Remove vector elements based on the index

Question

I wanted to remove the elements of the vector based on the index, say all the even indexed elements. I have read about the erase remove idiom but can't see how to apply it. This is what I tried:

    vector<int> line;
    line.reserve(10);
    for(int i=0;i<10;++i)
    {
      line.push_back(i+1);
    }
    for(unsigned int i=0;i<line.size();++i)
    {
      //remove the even indexed elements
      if(i%2 == 0)
      {
        remove(line.begin(),line.end(),line[i]);
      }
    }
line.erase( line.begin(),line.end() );

This erases the entire vector. I was hoping to only remove the elements that had been marked by the remove algorithm.

Then I tried this

for(unsigned int i=0;i<line.size();++i)
    {
      //remove the even indexed elements
      if(i%2 == 0)
      {
        line.erase( remove(line.begin(),line.end(),line[i]),line.end() );
      }
    }

This again doesn't work as there is a problem while removing, the indices seem to shift whilst iterating over the vector. What should be the correct approach to accomplish this.

Answer 1

By going from 0 to size, you'll end up skipping half of the elements because the indices change as you erase the elements. Make your for loop go from size() to 0:

for(unsigned int i = line.size(); i > 0; i--)
{

}

Answer 2

Online Demo:

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

/*Check if Index is Even or Odd*/ 
bool is_IndexEven(int i) 
{
   static int k = 1;

   /*Handle Index 0 as special case as per choice*/
   if(k == 1)
   {
       k++;
       return false;
   } 

   if(k++ % 2)
       return true;
   else 
       return false; 
 }

int main() 
{
    using namespace std;
    int elements[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    // create a vector that holds the numbers from 0-9.
    vector<int> v(elements, elements + 10); 

    /*Display elements before removal*/    
    vector<int>::const_iterator iter = v.begin();
    cout<<"Before\n";

    for(iter;iter!= v.end();++iter)
    {
        cout<<*iter;
    }

    /*Remove_if + Erase Algorithm for one step removal*/
    v.erase( remove_if(v.begin(), v.end(), is_IndexEven), v.end() ); 

    /*Display result after elements removed*/
    cout<<"\nAfter\n";
    iter = v.begin();
    for(iter;iter!= v.end();++iter)
    {
       cout<<*iter;
    }

    return 0;
}

Answer 3

Why don't you use remove_if? Use a static variable inside the function to signal the index for the current element.

Answer 4

An answer that generalizes not only upon the type of container to be handled but also upon the type of container that holds the indices to be removed, is given in : Erasing elements in stl::vector by using indexes

Answer 5

Here is how to use erase-remove method to remove odd numbers from a vector. I am not sure whether you can remove elements based on the index, because remove_if() applies the predicate on the values pointed by the iterators rather than the iterator itself.

See the following : http://cplusplus.com/reference/algorithm/remove_if/

#include <iostream>
#include <algorithm>
#include <vector>
#include <iterator>
using namespace std;

int main()
{
    vector<int> v;
    v.push_back(11);
    v.push_back(22);
    v.push_back(33);
    v.push_back(44);
    v.push_back(55);
    v.push_back(66);
    v.push_back(77);
    ostream_iterator<int> printit(cout, " ");

    cout << "Before removing odd numbers" << endl;
    copy(v.begin(), v.end(), printit);


    v.erase(remove_if(v.begin(), v.end(),
          [] (int e) { return e%2 == 1; }), v.end());

    cout << endl;
    cout << "After removing odd numbers" << endl;
    copy(v.begin(), v.end(), printit);
    cout << endl;
}