C++, std::transform replace items by indices

Question

There are 2 unsorted vectors of int v1 and v2, where v1 contains a subset of v2

v1: 8 12 4 17
v2: 6 4 14 17 9 0 5 12 8 

Is there any way, how to replace items of v1 by indices of its positions in v2?

v1: 8 7 1 3

It is no problem to write such an algorithm using 2 for cycles...

But is there any solution using using std::transform?

Answer 1

Combine std::transform with a function object which calls std::find:

#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>

struct find_functor
{
  std::vector<int> &haystack;

  find_functor(std::vector<int> &haystack)
    : haystack(haystack)
  {}

  int operator()(int needle)
  {
    return std::find(haystack.begin(), haystack.end(), needle) - haystack.begin();
  }
};

int main()
{
  std::vector<int> v1 = {8, 12,  4, 17};
  std::vector<int> v2 = {6,  4, 14, 17, 9, 0, 5, 12, 8};

  // in c++11:
  std::transform(v1.begin(), v1.end(), v1.begin(), [&v2](int x){
    return std::find(v2.begin(), v2.end(), x) - v2.begin();
  });

  // in c++03:
  std::transform(v1.begin(), v1.end(), v1.begin(), find_functor(v2));

  std::cout << "v1: ";
  std::copy(v1.begin(), v1.end(), std::ostream_iterator<int>(std::cout, " "));
  std::cout << std::endl;

  return 0;
}

The output:

$ g++ -std=c++0x test.cpp
$ ./a.out 
v1: 8 7 1 3 

Answer 2

You could use std::find() in the transformation:

std::transform(v1.begin(), v1.end(), v1.begin(), [&](int v)->int {
    return std::find(v2.begin(), v2.end(), v) - v2.begin());
});

Answer 3

std::transform takes a unary-function-like-object. You can therefore create a functor class that performs this operation efficiently, constructing it with the second vector, and then apply that functor to the first vector.

template <typename T>
class IndexSeeker{
    private:
        map<T, int> indexes;
    public:
        IndexSeeker(vector<T> source){
            for(int k = 0; k < t.size(); ++k){
                indexes[source[k]] = k;
            }
        }

        int operator()(const T& locateme){
            if(indexes.find(T) != indexes.end()){
                return indexes[T];
            }
            return -1;
        }
}

By cacheing the entire second list into a map, finding the index is efficient, instead of requiring a linear search. This requires type T to be sortable (and thus mappable). If T is not sortable, a less efficient approach requiring a brute force search is required.