Another Vector question

[Suchy]

This time I have a vector of names, some of them are duplicates. I am having troubles figuring out how to copy just the duplicates to a second vector and remove the them form the first vecotor. So far I got this:

    for ( int i = 0 ; i < friends.size () ; i ++ )
          {
            for ( int j = i - 1 ; j < friends.size () ; j++ )
              {

               if ( friends[i] == friends [j] )     //  if duplicate is found
                    {

                         cout << friends[i] << endl;
                         enemies.push_back ( friends [i] )                      
          
                     } 

              }

           }

But this code coppiec all elements from vector friends to vector enemies. How can I fix this ?

[ubergeek]

So in your inner loop, you need to check all the values in the vector except the current one -- the definition of a duplicate. I modified your code a bit:

    for ( int i = 0 ; i < friends.size () ; i ++ )
          {
            for ( int j = 0; j < friends.size () ; j++ ) //loop through the whole thing
              {
               if (j == i) continue; //THE KEY LINE -- skips the check if we are looking at the same element
               if ( friends[i] == friends [j] )     //  if duplicate is found
                    {

                         cout << friends[i] << endl;
                         enemies.push_back ( friends [i] )                      
          
                     } 

              }

           }

I haven't tested this, but it should work (famous last words).


EDIT: if you also need to remove the duplicates from the original, look into the vector::erase function.

[Suchy]

Thanks for the help, this works. Yet I have another problem

I have a vector in my public section of a class, it is a function (which transfers duplicates - see above).

The function starts like this :

vector <string> People::get_duplicates()
 {
     .....
 }

This compiles and works, but everytime the program gets to this function it freazes and I get that

vectors.exe has encountered a problem and needs to close. We are sorry for the inconvenience.

message. What is causing this problem ?

[zouyu1983]

It seems like it causes by the pointer. Maybe in somewhere your pointer is out of the bound (sorry , I come frome china, so my English is not good)

[Prelude]

>I am having troubles figuring out how to copy just the duplicates to a second
>vector and remove the them form the first vecotor.
Good news! That's a common problem. Better news! It's easily solvable with the standard library:

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

template <typename T, int N>
char (&array(T(&)[N]))[N];

int main()
{
  int a[] = {1,2,3,4,4,2,4,3,1,2,3,5,3,4,5,5,3,4,5};
  std::vector<int> v ( a, a + sizeof array ( a ) );
  std::vector<int>::iterator end;
  std::vector<int> save;

  // Original list
  copy ( v.begin(), v.end(), std::ostream_iterator<int> ( std::cout, " " ) );
  std::cout<<'\n';

  sort ( v.begin(), v.end() );
  end = unique ( v.begin(), v.end() );
  save.assign ( end, v.end() );
  v.erase ( end, v.end() );

  // Final list
  copy ( v.begin(), v.end(), std::ostream_iterator<int> ( std::cout, " " ) );
  std::cout<<'\n';

  // Saved duplicates
  copy ( save.begin(), save.end(), std::ostream_iterator<int> ( std::cout, " " ) );
  std::cout<<'\n';
}

[hk_mp5kpdw]

>I am having troubles figuring out how to copy just the duplicates to a second
>vector and remove the them form the first vecotor.
Good news! That's a common problem. Better news! It's easily solvable with the standard library:

Gahh!!! I was going to suggest that exact solution. Of course, if it's important to the OP to keep the original order of elements in the source vector then the sort kinda ruins that.

[Prelude]

>Of course, if it's important to the OP to keep the original order of elements in
>the source vector then the sort kinda ruins that.
I've been in that situation many times. Or so I thought at first. It turns out that in practice (most of the time, it seems), if you get to the point where you have duplicates stored and want to remove them, the current order isn't important.