Javascript: random pair of elements from an array without repeats

Question

I am trying to create a very simple "secret santa" generator as one of my first Javascript projects. I was looking for a watch to solve this problem, but so far nothing has worked that I found.

I have an array of names that need to be paired with each other. I successfully connect them with each other, but now someone can be attracted twice. I push randomly selected names to another array, but I cannot find a way to check randomly chosen names against already selected ones.

var names = ["Sean","Kyle","Emily","Nick","Cotter","Brian","Jeremy","Kimmy","Pat","Johnny"];

var used = [];
var picks = [];

if (names.length % 2 != 0) {
    alert("You must have an even number of names. You currently have " + names.length + " names.");
}

for( var i = 0; i < names.length; i++){

var random = Math.floor(Math.random()*names.length)

if(names[random] == names[i]) {
    names[random] = names[random++];
    picks.push(names[i] + " gets " + names[random]);
    used.push(names[random]);
} else {
    picks.push(names[i] + " gets " + names[random]);
    used.push(names[random]);
}

}

console.log("picked array: ")
for(var k=0; k<picks.length; k++) {
console.log(picks[k]);
}
console.log("used array: " + used);

Thanks in advance for any help.

Answer 1

Create two arrays with names, shuffle them and make sure that you do not select the same name from both arrays:

var names = ["Sean","Kyle","Emily","Nick","Cotter","Brian","Jeremy","Kimmy","Pat","Johnny"];

if (names.length % 2 != 0) {
    alert("You must have an even number of names. You currently have " + names.length + " names.");
} else {
    var arr1 = names.slice(), // copy array
        arr2 = names.slice(); // copy array again

    arr1.sort(function() { return 0.5 - Math.random();}); // shuffle arrays
    arr2.sort(function() { return 0.5 - Math.random();});

    while (arr1.length) {
        var name1 = arr1.pop(), // get the last value of arr1
            name2 = arr2[0] == name1 ? arr2.pop() : arr2.shift();
            //        ^^ if the first value is the same as name1, 
            //           get the last value, otherwise get the first

        console.log(name1 + ' gets ' + name2);
    }
}

FIDDLE

Answer 2

. Shuffle, split zip, :

var splitAt = function(i, xs) {
  var a = xs.slice(0, i);
  var b = xs.slice(i, xs.length);
  return [a, b];
};

var shuffle = function(xs) {
  return xs.slice(0).sort(function() {
    return .5 - Math.random();
  });
};

var zip = function(xs) {
  return xs[0].map(function(_,i) {
    return xs.map(function(x) {
      return x[i];
    });
  });
}

// Obviously assumes even array
var result = zip(splitAt(names.length/2, shuffle(names)));
//^
// [
//   [ 'Nick', 'Kimmy' ],
//   [ 'Sean', 'Johnny' ],
//   [ 'Kyle', 'Brian' ],
//   [ 'Cotter', 'Pat' ],
//   [ 'Emily', 'Jeremy' ]
// ]

Answer 3

.

, :

var names = ["Sean","Kyle","Emily","Nick","Cotter","Brian","Jeremy","Kimmy","Pat","Johnny"];
function getPicks(names) {
  return names.slice(0).sort(function(){ return Math.random()-0.5 }).map(function(name, index, arr){
    return name + " gets " + arr[(index+1)%arr.length];
  });
}
getPicks(names);

, , , . A- > B- > C- > A D- > E- > D.

, , , (. elclanrs).

, , , , . , , , .

var names = ["Sean","Kyle","Emily","Nick","Cotter","Brian","Jeremy","Kimmy","Pat","Johnny"];

var a = names.slice(0);
var b = names.slice(0);
var result = [];
while (a.length > 1) {
  var i = extractRandomElement(a);
  var j = extractRandomElement(b);

  while (i===j) {
    b.push(j);
    j = extractRandomElement(b);
  }
  result.push({ a:i, b:j });
}
if (a[0] === b[0]) {
  result.push({ a:a[0], b:result[0].b });
  result[0].b = a[0];
} else {
  result.push({ a:a[0], b:b[0] });
}
var pairs = result.map(function(item){ return item.a + ' gets ' + item.b});


function extractRandomElement(array) {
  return array.splice(Math.floor(Math.random()*array.length),1)[0];
}

Answer 4

If you don’t need to save the original array, you can delete the names as you select them, and each time you select a name, check that it is not an empty string before clicking on the next array.

Answer 5

I was a bit late, but thought I'd answer here. It essentially does the same as @adeneo, but uses the same base code as OP:

var names = ["Sean","Kyle","Emily","Nick","Cotter","Brian","Jeremy","Kimmy","Pat","Johnny"];
    pickpool = names.slice(0); // Slice the array at the first element to copy it by value

var used = [];
var picks = [];

if (names.length % 2 != 0) {
    alert("You must have an even number of names. You currently have " + names.length + " names.");
}

for( var i = 0; i < names.length; i++){

    var random = Math.floor(Math.random()*pickpool.length)

    if(names[random] == names[i]) {
        // names[random] = names[random++];
        picks.push(names[i] + " gets " + pickpool[random++]);
        pickpool.splice(random++,1);
    } else {
        picks.push(names[i] + " gets " + pickpool[random]);
        pickpool.splice(random,1);
    }
}
console.log("picked array: ");
for(var k=0; k<picks.length; k++) {
    console.log(picks[k]);
}

http://jsfiddle.net/SNJpC/