'Rcpp Unexpected NaN results
I have a complicated equation with lots of pow's and was getting a NaN result. So I broke it into pieces, called temp1 to temp4. temp4 which is pow(temp3, 0.25) is where I get a nan which then results in a NaN being returned.
#include <Rcpp.h>
#include <cmath>
using namespace Rcpp;
const double kVal = 273.15;
// [[Rcpp::export]]
double thermalRadiance(double tas, double wind, double Tg) {
double temp1 = pow(Tg + kVal, 4.0);
double temp2 = pow(wind, 0.6) * (Tg - tas);
double temp3 = (temp1 + 2.5e+8 * temp2);
double temp4 = pow(temp3, 0.25);
double tr = temp4 - kVal;
Rcpp::Rcout << "tas " << tas << " wind " << wind << " Tg " << Tg << " temp1 " << temp1 << " temp2 " << temp2 << " temp3 " << temp3 << " temp4 " << temp4 << " tr " << tr << std::endl;
return tr;
}
/*** R
thermalRadiance(29., 4.5, 17.4)
# test of temp4 in R
-2.37018e+07^.25
*/
tas 29 wind 4.5 Tg 17.4 temp1 7.12662e+09 temp2 -28.6013 temp3 -2.37018e+07 temp4 nan tr nan
[1] NaN
> # test of temp4 in R
> -2.37018e+07^.25
[1] -69.77427
Seems really straightforward until I get to the temp4 value. Any assistance greatly appreciated!
Solution 1:[1]
You are trying to take the 4th root of a negative number in C++. Perhaps you should replace pow(temp3, 0.25) with -pow(abs(temp3), 0.25)
#include <Rcpp.h>
#include <cmath>
using namespace Rcpp;
const double kVal = 273.15;
// [[Rcpp::export]]
double thermalRadiance(double tas, double wind, double Tg) {
double temp1 = pow(Tg + kVal, 4.0);
double temp2 = pow(wind, 0.6) * (Tg - tas);
double temp3 = (temp1 + 2.5e+8 * temp2);
double temp4 = -pow(abs(temp3), 0.25);
double tr = temp4 - kVal;
Rcpp::Rcout << "tas " << tas << " wind " << wind << " Tg " << Tg << " temp1 "
<< temp1 << " temp2 " << temp2 << " temp3 " << temp3 << " temp4 "
<< temp4 << " tr " << tr << std::endl;
return tr;
}
Which in R gives you:
thermalRadiance(29., 4.5, 17.4)
tas 29 wind 4.5 Tg 17.4 temp1 7.12662e+09 temp2 -28.6013 temp3 -2.37018e+07 temp4 -69.7743 tr -342.924
[1] -342.9243
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Allan Cameron |
