比较元组中的和分类元素

Question

我有一个 vector <tuple<int a, int b, Vec4i c>>，我已经根据a对元组进行了分类。结构就是这样。

vect = { 42,324,{}; //[0]
         43,231,{};
         45,97 ,{};
         73,32 ,{}; //[1]
         112,87,{};
         114,249,{}; //[2]
        }

我试图比较" a"，如果元素之间的差异小于5。 if(a[i+1]-a[i] >= 5

在每个组内，发现B的最大元素，相关的C是将_back推到新向量的。

元组是通过以下以下来实现的：

    vector<Vec4i> horiz;
    vector<int> ly, lx;                                                 
    using tuple_t = std::tuple<int, int, Vec4i>;
    vector <tuple_t> vect;
    int n = ly.size();
    auto sort_A = [&](tuple_t lhs, tuple_t rhs)
    { return (get<0>(lhs) < get<0>(rhs)); };
    for (int i = 0; i < n; i++) 
        vect.push_back(make_tuple(ly[i],lx[i],horiz[i]));
    sort(vect.begin(), vect.end(), sort_A);

Answer 1

您不需要复制到子向量中，您只需要记录迭代器的范围。

#include <vector>
#include <tuple>
#include <algorithm>
#include <numeric>
struct Vec4i {};
int main()
{
    using tuple_t = std::tuple<int, int, Vec4i>;
    std::vector<tuple_t> tuples;
    tuples.push_back(std::make_tuple( 42, 32,Vec4i{}));
    tuples.push_back(std::make_tuple( 43,231,Vec4i{}));
    tuples.push_back(std::make_tuple( 45, 97,Vec4i{}));
    tuples.push_back(std::make_tuple( 73,324,Vec4i{}));
    tuples.push_back(std::make_tuple(112, 87,Vec4i{}));
    tuples.push_back(std::make_tuple(114,249,Vec4i{}));
  // Finds the end of a group
    auto next_group = [](const tuple_t & lhs, const tuple_t & rhs) 
    { return (std::get<0>(rhs) - std::get<0>(lhs)) > 5; };
    using iter_t = std::vector<tuple_t>::iterator;
  // Collect up the groups as iterators
    std::vector<std::pair<iter_t, iter_t>> iters;
    for (iter_t it = tuples.begin(), next; it != tuples.end(); it = next)
    {
      // Original grouping, long runs are one group
        next = std::adjacent_find(it, tuples.end(), next_group);
      // only advance next if it is not end
        next += (next != tuples.end());
        iters.emplace_back(it, next);
      // Variant grouping, compares to first element of group
        next = it;
        while(next != tuples.end() && !next_group(*it, *next)) 
        { ++next; }
        iters.emplace_back(it, next);
    }
    auto compare_b = [](const tuple_t & lhs, const tuple_t & rhs) 
    { return std::get<1>(lhs) < std::get<1>(rhs); };
  // Operate on a pair of iter_t as a range, finding the maximum b 
    auto get_max_c = [&](std::pair<iter_t, iter_t> pair) 
    { return std::get<2>(*std::max_element(pair.first, pair.second, compare_b)); };
    std::vector<Vec4i> results;
    std::transform(iters.begin(), iters.end(), std::back_inserter(results), get_max_c);
}

Answer 2

在元组复制列表限制中不起作用，您必须求助于使用make_tuple。您可以使用std :: tie与元组交往。我无法完全理解您的问题，但是我给您一个伪代码，以在" vect"中找到最高元素(根据参数'b'(，例如：

#include <iostream>
#include <vector>
#include <tuple>
#include <algorithm>
using namespace std;
struct Vec4i
{
    int i1;
    int i2;
    int i3;
    int i4;
    Vec4i() { i1=i2=i3=i4 = 0;}
    Vec4i(int i11,int i21,int i31,int i41) { i1=i11; i2=i21; i3=i31; i4 = i41;}    
};
int main()
{
    typedef tuple<int, int, Vec4i> foo_t;
    vector <foo_t> vect;
    vect.push_back( make_tuple( 42, 32,Vec4i(1,2,3,4)));
    vect.push_back( make_tuple( 43,231,Vec4i(2,3,4,5)));
    vect.push_back( make_tuple( 45, 97,Vec4i(3,4,5,6)));
    vect.push_back( make_tuple( 73,324,Vec4i(4,5,6,7)));
    vect.push_back( make_tuple(112, 87,Vec4i(5,6,7,8)));
    vect.push_back( make_tuple(114,249,Vec4i(6,7,8,9)));

    auto result = std::max_element(vect.begin(),vect.end(),
                                    [](const foo_t& lhs,const foo_t& rhs)
                                    {
                                                int a1,b1,a2,b2;
                                                Vec4i v1,v2;
                                                tie(a1,b1,v1) = lhs;
                                                tie(a2,b2,v2) = rhs;
                                                return b1<b2;
                                    } );
        int a,b;
        Vec4i v;
        tie(a,b,v) = *result;
        cout <<"a ="<<  a << " "<<"b ="<<  b << " "<<"i1 ="<<  v.i1 << " "<<"i2 ="<<  v.i2 << " "<<"i3 ="<<  v.i3 << " "<<"i4 ="<<  v.i4 <<endl;

}

结果是：

a =73 b =324 i1 =4 i2 =5 i3 =6 i4 =7