这个排序两个向量作为参考传递给函数

Question

我尝试使用另一个值对一个向量进行排序，
这是代码：

template <typename T>
    void ordered_by_old_indices(std::vector<T> const& values, std::vector<size_t> &indices) {
        //std::vector<size_t> indices(values.size());
        //std::iota(begin(indices), end(indices), static_cast<size_t>(0));
        //auto x = indices.begin();
        //std::cout << *x << std::endl;
        //auto y = indices.end()-1;
        //std::cout << *y << std::endl;

        std::sort(
            indices.begin(), indices.end(),
            [&](size_t a, size_t b) { return values[a] < values[b]; }
        );
        //return indices;
    }

但它给了我错误：

vector subscript out of range

这里有什么问题，它只是排序的包装器？？？

更新： 问题是 sort 给出了要排序的向量元素的比较函数值，我想要一种方法让 sort 传递元素的位置而不是它的值。这意味着传递 01234567.... 但在给定的向量索引上工作

Answer 1

我认为这应该可行。还在努力中

template <typename T>
    void ordered_by_old_indices(std::vector<T> const& values, std::vector<size_t> &indices) {
        std::vector<size_t> indices_1(indices.size());
        std::iota(begin(indices_1), end(indices_1), static_cast<size_t>(0));
        std::sort(
            begin(indices_1), end(indices_1),
            [&](size_t a, size_t b) { return values[a] < values[b]; }
        );
        assert(values.size() == indices_1.size());
        //assert([&]() { for (size_t i : indices_1) { if (i >= values.size()) return false; } return true; }());

        //std::sort(
        //  begin(indices), end(indices),
        //  [&](size_t a, size_t b) { return indices_1[a] < indices_1[b]; }
        //);
        //now I should use sorted indices_1 to pushback original indices into indices
        for (std::vector<size_t>::iterator indices_1_it = indices_1.begin(); indices_1_it != indices_1.end(); indices_1_it ++)
        {
            indices.push_back(values[indices_1[]]);
        }
        //return indices;
    }

Answer 2

您的函数有效，见下文，（正如我认为您想要的那样）所以@Scheff 说的是您的索引超出了值的范围。所以试着在你的函数中打印你的索引，看看是否有任何索引[x]会导致值[indices[x]]超出范围。

#include <vector>
#include <algorithm>
#include <numeric>
template <typename T>
void ordered_by_old_indices(std::vector<T> const& values, std::vector<size_t> &indices) {
    std::sort(
        indices.begin(), indices.end(),
        [&](size_t a, size_t b) { return values[a] < values[b]; }
    );
    //return indices;
}

void TestIndices() {
  std::vector<size_t> values(50);
  std::iota(begin(values), end(values), static_cast<size_t>(0));
  std::random_shuffle(values.begin(), values.end());
  std::vector<size_t> indices(values.size());
  std::iota(begin(indices), end(indices), static_cast<size_t>(0));
  auto x = indices.begin();
  std::cout << *x << std::endl;
  auto y = indices.end()-1;
  std::cout << *y << std::endl;
  ordered_by_old_indices(values, indices);
  std::cout << values.size() << indices.size() << "\n";
  for(int i = 0; i < values.size(); ++i) // print the values ordered by indices
    std::cout << indices[i] << " : " << values[indices[i]] << "\n";
}