Remove uncommon elements from vector

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I have two vectors. I need to remove second vector elements that does not matched with first vector element.

Example:

C++

vector < string > v1 ={"str1"};
vector < string > v2 ={"str2", "str1", "str3"};

So, I need to remove v2 elements that is not common to v1 elemnts.

I want to make v2 now as v2 = {"str1"};

What I have tried:

C++

#include<iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main ()
{
  vector < string > v1 ={"str1"};
  vector < string > v2 ={"str2", "str1", "str3"};

  for (auto it = v1.begin (); it != v1.end (); it++)
    {
        auto fit = std::find (v2.begin (), v2.end (), *it);
        if (fit != v2.end ())
	    {
	        std::cout << "Element " << *fit << " found at position : ";
	        std::cout << fit - v2.begin() << " (counting from zero) \n" ;
	    }
	    else
	    {
          std::cout << "Element not found.\n\n";
          //remove v2 elements not matched with v1 elements
	    }
      }
   return 0; 
}

Posted 11-Jan-22 20:57pm

Member 14036158

Updated 11-Jan-22 22:51pm

CPallini

v2

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Comments

Richard MacCutchan 12-Jan-22 3:53am

You just need to check each element of v2 to see if it exists in v1. If it does not then remove it from v2.

Member 14036158 12-Jan-22 4:07am

i am doing in else part
if (fit != v2.end ())
{
std::cout << "Element " << *fit << " found at position : ";
std::cout << fit - v2.begin() << " (counting from zero) \n" ;
}
else
{
v2.erase(fit);
}
But, its not working.

Richard MacCutchan 12-Jan-22 4:24am

Your outer loop is looking for elements of v1 that exist in v2. And since v1 only contains a single element, the loop ends after one iteration. You need to iterate over the elements of v2, looking for matches in v1, as I suggested above.

Richard MacCutchan 12-Jan-22 4:38am

~~See my (partial) solution below.~~
Forget that, the solution provided by CPallini is the perfect answer.

Rick York 12-Jan-22 12:04pm

In general, a vector is not a good container to use if deletions of this nature are required. A list is better if this is necessary.

3 solutions

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_Asif_ · Accepted Answer · 2022-01-11T22:19:00

Your logic is correct, however instead of v1 searching in v2, we should search v2 in v1 and remove items not present in v1.

for (auto it = v2.begin (); it != v2.end (); it++)
  {
      auto fit = std::find (v1.begin (), v1.end (), *it);
      if (fit != v1.end ())
      {
          std::cout << "Element " << *fit << " found at position : ";
          std::cout << fit - v2.begin() << " (counting from zero) \n" ;
      }
      else
      {
        std::cout << "Element not found.\n\n";
        remove(v2.begin(), v2.end(), *fit);
      }
    }
 return

Richard MacCutchan · Accepted Answer · 2022-01-11T22:37:00

This is the code you need to find all the elements:

C++

for (auto it = v2.begin (); it != v2.end (); it++) // iterate the elements of v2
{
    auto fit = std::find (v1.begin (), v1.end (), *it); // is the element in v1?
    if (fit != v1.end ())
    {
        std::cout << "Element " << *fit << " found at position : ";
        std::cout << it - v2.begin() << " (counting from zero) \n" ;
    }
    else
    {
        std::cout << "Element " << *it << " not found.\n";

        //remove v2 elements not matched with v1 elements
    }
}

Note that you cannot use v2.erase in the else clause as that will destroy the iterator. You need to make a copy of all the positions that need to be removed, and delete them later when this loop has completed.

CPallini · Accepted Answer · 2022-01-11T22:51:00

Solution 3

Try also

C++

#include<iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main ()
{
  vector < string > v1 ={"str1"};
  vector < string > v2 ={"str2", "str1", "str3"};

  auto not_in_v1 = [&v1](const string & s) { return find(v1.begin(), v1.end(), s ) == v1.end(); };

  v2.erase( remove_if( v2.begin(), v2.end(), not_in_v1), v2.end() );

  for (auto s : v2)
    cout << s << " " << endl;

   return 0;
}

Posted 11-Jan-22 22:51pm

CPallini

Comments

Richard MacCutchan 12-Jan-22 5:32am

+5 for an elegant solution.

CPallini 12-Jan-22 5:39am

Thank you!

_Asif_ 12-Jan-22 6:04am

+5

Member 14036158 12-Jan-22 6:15am

Thanks for the explanation.
It's working.

CPallini 12-Jan-22 6:29am

You are welcome.
"It's working"
Did you really doubt? :-D

Member 14036158 12-Jan-22 6:50am

yes, got it.

jeron1 12-Jan-22 10:57am

Very nice!

CPallini 13-Jan-22 2:49am