R help - May 2018 - 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))
> 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

???

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.

The printed output is an issue (I've emailed the maintainer about it)
however, I think I figured it out.

It seems that the function is not vectorized, making me need to do the
following:

mutate(SG=map2(T,p,function(x,y) DTp(x,y))

This was able to get the correct results.  Like usual, the error was between the
head and the keyboard.

Thanks!

Shawn Way

-----Original Message-----
From: Ista Zahn <istazahn at gmail.com> 
Sent: Wednesday, May 30, 2018 2:28 PM
To: Shawn Way <SWay at meco.com>
Cc: r-help at r-project.org
Subject: Re: [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.