replace items which appear in both arrays in javascript

Question

I have two arrays and want to remove from one all elements which exist in the other as well.

Can this be done with native JS?
Is there a jQuery function to do it?
What are best practices to do so (the faster the better)

p.s.: just post code in other languages too, maybe I can port it to Javascript

Update, after accepting answer to help the JS dudes ;-)

// Array Remove - By John Resig (MIT Licensed)
Array.prototype.remove = function(from, to) {
    var rest = this.slice((to || from) + 1 || this.length);
    this.length = from < 0 ? this.length + from : from;
    return this.push.apply(this, rest);
};
// Array Contains - By Helmuth Lammer (TU Vienna)
Array.prototype.contains = function(key){

    for(var i = 0; i<this.length; i++){
        if(this[i]  == key) return i;
    }

    return false;
}

(There is a native JS method named contains too, but it should work)

@galam He wants to make A and B disjoint, leaving A unchanged. — corsiKa
– corsiKa, Commented Jul 6, 2010 at 17:13
now I really don't know what you mean. Do you mean "remove" so that the items in A don't appear in B as @glowcoder suggests? — tvanfosson
– tvanfosson, Commented Jul 6, 2010 at 17:21
damn my english is too bad. remove is the right thing. sorry for that. trying out @glowcoders solution in some minutes ... — helle
– helle, Commented Jul 6, 2010 at 17:30

corsiKa · Accepted Answer · 2010-07-06 17:12:07Z

1

Given two sets a and b, to remove all elements present in a from b. Translation to array formatting and javascript left as an exercise to the reader. Optimizations exist for sorted arrays.

for(element e : b) {
    if(a.contains(e)) {
         b.remove(e);
    }
}

answered Jul 6, 2010 at 17:12

corsiKa

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2 Comments

helle Over a year ago

well works not in JS with this syntax, but the idea is clear. i think it is not very fast. the contains() and the remove() methode have to run through all items each time ... or am i absolutely wrong?

corsiKa Over a year ago

The idea was to take the pseudocode and apply to any language. The contains is necessary, and is O(n). The remove can be optimized by using a secondary structure (ArrayList-esque comes to mind) to build a new B array, and adding elements to it, rather than removing. Not being a Javascript guy, I don't know what mechanisms exist for such operations.

lawnsea · Accepted Answer · 2010-07-06 17:46:43Z

1

var foo = [1, 2, 3, 4, 5];

function removeAll(a, b) {
    var i = 0,
        j;

    while (i < b.length) {
        for (j = 0; j < a.length; j++) {
            if (a[j] === b[i]) {
                b.splice(i, 1);
                i--;
                break;
            }
        }
        i++;
    }
}

removeAll([2, 4], foo);
// foo === [1, 3, 5]

answered Jul 6, 2010 at 17:46

lawnsea

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Comments

gblazex · Accepted Answer · 2010-07-06 17:38:38Z

0

[See it in action]

// two arrays
var arr1 = [1,2,3,4,5];
var arr2 = [4,5,6,7,8];

// returns the element's index in an array
// or -1 if there isn't such an element
function indexOf( arr, el ) {
  for ( var i = arr.length; i--; ) {
    if ( arr[i] === el ) return i;
  }
  return -1;
}

// go through array1 and remove all
// elements which is also in array2
for ( var i = arr1.length; i--; ) {
  if ( indexOf( arr2, arr1[i] ) !== -1 ) {
    arr1.splice(i, 1);
  }
}

// result should be: arr1 = [1,2,3] 
alert( arr1 );

answered Jul 6, 2010 at 17:38

gblazex

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Comments

Cesar Canassa · Accepted Answer · 2010-07-07 22:55:31Z

0

I had to write a code once where an Ajax call would return an array containing a list of element which must be deleted from a local (client-side) array.

I ended implementing it this way:

// The filter method creates a new array with all elements that pass the
// test implemented by the provided function.
local_array = local_array.filter( function (element) {
    // Now I check if this element is present in the "delete" array. In order to do 
    // that I use the "some" method which runs a callback function for each of the
    // array elements and returns true or false.
    return !items_to_delete.some( function(item) {
        return item.pk === element.pk;
    });
});

Edit: In order to make it cross-browser (I am talking about IE here). You will have to define the some and filter functions. Just put this somewhere in your code:

if (!Array.prototype.some)
{
  Array.prototype.some = function(fun /*, thisp*/)
  {
    var i = 0,
        len = this.length >>> 0;

    if (typeof fun != "function")
      throw new TypeError();

    var thisp = arguments[1];
    for (; i < len; i++)
    {
      if (i in this &&
          fun.call(thisp, this[i], i, this))
        return true;
    }

    return false;
  };
}

if (!Array.prototype.filter)
{
  Array.prototype.filter = function(fun /*, thisp*/)
  {
    var len = this.length >>> 0;
    if (typeof fun != "function")
      throw new TypeError();

    var res = [];
    var thisp = arguments[1];
    for (var i = 0; i < len; i++)
    {
      if (i in this)
      {
        var val = this[i]; // in case fun mutates this
        if (fun.call(thisp, val, i, this))
          res.push(val);
      }
    }

    return res;
  };
}

edited Jul 7, 2010 at 22:55

answered Jul 6, 2010 at 21:57

Cesar Canassa

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2 Comments

gblazex Over a year ago

nice but not unfortunately cross-browser :(

Cesar Canassa Over a year ago

it's cross-browser. but you will have to define the some and filter methods if the client is using IE

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