CODEWITHSUNDEEP
java
user1568840
user1568840

Reputation: 3

How to get distinct random values in java?

Here is my problem: I have an array of integers, I would recover 20% of the array elements in a random manner. Is there a function in java to have more than a random value? I tested for only one value:

Random rand=new Random();
int min=0, max=10;
int valeur = rand.nextInt(max - min + 1) + min;

Upvotes: 0

Views: 968

Answers (6)

baraber
baraber

Reputation: 3356

private Integer[] recover(double percentage, Integer[] ints) {
    int numberOfElementsToRecover = (int) (ints.length*percentage);
    List<Integer> list = Arrays.asList(ints);
    Collections.shuffle(list);
    return Arrays.copyOfRange(list.toArray(new Integer[]{}), 0, numberOfElementsToRecover);
}

Upvotes: 0

Peter Lawrey
Peter Lawrey

Reputation: 533580

The simplest way to get random but distinct values is to use shuffle.

List<Integer> list = new ArrayList<>();
for(int i = min; i <= max; i++) list.add(i);
Collections.shuffle(list);

You can iterate over list and get unique numbers in a random order.

You can even using this approach if you want not more than two of something and three of something else by adding the number you wan to the list.

BTW: This approach is O(n) for n numbers whereas generating random numbers and checking if they are there already is O(n^2 * log(n)) which is much slower.

Upvotes: 2

Byter
Byter

Reputation: 1132

I didnt quite understand the question..

from what i understood you need something like this

List<Integer> listOfRandomNumbers = new ArrayList<Integer>();
//Initialize this List with required numbers

int numberOfRandomNumbersNeeded = listOfRandomNumbers/5;
int min = 0; 
int max = listOfRandomNumbers.size() - 1;

Random rand = new Random();
Set<Integer> randomNumbers = new HashSet<Integer>();
while(randomNumbers.size() < numberOfRandomNumbersNeeded) {
int value = rand.nextInt(max - min + 1) + min;
randomNumbers.add(listOfRandomNumbers.get(value));
}

System.out.println(randomNumbers);

Upvotes: 0

Reimeus
Reimeus

Reputation: 159804

This will prevent numbers being used twice and will get 20%:

    int[] nums = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

    Random random = new Random();

    int numsFound = 0;
    int twentyPercent = (int)(nums.length/5F);
    List<Integer> usedIndices = new ArrayList<Integer>();
    while (numsFound < twentyPercent) {
        int index = random.nextInt(nums.length);

        if (!usedIndices.contains(index)) { // not already used?
            System.out.println(nums[index]);
            usedIndices.add(index);
            numsFound++;
        }
    }

Upvotes: 1

gmustudent
gmustudent

Reputation: 2209

Try This.

import java.util.Random;
public class Dice 
{
    public static void main(String[] args)
    {
        // Initialize Variables and Array
        Random rand = new Random();
        int[] numbers = new int[10];
        int min = 0, max = 10;


        // Input Random Number Into 10 Elements Of Array
        for (int i = 0; i < numbers.length; i++)
            numbers[i] = rand.nextInt(max - min + 1) + min;


        // Output 10 Array Cells
        for (int i = 0; i < numbers.length; i++)
            System.out.println(numbers[i]);
    }
}

Upvotes: 0

Baz
Baz

Reputation: 36894

If I understood your question right, you are trying to obtain a "random" array. If so, you can use the following:

int nrOfValues = 20;
int[] values = new int[nrOfValues];
for(int i = 0; i < values.length; i++)
{
    values[i] = rand.nextInt(max - min + 1) + min;
}

If you would like to obtain distinct random values, try the following:

ArrayList<Integer> values = new ArrayList<Integer>();
int nrOfValues = 20;
for(int i = 0; i < values.length; i++)
{
    int value = rand.nextInt(max - min + 1) + min;
    while(values.contains(value))
        value = rand.nextInt(max - min + 1) + min;
    values.add(value);
}

Of course, if min and max are poorly chosen, you might end up with an infinite loop...

Upvotes: 1

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