How to ensure two different vectors are shuffled in the same order in C++? [duplicate]

Question

I have two vectors:

vector1 = [1 2 3 4 5 6 7 8 9]

vector2 = [1 2 3 4 5 6 7 8 9]

I want to ensure, that when I shuffle both using random_shuffle they should be shuffled in the same corresponding order. For example:

Output after shuffling should be like:

vector1 = [1 9 3 4 2 7 8 5 6]

vector2 = [1 9 3 4 2 7 8 5 6]

But I am getting output like:

vector1 = [5 1 7 4 2 3 9 8 6]

vector2 = [3 4 1 9 8 2 5 7 6]

Heres my code:

int main () 
{
  std::srand ( unsigned ( std::time(0) ) );
  std::vector<int> vector1, vector2;

  // set some values:
  for (int i=1; i<10; ++i)
  {
    vector1.push_back(i);
    vector2.push_back(i);
  }

  // using built-in random generator:
  std::random_shuffle ( vector1.begin(), vector1.end() );
  std::random_shuffle ( vector2.begin(), vector2.end() );

  // print out content:
  std::cout << "vector1 contains:";
  for ( std::vector<int>::iterator it1 = vector1.begin(); it1 != vector1.end(); ++it1 )
    std::cout << ' ' << *it1;

  std::cout << '\n';
  std::cout << '\n';

  std::cout << "vector2 contains:";
  for ( std::vector<int>::iterator it2 = vector2.begin(); it2 != vector2.end(); ++it2 )
    std::cout << ' ' << *it2;

  std::cout << '\n';
  std::cout << '\n';

  return 0;
}

EDIT This is an example case that I tried to implement. In practise, I have one vector of images and one vector of corresponding labels. I need them to be shuffled in the same manner. Could anybody please help...... thanks a lot!!

Answer 1

Instead of shuffling the vectors themselves, shuffle a vector of indexes into the other vectors. Since you'll be using the same indexes for both, they're guaranteed to be in the same order.

std::vector<int> indexes;
indexes.reserve(vector1.size());
for (int i = 0; i < vector1.size(); ++i)
    indexes.push_back(i);
std::random_shuffle(indexes.begin(), indexes.end());

std::cout << "vector1 contains:";
for ( std::vector<int>::iterator it1 = indexes.begin(); it1 != indexes.end(); ++it1 )
    std::cout << ' ' << vector1[*it1];

Answer 2

Make sure you use the same seed for both calls to random_shuffle():

auto seed = unsigned ( std::time(0) );

// ...

std::srand ( seed );
std::random_shuffle ( vector1.begin(), vector1.end() );

std::srand ( seed );
std::random_shuffle ( vector2.begin(), vector2.end() );

Notice, however, that the Standard does not specify that random_shuffle() should use the rand() function to generate a random permutation - this is implementation-defined. Therefore, srand() will not affect the result of random_shuffle() on implementations that do not use rand().

Paragraph 25.3.12/4 of the C++11 Standard on random_shuffle() specifies:

Remarks: To the extent that the implementation of these functions makes use of random numbers, the implementation shall use the following sources of randomness:

The underlying source of random numbers for the first form of the function is implementation-defined. An implementation may use the rand function from the standard C library. [...]

Therefore, if you want to make sure you are writing portable code, use the version of random_shuffle() that accepts a random number generator as a third argument, so that you have control over the seeding.