I'm having a problem while trying to learn RcppParallel. I tried to modify the code form https://rcppcore.github.io/RcppParallel/ for the vector summation to calculate the mean of a vector instead, to see if I understand the general principle.
My code is shown below. Using the function parallelVectorMean() on c(1,2,3,4,5) delivers inconsistent and most often incorrect results. I assume this has to do with me not understanding how to access begin and end correctly to scale my partial means accordingly during the join.
// [[Rcpp::depends(RcppParallel)]]
#include <RcppParallel.h>
#include <Rcpp.h>
using namespace RcppParallel;
struct Mean : public Worker
{
// source vector
const RVector<double> input;
// accumulated value
double value;
// number of elements
double num;
// constructors
Mean(const Rcpp::NumericVector input) : input(input), value(0), num(0) {}
Mean(const Mean& mean, Split) : input(mean.input), value(0), num(0) {}
// accumulate just the element of the range I've been asked to
void operator()(std::size_t begin, std::size_t end) {
num = (double) end - (double) begin;
value += (std::accumulate(input.begin() + begin, input.begin() + end, 0.0) / num);
}
// join my value with that of another Mean
void join(const Mean& rhs) {
value = (num*value + rhs.num*rhs.value)/(num + rhs.num);
num = num + rhs.num;
}
};
// [[Rcpp::export]]
double parallelVectorMean(Rcpp::NumericVector x) {
// declare the MeanBody instance
Mean mean(x);
// call parallel_reduce to start the work
RcppParallel::parallelReduce(0, x.length(), mean);
// return the computed mean
return mean.value;
}
I'm looking forward to learning from you guys.
Accessing begin and end is fine, but your operator function's logic isn't correct.
Check this:
void operator()(std::size_t begin, std::size_t end) {
double temp_num = (double) end - (double) begin;
double temp_value = (std::accumulate(input.begin() + begin, input.begin() + end, 0.0) / temp_num);
value = (num*value + temp_num*temp_value)/(num + temp_num);
num = num + temp_num;
}
A thread may continue running on a new range without joining, so you have to account for that possibility.