如果您只需要计算最小值以上的元素,这将很简单
vex::Reductor<int, vex::SUM> sum(ctx);
int amount = sum( vec >= minimum );
该vec >= minimum
表达式产生一系列 1 和 0,sum
然后计算 1。
现在,由于您还需要获得高于最小值的元素的位置,所以它变得有点复杂:
#include <iostream>
#include <vexcl/vexcl.hpp>
int main() {
vex::Context ctx(vex::Filter::Env && vex::Filter::Count(1));
// Input vector
vex::vector<int> vec(ctx, {1, 3, 5, 2, 6, 8, 0, 2, 4, 7});
int n = vec.size();
int minimum = 5;
// Put result of (vec >= minimum) into key, and element indices into pos:
vex::vector<int> key(ctx, n);
vex::vector<int> pos(ctx, n);
key = (vec >= minimum);
pos = vex::element_index();
// Get number of interesting elements in vec.
vex::Reductor<int, vex::SUM> sum(ctx);
int amount = sum(key);
// Sort pos by key in descending order.
vex::sort_by_key(key, pos, vex::greater<int>());
// First 'amount' of elements in pos now hold indices of interesting
// elements. Lets use slicer to extract them:
vex::vector<int> indices(ctx, amount);
vex::slicer<1> slice(vex::extents[n]);
indices = slice[vex::range(0, amount)](pos);
std::cout << "indices: " << indices << std::endl;
}
这给出了以下输出:
indices: {
0: 2 4 5 9
}