How to select a value from a list with non-uniform probabilities?

问题

I am looking at the k-means++ initialization algorithm. The following two steps of the algorithm give rise to non-uniform probabilities:

For each data point x, compute D(x), the distance between x and the nearest center that has already been chosen.

Choose one new data point at random as a new center, using a weighted probability distribution where a point x is chosen with probability proportional to D(x)^2.

How can I select with this stated weighted probability distribution in C++?

回答1:

With a finite set of individual data points X, this calls for a discrete probability distribution.

The easiest way to do this is to enumerate the points X in order, and calculate an array representing their cumulative probability distribution function: (pseudocode follows)

/* 
 * xset is an array of points X,
 * cdf is a preallocated array of the same size
 */
function prepare_cdf(X[] xset, float[] cdf)
{
   float S = 0;
   int N = sizeof(xset);
   for i = 0:N-1
   {
      float weight = /* calculate D(xset[i])^2 here */
      // create cumulative sums and write to the element in cdf array
      S += weight;
      cdf[i] = S;
   }

   // now normalize so the CDF runs from 0 to 1
   for i = 0:N-1
   {
      cdf[i] /= S;
   }
}

function select_point(X[] xset, float[] cdf, Randomizer r)
{
   // generate a random floating point number from a 
   // uniform distribution from 0 to 1
   float p = r.nextFloatUniformPDF();
   int i = binarySearch(cdf, p);
   // find the lowest index i such that p < cdf[i]

   return xset[i];
}

You call prepare_cdf once, and then call select_point as many times as you need to generate random points.

回答2:

Discrete distributions is a lot easier to do in C++11 with the random header and using std::discrete_distribution. This is example:

#include <iostream>
#include <map>
#include <random>

int main()
{
    std::random_device rd;
    std::mt19937 gen(rd());
    std::discrete_distribution<> d({20,30,40,10});
    std::map<int, int> m;
    for(int n=0; n<10000; ++n) {
        ++m[d(gen)];
    }
    for(auto p : m) {
        std::cout << p.first << " generated " << p.second << " times\n";
    }
}

and this is a sample of the output:

0 generated 2003 times
1 generated 3014 times
2 generated 4021 times
3 generated 962 times

回答3:

I'd take the following approach:

iterate over the data-points, storing their D-squared's in a double distance_squareds[] or std::vector<double> distance_squareds or whatnot, and storing the sum of their D-squared's in a double sum_distance_squareds.
use the drand48 function to choose a random number in [0.0, 1.0), and multiply it by sum_distance_squareds; store the result in random_number.
iterate over distance_squareds, adding together the values (again), and as soon as the running total meets or exceeds random_number, return the data-point corresponding to the D-squared that you'd just added.
due to round-off error, it's remotely possible that you'll finish the loop without having returned; if so, just return the first data-point, or the last one, or whatever. (But don't worry, this should be a very rare edge case.)

回答4:

Here you have something that may help you, using (numbers..) array with given probability distribution (prob..) it will generate for you (numbers) with those probabilities (here it will count them).

#include <iostream>
#include <cmath>
#include <time.h>
#include <stdlib.h>
#include <map>
#include <vector>
using namespace std;
#define ARRAY_SIZE(array) (sizeof(array)/sizeof(array[0]))

int checkDistribution(double random, const map<double, vector<int> > &distribution_map)
{
    int index = 0;
    map<double, vector<int> >::const_iterator it = distribution_map.begin();
    for (; it!=distribution_map.end(); ++it)
    {
        if (random < (*it).first)
        {
                int randomInternal = 0;
                if ((*it).second.size() > 1)
                    randomInternal = rand() % ((*it).second.size());
                index = (*it).second.at(randomInternal);
                break;
        }
    }
    return index;
}

void nextNum(int* results, const map<double, vector<int> > &distribution_map)
{
    double random  = (double) rand()/RAND_MAX;
    int index = checkDistribution(random,distribution_map);
    results[index]+=1;
}

int main() {

    srand (time(NULL));
    int results [] = {0,0,0,0,0};
    int numbers [] = {-1,0,1,2,3};
    double prob [] =  {0.01, 0.3, 0.58, 0.1, 0.01};
    int size = ARRAY_SIZE(numbers);
    // Building Distribution
    map<double, vector<int> > distribution_map;
    map<double, vector<int> >::iterator it;
    for (int i = 0; i < size; i++)
    {
        it = distribution_map.find(prob[i]);
        if (it!=distribution_map.end())
            it->second.push_back(i);
        else
        {
            vector<int> vec;
            vec.push_back(i);
            distribution_map[prob[i]] = vec;
        }
    }
    // PDF to CDF transform
    map<double, vector<int> > cumulative_distribution_map;
    map<double, vector<int> >::iterator iter_cumulative;
    double cumulative_distribution = 0.0;
    for (it=distribution_map.begin();it!=distribution_map.end();++it)
    {
        cumulative_distribution += ((*it).second.size() * (*it).first);
        cumulative_distribution_map[cumulative_distribution] = (*it).second;
    }

    for (int i = 0; i<100; i++)
    {
        nextNum(results, cumulative_distribution_map);
    }
    for (int j = 0; j<size; j++)
        cout<<" "<<results[j]<<" ";
    return 0;
}

来源：https://stackoverflow.com/questions/8568203/how-to-select-a-value-from-a-list-with-non-uniform-probabilities

标签

c++

probability