我正在尝试使用RcppParallel并行处理(大型)向量的添加。这就是我想出来的。
// [[Rcpp::depends(RcppParallel)]]
#include <RcppParallel.h>
#include <Rcpp.h>
#include <assert.h>
using namespace RcppParallel;
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector directVectorAddition(NumericVector first, NumericVector second) {
assert (first.length() == second.length());
NumericVector results(first.length());
results = first + second;
return results;
}
// [[Rcpp::export]]
NumericVector loopVectorAddition(NumericVector first, NumericVector second) {
assert (first.length() == second.length());
NumericVector results(first.length());
for(unsigned i = 0; i != first.length(); i++)
results[i] = first[i] + second[i];
return results;
}
struct VectorAddition : public Worker
{
const RVector<double> first, second;
RVector<double> results;
VectorAddition(const NumericVector one, const NumericVector two, NumericVector three) : first(one), second(two), results(three) {}
void operator()(std::size_t a1, std::size_t a2) {
std::transform(first.begin() + a1, first.begin() + a2,
second.begin() + a1,
results.begin() + a1,
[](double i, double j) {return i + j;});
}
};
// [[Rcpp::export]]
NumericVector parallelVectorAddition(NumericVector first, NumericVector second) {
assert (first.length() == second.length());
NumericVector results(first.length());
VectorAddition myVectorAddition(first, second, results);
parallelFor(0, first.length(), myVectorAddition);
return results;
}这似乎是可行的,但并没有加快速度(至少在4核机器上是不行的)。
> v1 <- 1:1000000
> v2 <- 1000000:1
> all(directVectorAddition(v1, v2) == loopVectorAddition(v1, v2))
[1] TRUE
> all(directVectorAddition(v1, v2) == parallelVectorAddition(v1, v2))
[1] TRUE
> result <- benchmark(v1 + v2, directVectorAddition(v1, v2), loopVectorAddition(v1, v2), parallelVectorAddition(v1, v2), order="relative")
> result[,1:4]
test replications elapsed relative
1 v1 + v2 100 0.206 1.000
4 parallelVectorAddition(v1, v2) 100 0.993 4.820
2 directVectorAddition(v1, v2) 100 1.015 4.927
3 loopVectorAddition(v1, v2) 100 1.056 5.126这能否更有效地执行?
提前谢谢你,
mce
发布于 2018-11-09 21:41:41
)将其定义为Rcpp::NumericVector,但创建通过序列运算符创建的数据。这将创建整数值,因此您将强制将一个副本复制到所有函数上!
搞定
v1 <- as.double(1:1000000)
v2 <- as.double(1000000:1)相反,在一台有许多核心的机器上(在工作中),我看到
R> result[,1:4]
test replications elapsed relative
4 parallelVectorAddition(v1, v2) 100 0.301 1.000
2 directVectorAddition(v1, v2) 100 0.424 1.409
1 v1 + v2 100 0.436 1.449
3 loopVectorAddition(v1, v2) 100 0.736 2.445这个例子仍然不那么令人印象深刻,因为相关的操作是“廉价的”,而并行方法需要分配内存、将数据复制到工作人员、再次收集等等pp。
但好消息是,您正确地编写了并行代码。这可不是个小任务。
https://stackoverflow.com/questions/53233512
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