Creating a random std::vector with ones and zeros C++

Question

I want to create a vector of random 1's and 0's in a proportion set by me (in the program I called it dropout) The vector is the same size of a previously created vector CSUM.

in MATLAB it would be

dropout=0.9;
n_elements=size(CSUM)
drpoutmask = (rand(n_elements) > dropout); 

in C++ I have

size_t elements = Csum.size();
std::vector<float> y(elements);
std::uniform_real_distribution<float> distribution(0.0f, 1.0f); 
std::mt19937 engine; // Mersenne twister MT19937
auto generator = std::bind(distribution, engine);
std::generate_n(y.begin(), elements, generator);
std::vector<int> dropoutmask(elements,0);
float dropout=0.9;

for(int i=0; i<elements; i++)
  {
  if(y.at(i)>dropout)
    {
    dropoutmask.at(i)=1;
    }
  }
}

which works but for huge vectors is very very slow, is there a faster way to do this? I am very new at C++.

Any help will be much appreciated

Answer 1

1. You do know about bernoulli distribution, right? You can use it to generate your integer vector directly.
   Example:

   #include <iostream>
   #include <algorithm>
   #include <string>
   #include <random>
   
   int main()
   {
       constexpr double dropout = 0.9; // Chance of 0
       constexpr size_t size = 1000;
       std::random_device rd;
       std::mt19937 gen(rd());
       std::bernoulli_distribution dist(1 - dropout); // bernoulli_distribution takes chance of true n constructor
   
       std::vector<int> dropoutmask(size);
       std::generate(dropoutmask.begin(), dropoutmask.end(), [&]{ return dist(gen); });
       size_t ones = std::count(dropoutmask.begin(), dropoutmask.end(), 1);
       std::cout << "vector contains " << ones << " 1's, out of " << size << ". " << ones/double(size) << "%\n";
       std::cout << "vector contains " << size - ones << " 0's, out of " << size << ". " << (size - ones)/double(size) << "%\n";
   }
   

   Live example: http://coliru.stacked-crooked.com/a/a160743185ded5c5

2. Alternatively, you can create a integer vector of desired size (This will set all elements to 0), set first N elements to 1, where n is (1 - dropout) * size (You said you want a proportion, not random amount close to proportion) and then shuffle vector.

   #include <iostream>
   #include <algorithm>
   #include <string>
   #include <random>
   
   int main()
   {
       constexpr double dropout = 0.9; // Chance of 0
       constexpr size_t size = 77;
       std::random_device rd;
       std::mt19937 gen(rd());
   
       std::vector<int> dropoutmask(size);
       std::fill_n(dropoutmask.begin(), dropoutmask.size() * (1 - dropout), 1);
       std::shuffle(dropoutmask.begin(), dropoutmask.end(), gen);
   
       size_t ones = std::count(dropoutmask.begin(), dropoutmask.end(), 1);
       std::cout << "vector contains " << ones << " 1's, out of " << size << ". " << ones/double(size) << "%\n";
       std::cout << "vector contains " << size - ones << " 0's, out of " << size << ". " << (size - ones)/double(size) << "%\n";
   
       for (auto i :dropoutmask) {
           std::cout << i << ' ';   
       }
       std::cout << '\n';
   }
   

Live example: http://coliru.stacked-crooked.com/a/0a9dacd7629e1605