生成具有非均匀分布的随机整数数组

Question

I want to write Java code to produce an array of random integers in the range [1,4]. The array's length is N, which is provided at run time. The problem is that the range [1,4] is not uniformly distributed:

It means that if I create arrays with N=100, the number '1' will appear averagely 40 times in an array, number '2' 10 times, and so on.

For now I am using this code to generate uniform-distributed random numbers in range [1,4]:

public static void main(String[] args)
    {
        int N;
        System.out.println();
        System.out.print("Enter an integer number: ");
        N = input.nextInt();
        int[] a = new int[N];
        Random generator = new Random();
        for(int i = 0; i < a.length; i++)
        {
            a[i] = generator.nextInt(4)+1;
        }
    }

How do I implement it with a the non-uniform distribution as shown in the graph above?

Answer 1

Here's a way to do it, starting from your code:

public static void main(String[] args){
    int N;
    System.out.println();
    System.out.print("Enter an integer number: ");
    N = input.nextInt();
    int[] a = new int[N];
    Random generator = new Random();
    for (int i = 0; i < a.length; i++) {
        float n = generator.nextFloat();
        if (n <= 0.4) {
            a[i] = 1;
        } else if (n <= 0.7) {
            a[i] = 3;
        } else if (n <= 0.9) {
            a[i] = 4;
        } else {
            a[i] = 2;
        }
    }
}

UPDATE: at @pjs' suggestion, select numbers in order of desdencing probability so you tend to exit the if block earlier

Answer 2

Another easy solution is to use nextDouble() which generates a random double in [0,1). If the value is < .4 choose 1, else if it is < (.4 + .2) choose 2, etc, with the last branch always choosing the last choice. This is easily generalized using a for loop.

Answer 3

For a more generic approach, you can populate a NavigableMap with the distribution probability:

double[] probs = {0.4, 0.1, 0.2, 0.3};
NavigableMap<Double, Integer> distribution = new TreeMap<Double, Integer>();
for(double p : probs) {
    distribution.put(distribution.isEmpty() ? p : distribution.lastKey() + p, distribution.size() + 1);
}

and later query the map with a uniformly distributed random key in the range [0, 1>:

Random rnd = new Random();
for(int i=0; i<20; i++) {
    System.out.println(distribution.ceilingEntry(rnd.nextDouble()).getValue());
}

This will populate the map with the following key/value pairs:

0.4 -> 1
0.5 -> 2
0.7 -> 3
1.0 -> 4

To query the map, you first generate a uniformly distributed double in the range 0 to 1. Querying the map using the ceilingEntry method and passing the random number will return the "mapping associated with the least key greater than or equal to the given key" , so eg passing a value in the range <0.4, 0.5] will return the entry with the mapping 0.5 -> 2 . Using getValue() on the returned map entry will hence return 2.

Answer 4

Let a1, a2, a3 and a4 be doubles that specify the relative probabilities and s = a1+a2+a3+a4 That means the probability for 1 is a1/s , the probability for 2 is a2/s , ...

Then create a random double d using generator.nextDouble() .

If 0 <= d < a1/s then the integer should be 1,

if a1/s <= d < (a1+a2)/s then the integer should be 2

if (a1+a2)/s <= d < (a1+a2+a3)/s then the integer should be 3

if (a1+a2+a3)/s <= d < 1 then the integer should be 4

Answer 5

a slightly more extensible version of Miquel's (and also what Teresa suggested):

    double[] distro=new double[]{.4,.1,.3,.2};        
    int N;
    System.out.println();
    System.out.print("Enter an integer number: ");
    Scanner input = new Scanner(System.in);
    N = input.nextInt();
    int[] a = new int[N];
    Random generator = new Random();
    outer:
    for(int i = 0; i < a.length; i++)
    {
        double rand=generator.nextDouble();
        double val=0;
        for(int j=1;j<distro.length;j++){
            val+=distro[j-1];
            if(rand<val){
                a[i]=j;
                continue outer;
            }
        }
        a[i]=distro.length;
    }

Answer 6

For the specific problem you gave above, the solutions provided by others work very well and the alias method would be overkill. However, you said in a comment that you were actually going to use this in a distribution with a much larger range. In that case, the overhead of setting up an alias table may be worthwhile to get the O(1) behavior for actually generating values.

