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How to get a number of random elements from an array?

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How do I print a random number from an array?

Using it you can do the following: Random rnd = new Random(); int randomNumberFromArray = array[rnd. nextInt(3)]; Hope this helps!

Can we access elements randomly in an array?

Random(direct) access implies the ability to access any entry in a array in constant time (independent of its position in the array and of array's size). And that is big advantage. It is typically contrasted to sequential access.

How do you select a random number from an array in Python?

Use the numpy. random. choice() function to generate the random choices and samples from a NumPy multidimensional array. Using this function we can get single or multiple random numbers from the n-dimensional array with or without replacement.

How do you find the element number in an array?

To find the position of an element in an array, you use the indexOf() method. This method returns the index of the first occurrence the element that you want to find, or -1 if the element is not found. The following illustrates the syntax of the indexOf() method.


Just two lines :

// Shuffle array
const shuffled = array.sort(() => 0.5 - Math.random());

// Get sub-array of first n elements after shuffled
let selected = shuffled.slice(0, n);

DEMO:


Try this non-destructive (and fast) function:

function getRandom(arr, n) {
    var result = new Array(n),
        len = arr.length,
        taken = new Array(len);
    if (n > len)
        throw new RangeError("getRandom: more elements taken than available");
    while (n--) {
        var x = Math.floor(Math.random() * len);
        result[n] = arr[x in taken ? taken[x] : x];
        taken[x] = --len in taken ? taken[len] : len;
    }
    return result;
}

There is a one-liner unique solution here

 array.sort(() => Math.random() - Math.random()).slice(0, n)

Porting .sample from the Python standard library:

function sample(population, k){
    /*
        Chooses k unique random elements from a population sequence or set.

        Returns a new list containing elements from the population while
        leaving the original population unchanged.  The resulting list is
        in selection order so that all sub-slices will also be valid random
        samples.  This allows raffle winners (the sample) to be partitioned
        into grand prize and second place winners (the subslices).

        Members of the population need not be hashable or unique.  If the
        population contains repeats, then each occurrence is a possible
        selection in the sample.

        To choose a sample in a range of integers, use range as an argument.
        This is especially fast and space efficient for sampling from a
        large population:   sample(range(10000000), 60)

        Sampling without replacement entails tracking either potential
        selections (the pool) in a list or previous selections in a set.

        When the number of selections is small compared to the
        population, then tracking selections is efficient, requiring
        only a small set and an occasional reselection.  For
        a larger number of selections, the pool tracking method is
        preferred since the list takes less space than the
        set and it doesn't suffer from frequent reselections.
    */

    if(!Array.isArray(population))
        throw new TypeError("Population must be an array.");
    var n = population.length;
    if(k < 0 || k > n)
        throw new RangeError("Sample larger than population or is negative");

    var result = new Array(k);
    var setsize = 21;   // size of a small set minus size of an empty list

    if(k > 5)
        setsize += Math.pow(4, Math.ceil(Math.log(k * 3) / Math.log(4)))

    if(n <= setsize){
        // An n-length list is smaller than a k-length set
        var pool = population.slice();
        for(var i = 0; i < k; i++){          // invariant:  non-selected at [0,n-i)
            var j = Math.random() * (n - i) | 0;
            result[i] = pool[j];
            pool[j] = pool[n - i - 1];       // move non-selected item into vacancy
        }
    }else{
        var selected = new Set();
        for(var i = 0; i < k; i++){
            var j = Math.random() * n | 0;
            while(selected.has(j)){
                j = Math.random() * n | 0;
            }
            selected.add(j);
            result[i] = population[j];
        }
    }

    return result;
}

Implementation ported from Lib/random.py.

Notes:

  • setsize is set based on characteristics in Python for efficiency. Although it has not been adjusted for JavaScript, the algorithm will still function as expected.
  • Some other answers described in this page are not safe according to the ECMAScript specification due to the misuse of Array.prototype.sort. This algorithm however is guaranteed to terminate in finite time.
  • For older browsers that do not have Set implemented, the set can be replaced with an Array and .has(j) replaced with .indexOf(j) > -1.

Performance against the accepted answer:

  • enter image description here
  • https://jsperf.com/pick-random-elements-from-an-array
  • The performance difference is the greatest on Safari.

lodash _.sample and _.sampleSize.

Gets one or n random elements at unique keys from collection up to the size of collection.

_.sample([1, 2, 3, 4]);
// => 2

_.sampleSize([1, 2, 3], 2);
// => [3, 1]
 
_.sampleSize([1, 2, 3], 3);
// => [2, 3, 1]

create a funcion which does that:

var getMeRandomElements = function(sourceArray, neededElements) {
    var result = [];
    for (var i = 0; i < neededElements; i++) {
        result.push(sourceArray[Math.floor(Math.random()*sourceArray.length)]);
    }
    return result;
}

you should also check if the sourceArray has enough elements to be returned. and if you want unique elements returned, you should remove selected element from the sourceArray.


Getting 5 random items without changing the original array:

const n = 5;
const sample = items
  .map(x => ({ x, r: Math.random() }))
  .sort((a, b) => a.r - b.r)
  .map(a => a.x)
  .slice(0, n);

(Don't use this for big lists)