Weighted random numbers

Question

I\'m trying to implement a weighted random numbers. I\'m currently just banging my head against the wall and cannot figure this out.

In my project (Hold\'em hand-ran

Answer 1

If your weights change more slowly than they are drawn, C++11 discrete_distribution is going to be the easiest:

#include 
#include 
std::vector weights{90,56,4};
std::discrete_distribution dist(std::begin(weights), std::end(weights));
std::mt19937 gen;
gen.seed(time(0));//if you want different results from different runs
int N = 100000;
std::vector samples(N);
for(auto & i: samples)
    i = dist(gen);
//do something with your samples...

Note, however, that the c++11 discrete_distribution computes all of the cumulative sums on initialization. Usually, you want that because it speeds up the sampling time for a one time O(N) cost. But for a rapidly changing distribution it will incur a heavy calculation (and memory) cost. For instance if the weights represented how many items there are and every time you draw one, you remove it, you will probably want a custom algorithm.

Will's answer https://stackoverflow.com/a/1761646/837451 avoids this overhead but will be slower to draw from than the C++11 because it can't use binary search.

To see that it does this, you can see the relevant lines (/usr/include/c++/5/bits/random.tcc on my Ubuntu 16.04 + GCC 5.3 install):

  template
    void
    discrete_distribution<_IntType>::param_type::
    _M_initialize()
    {
      if (_M_prob.size() < 2)
        {
          _M_prob.clear();
          return;
        }

      const double __sum = std::accumulate(_M_prob.begin(),
                                           _M_prob.end(), 0.0);
      // Now normalize the probabilites.
      __detail::__normalize(_M_prob.begin(), _M_prob.end(), _M_prob.begin(),
                            __sum);
      // Accumulate partial sums.
      _M_cp.reserve(_M_prob.size());
      std::partial_sum(_M_prob.begin(), _M_prob.end(),
                       std::back_inserter(_M_cp));
      // Make sure the last cumulative probability is one.
      _M_cp[_M_cp.size() - 1] = 1.0;
    }