Here's source in Java. It's easy to revert it back to using Java's stock Random if you don't want to grab Mersenne Twister:

/*
 * Created on Mar 12, 2007
 *    Feb 13, 2011: Updated to use Mersenne Twister - pjs
 */
package edu.nps.or.simutils;

import java.lang.IllegalArgumentException;
import java.text.DecimalFormat;
import java.util.Comparator;
import java.util.Stack;
import java.util.PriorityQueue;
import java.util.Random;

import net.goui.util.MTRandom;

public class AliasTable<V> {
   private static Random r = new MTRandom();
   private static DecimalFormat df2 = new DecimalFormat(" 0.00;-0.00");

   private V[] primary;
   private V[] alias;
   private double[] primaryP;
   private double[] primaryPgivenCol;

   private static boolean notCloseEnough(double target, double value) {
      return Math.abs(target - value) > 1E-10;
   }

   /**
    * Constructs the AliasTable given the set of values
    * and corresponding probabilities.
    * @param value
    *   An array of the set of outcome values for the distribution. 
    * @param pOfValue
    *   An array of corresponding probabilities for each outcome.
    * @throws IllegalArgumentException
    *   The values and probability arrays must be of the same length,
    *   the probabilities must all be positive, and they must sum to one.
    */
   public AliasTable(V[] value, double[] pOfValue) {
      super();      
      if (value.length != pOfValue.length) {
         throw new IllegalArgumentException(
               "Args to AliasTable must be vectors of the same length.");
      }
      double total = 0.0;
      for (double d : pOfValue) {
         if (d < 0) {
            throw new
               IllegalArgumentException("p_values must all be positive.");
         }
         total += d;
      }
      if (notCloseEnough(1.0, total)) {
         throw new IllegalArgumentException("p_values must sum to 1.0");
      }

      // Done with the safety checks, now let's do the work...

      // Cloning the values prevents people from changing outcomes
      // after the fact.
      primary = value.clone();
      alias = value.clone();
      primaryP = pOfValue.clone();
      primaryPgivenCol = new double[primary.length];
      for (int i = 0; i < primaryPgivenCol.length; ++i) {
         primaryPgivenCol[i] = 1.0;
      }
      double equiProb = 1.0 / primary.length;

      /*
       * Internal classes are UGLY!!!!
       * We're what you call experts.  Don't try this at home!
       */
      class pComparator implements Comparator<Integer> {
         public int compare(Integer i1, Integer i2) {
            return primaryP[i1] < primaryP[i2] ? -1 : 1;
         }
      }

      PriorityQueue<Integer> deficitSet =
         new PriorityQueue<Integer>(primary.length, new pComparator());
      Stack<Integer> surplusSet = new Stack<Integer>();

      // initial allocation of values to deficit/surplus sets
      for (int i = 0; i < primary.length; ++i) {
         if (notCloseEnough(equiProb, primaryP[i])) {
            if (primaryP[i] < equiProb) {
               deficitSet.add(i);
            } else {
               surplusSet.add(i);
            }
         }
      }

      /*
       * Pull the largest deficit element from what remains.  Grab as
       * much probability as you need from a surplus element.  Re-allocate
       * the surplus element based on the amount of probability taken from
       * it to the deficit, surplus, or completed set.
       * 
       * Lather, rinse, repeat.
       */
      while (!deficitSet.isEmpty()) {
         int deficitColumn = deficitSet.poll();
         int surplusColumn = surplusSet.pop();
         primaryPgivenCol[deficitColumn] = primaryP[deficitColumn] / equiProb;
         alias[deficitColumn] = primary[surplusColumn];
         primaryP[surplusColumn] -= equiProb - primaryP[deficitColumn];
         if (notCloseEnough(equiProb, primaryP[surplusColumn])) {
            if (primaryP[surplusColumn] < equiProb) {
               deficitSet.add(surplusColumn);
            } else {
               surplusSet.add(surplusColumn);
            }
         }
      }
   }

   /**
    * Generate a value from the input distribution.  The alias table
    * does this in O(1) time, regardless of the number of elements in
    * the distribution.
    * @return
    *   A value from the specified distribution.
    */
   public V generate() {
      int column = (int) (primary.length * r.nextDouble());
      return r.nextDouble() <= primaryPgivenCol[column] ?
                  primary[column] : alias[column];
   }

   public void printAliasTable() {
      System.err.println("Primary\t\tprimaryPgivenCol\tAlias");
      for(int i = 0; i < primary.length; ++i) {
         System.err.println(primary[i] + "\t\t\t"
            + df2.format(primaryPgivenCol[i]) + "\t\t" + alias[i]);
      }
      System.err.println();
   }
}