我有一个std::vector<double>
,其中可能包含几个NAN
值。我想在 vector 中找到最大的元素。如何在比较中有效地跳过NAN
?我想避免必须在每个元素上调用isnan
。有任何想法吗?
// std::max_element([NAN,NAN,NAN,-31,-89]) = NAN
// because NAN > -31 returns NAN.
// how can I skip all NANs in the comparison?
// test 2 below is my use case.
#include <vector>
#include <iostream>
#include <cmath>
void vector_max(std::vector<double> v, double &max, int &imax){
std::vector<double>::iterator v_iter;
v_iter = std::max_element(v.begin(),v.end());
imax = std::distance(v.begin(), v_iter);
max = *v_iter;
}
int main(){
std::vector<double> v_vec;
std::vector<double>::iterator v_vec_iter;
int imax;
double val;
std::cout << "test 1. " << std::endl;
v_vec.push_back( -33.0 );
v_vec.push_back( -124.0 );
v_vec.push_back( -31.0 );
v_vec.push_back( 18.4 );
vector_max(v_vec,val,imax);
std::cout << "max(v_vec) = " << val << std::endl;
std::cout << "indmax(v_vec) = " << imax << std::endl;
std::cout << "test 2: my case. " << std::endl;
v_vec.clear();
v_vec.push_back( NAN );
v_vec.push_back( NAN );
v_vec.push_back( NAN );
v_vec.push_back( -33.0 );
v_vec.push_back( -124.0 );
v_vec.push_back( -31.0 );
v_vec.push_back( 31.0 );
vector_max(v_vec,val,imax);
std::cout << "max(v_vec) = " << val << std::endl;
std::cout << "indmax(v_vec) = " << imax << std::endl;
};
这将返回:
test 1.
max(v_vec) = 18.4
indmax(v_vec) = 3
test 2.
max(v_vec) = nan
indmax(v_vec) = 0
最佳答案
我会尝试这样的事情:
void vector_max(std::vector<double> v, double &max, int &imax){
std::vector<double>::size_type p=0;
imax = -1;
max = std::numeric_limits<double>::lowest();
for (auto &val : v)
{
if (!std::isnan(val) && val>max)
{
imax = p;
max = val;
}
p++;
}
}