4

I've got two arrays

var mp3 = ['sing.mp3','song.mp3','tune.mp3','jam.mp3',etc];
var ogg = ['sing.ogg','song.ogg','tune.ogg','jam.ogg',etc];

i need to shuffle both arrays so that they come out the same way, ex:

var mp3 = ['tune.mp3','song.mp3','jam.mp3','sing.mp3',etc];
var ogg = ['tune.ogg','song.ogg','jam.ogg','sing.ogg',etc];

there's a few posts on stackoverflow that shuffle arrays in different ways--this one is pretty great--but none of them demonstrate how to shuffle two arrays in the same exact way.

thnx!

Community
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Nick Briz
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    if the filenames are the same for both the arrays, why can't you use one array instead of two? var files = ['tunes', 'jam', 'sing']; And, you can call tunes.mp3 or tunes.ogg depending on your application. – Srikanth AD Aug 12 '13 at 19:11
  • the arrays (which are fairly large) are dynamically generated from a directory of files, so they include the file extensions (.mp3 or .ogg) so there is no array that contains an extensionless list like ['tunes', 'jam', 'sing']... but are you suggesting that it might be easier to create a third array by somehow automatically stripping off the extensions? – Nick Briz Aug 12 '13 at 19:15
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    In general, it would be easier to just shuffle one array and use the output rather than shuffling 2 arrays - specifically if you are expecting the same output and the arrays are fairly large. – Srikanth AD Aug 12 '13 at 19:18

5 Answers5

7

Add an extra argument to the Fisher-Yates shuffle. (assumes that your arrays are equal length)

var mp3 = ["sing.mp3", "song.mp3"];
var ogg = ["sing.ogg", "song.ogg"];

function shuffle(obj1, obj2) {
  var index = obj1.length;
  var rnd, tmp1, tmp2;

  while (index) {
    rnd = Math.floor(Math.random() * index);
    index -= 1;
    tmp1 = obj1[index];
    tmp2 = obj2[index];
    obj1[index] = obj1[rnd];
    obj2[index] = obj2[rnd];
    obj1[rnd] = tmp1;
    obj2[rnd] = tmp2;
  }
}

shuffle(mp3, ogg);

console.log(mp3, ogg);

UPDATE:

If you are going to support more arrays (as suggested in the comments), then you could modify the Fisher-Yates as follows (aswell as perform some checks to make sure that the arguments are of Array and that their lengths match).

var isArray = Array.isArray || function(value) {
  return {}.toString.call(value) !== "[object Array]"
};

var mp3 = ["sing.mp3", "song.mp3", "tune.mp3", "jam.mp3"];
var ogg = ["sing.ogg", "song.ogg", "tune.ogg", "jam.ogg"];
var acc = ["sing.acc", "song.acc", "tune.acc", "jam.acc"];
var flc = ["sing.flc", "song.flc", "tune.flc", "jam.flc"];

function shuffle() {
  var arrLength = 0;
  var argsLength = arguments.length;
  var rnd, tmp;

  for (var index = 0; index < argsLength; index += 1) {
    if (!isArray(arguments[index])) {
      throw new TypeError("Argument is not an array.");
    }

    if (index === 0) {
      arrLength = arguments[0].length;
    }

    if (arrLength !== arguments[index].length) {
      throw new RangeError("Array lengths do not match.");
    }
  }

  while (arrLength) {
    rnd = Math.floor(Math.random() * arrLength);
    arrLength -= 1;
    for (argsIndex = 0; argsIndex < argsLength; argsIndex += 1) {
      tmp = arguments[argsIndex][arrLength];
      arguments[argsIndex][arrLength] = arguments[argsIndex][rnd];
      arguments[argsIndex][rnd] = tmp;
    }
  }
}

shuffle(mp3, ogg, acc, flc);

console.log(mp3, ogg, acc, flc);
Xotic750
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1

From that example, simply add a second parameter (your second array) and perform the operation on both arrays. You will just need to add and use a second temp, so you aren't overwriting your temps.

This should do the trick ASSUMING THE ARRAYS ARE THE SAME LENGTH:

function shuffle(array, array2) {
    var counter = array.length, temp, temp2, index;

    // While there are elements in the array
    while (counter > 0) {
        // Pick a random index
        index = Math.floor(Math.random() * counter);

        // Decrease counter by 1
        counter--;

        // And swap the last element with it
        temp = array[counter];
        temp2 = array2[counter];

        array[counter] = array[index];
        array2[counter] = array2[index];

        array[index] = temp;
        array2[index] = temp2;
    }
}
Tricky12
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  • Note: After passing `mp3` and `ogg` to the function, those variables will be themselves shuffled, no need for the return and further assignments. – Xotic750 Aug 12 '13 at 19:23
1

I would seriously consider restructuring the way you're keeping track of the information, but in general you can separate out the shuffle itself from the stuff being shuffled. You need a function to generate a random permutation, and then a function to apply a permutation to an array.

function shuffle(o) { //v1.0
    for(var j, x, i = o.length; i; j = Math.floor(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

function permutation( length ) {
  var p = [], i;
  for (i = 0; i < length; ++i) p[i] = i;
  return shuffle(p);
}

function permute( a, p ) {
  var r = [];
  for (var i = 0; i < a.length; ++i)
    r.push(a[p[i]]);
  for (i = 0; i < a.length; ++i)
    a[i] = r[i];
}

Then you can create a single random permutation and apply it to any list (of the right length) you want.

var p = permutation( mp3.length );
permute(mp3, p);
permute(ogg, p);
permute(aac, p);
// etc

(Shuffle function taken from the SO question linked in the OP.)

Pointy
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0

If you have two arrays of length 2, @Xotic750s function always returns the same value. UseMath.round instead of Math.floor.

function shuffle_two_arrays_identically(arr1, arr2){
"use strict";
var l = arr1.length,
    i = 0,
    rnd,
    tmp1,
    tmp2;

while (i < l) {
    rnd = Math.round(Math.random() * i)
    tmp1 = arr1[i]
    tmp2 = arr2[i]
    arr1[i] = arr1[rnd]
    arr2[i] = arr2[rnd]
    arr1[rnd] = tmp1
    arr2[rnd] = tmp2
    i += 1
}}
andyw
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0

<script>
var arrayList= ['a','b','c','d','e','f','g'];
arrayList.sort(function(){
    return 0.5 - Math.random()
})

document.getElementById("output").innerHTML  = arrayList;

<script>
dipenparmar12
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