Shawn Way > > > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.