can you remove one array from another in javascript or jquery

Question

i have a three arrays in javascript

var array1 = new Array (1,2,3,4,5);
var array2 = new Array ("a", "b", "c", "d", "e");
var array3 = new Array ("a", "c", "d");

and i basically want to:

1. Create a new array with array2 minus items in array3. So it should result in

   var array4 = new Array {"b", "e"};

2. Create another array with the corresponding index of array1 that aligns with array 4, so in this case i would also want to be able to generate

   var array5 = new Array {2, 5}

i know in dotnet 3.5 there are lots of simple methods to do this operation but wasn't sure if javascript had anything similar.

Answer 1

Ok, for starters, array declaration takes one of two forms:

var myArr = new Array(1, 2, 3);
// or
var myArr = [1, 2, 3];

The second option is better, btw. See here for reasons .

What you're actually looking for is something called difference (as opposed to intersection or union). There's some other questions on Stack Overflow about array difference, like this one , however the accepted solution uses for .. in on an array, which is not a good idea. I heard Doug Crockford literally punches you in the face through the Internet every time you do that. Here's a more technical discussion about it if you're interested.

The answers to this question might suit you better.

Answer 2

This is made much simpler using Javascript 1.6 Array functions:

Array.prototype.remove = function(set){return this.filter(
    function(e,i,a){return set.indexOf(e)<0}
)};
Array.prototype.mapTo = function(set,to){return this.map(
    function(e,i,a){return to[set.indexOf(e)]}
)};

var array1 = [1,2,3,4,5];
var array2 = ["a", "b", "c", "d", "e"];
var array3 = ["a", "c", "d"];

var array4 = array2.remove(array3);
var array5 = array4.mapTo(array2, array1);

Answer 3

Using jQuery:

var array4 = $.grep(array2, function(n, i){
  return $.inArray(n, array3) == -1;
});

For pure JavaScript, see these questions:

• Javascript array difference
• What is the fastest or most elegant way to compute a set difference using Javascript arrays?

Answer 4

一个非常简短有效的方法是使用：

_.difference(array1,array2)

Answer 5

I recommend implementing these: http://www.dustindiaz.com/sugar-arrays/

Makes working with arrays a whole lot easier. The solution to your problem would then look like this:

var array1 = [1, 2, 3, 4, 5],
  array2 = ["a", "b", "c", "d", "e"],
  array3 = ["a", "c", "d"],
  array5 = [],
  array4 = array2.filter(function(item, i) {
        var ok = array3.indexOf(item) === -1;
        if (ok) {
              array5.push(i + 1);
        }
        return ok;
  });

  alert(array4);
  alert(array5);

Answer 6

Underscore.js includes difference & map methods to simplify to this:

var array4 = _.difference(array2, array3);
var array5 = _.map(array4, function(i){ return array1[array2.indexOf(i)] });

Answer 7

This really helped me:

 var a = [1, 2, 3, 4]; var b = [2, 3, 4, 5]; var c = _.without.apply(_, [a].concat(b)); 

 Result: c = [1]

Answer 8

This solution works only if array2 and array3 have only strings or numbers.

Step 1. Convert array3 into an object with properties (for faster lookups)

var obj = {}, i;
for(i=0;i<array3.length;i++) {
    obj[array3[i]] = true;
}

step 2: loop thru the array2 and get only the elements not in array3

var array4 = [], array5 = [];
for(i=0; i<array2.length;i++) {
    if (!obj.hasOwnProperty(array2[i])) {
        array4.push(array2[i]);
        array5.push(array1[i]);
    }
}

Answer 9

javascript is a compact language, by design.

if you don't have a library, and want a convenient method to work in most browsers, you need to write some code.

Array.prototype.lastIndex= function(what){
 var L= this.length;
 while(L){
  if(this[--L]=== what) return L;
 }
 return -1;
}
Array.prototype.remove= function(arg){
 var what, L= arg.length, ax;
 while(L && this.length){
  what= arg[--L];
  while((ax= this.lastIndex(what))!= -1){
   this.splice(ax, 1);
  }
 }
 return this;
}


var array2 = ["a", "b", "c", "d", "e"], array3 = ["a", "c", "d"];
alert(array2.remove(array3))