Math.random to never generate same number as other Math.random

Question

I got 4 Math.random generators. Each picking 1 of the X objects from the array.

  var randomItem1 = projects[Math.floor(Math.random()*projects.length)];
  var randomItem2 = projects[Math.floor(Math.random()*projects.length)];
  var randomItem3 = projects[Math.floor(Math.random()*projects.length)];
  var randomItem4 = projects[Math.floor(Math.random()*projects.length)];

How can I write a function that prevents the Math.random to generate the same number as other Math.random generator.

My guess: Creating a loop that loops through the var randomItem 1 till 4. If it finds 1 or more outputs to be the same, it will regenerate a new output for 1 or more the duplicated outputs.

Any other suggestions?

Edit: This is for a website.

Shuffle the array (or an array of indices) and pick 4. – Bergi Jul 08 '18 at 15:54 — Bergi, Jul 08 '18 at 15:54
or take a set and check against. – Nina Scholz Jul 08 '18 at 15:54 — Nina Scholz, Jul 08 '18 at 15:54
Once you pick something, remove it from the array. – Salman A Jul 08 '18 at 22:16 — Salman A, Jul 08 '18 at 22:16

Oscar · Answer 1 · 2018-07-08T21:58:44.907

Thanks for the interesting problem. I always used a library for this sort of thing so it was fun figuring it out.

var projects
var randomProjects

function getRandomProjects(projects, sampleSize) {
  var projectsClone = projects.slice();
  var randomItems = [];
  while (sampleSize--) {
    randomItems.push(
      projectsClone.splice(Math.floor(Math.random()*projectsClone.length), 1)[0]
    );
  }
  return randomItems;
}

projects = ['1st project', '2nd project', '3rd project', '4th project', '5th project', '6th project'];
randomProjects = getRandomProjects(projects, 4);

randomProjects.forEach(function(randomProject) {
  console.log(randomProject);
});

The projectsClone.splice(...) deletes a random project from projectsClone and returns it as a single item in an array ([<project>]). Thus in the next iteration of the loop that value (<project>) can no longer be chosen.

However I would suggest using a library if you are using this in production code. For example losdash's _.sampleSize(projects, 4)

zer00ne · Answer 2 · 2018-07-08T22:23:14.803

Update

Removed unneeded parts like ~~Set, for loop~~, and added splice(). This function will mutate the original array due to splice(). Using splice() on an array of consecutive ordered numbers (even after shuffling) guarantees a set of unique numbers.

/**
 * genRan(quantity, length)
 *
 * Need a given amount of unique random numbers.
 *
 * Given a number for how many random numbers are to be returned (@quantity) and a 
 * number that represents the range of consecutive numbers to extract the random 
 * numbers from (@length), this function will:
 *  - generate a full array of numbers with the length of @length
 *  - use shuffle() to shuffle the array to provide an array of randomly ordered numbers 
 *  - splices the first @quantity numbers of the shuffled array
 *  - returns new array of unique random numbers
 *
 * @param {Number} quantity      length of returned array.
 * @param {Number} length        length of an array of numbers. 
 * 
 * @return {Array} An array of unique random numbers.
 */

Demo

Details commented in Demo

var qty = 4
var len = 100;

function genRan(quantity, length) {
  // Ensure that quantity is never greater than length
  if (quantity > length) {
    quantity = length;
  }
  // Create an array of consecutive numbers from 1 to N
  var array = Array.from({
    length: length
  }, (value, key = 0) => key + 1);
  // Run shuffle get the returned shuffled array
  var shuffled = shuffle(array);
  // return an array of N (quantity) random numbers 
  return shuffled.splice(0, quantity);
}

console.log(genRan(qty, len));

/* shuffle utility 
|| Uses Fisher-Yates algorithm
*/
function shuffle(array) {
  var i = 0;
  var j = 0;
  var temp = null;
  for (i = array.length - 1; i > 0; i -= 1) {
    j = Math.floor(Math.random() * (i + 1))
    temp = array[i];
    array[i] = array[j];
    array[j] = temp;
  }
  return array;
}

Math.random to never generate same number as other Math.random

2 Answers2

Update

Demo