tags:

views:

1950

answers:

6

Is there a way to return the difference between two array in Javascript?

For example:

array1 = "test1", "test2","test3", "test4"
array2 = "test1", "test2","test3","test4", "test5", "test6"
new array = test5, test6

Any advice greatly appreciated.

+4  A: 

You could use a Set in this case. It is optimized for this kind of operation (union, intersection, difference).

Make sure it applies to your case, once it allows no duplicates.

var a = new JS.Set([1,2,3,4,5,6,7,8,9]);
var b = new JS.Set([2,4,6,8]);

a.difference(b)
// -> Set{1,3,5,7,9}
Samuel Carrijo
That looks like a nice library! What a shame that you can't download just the `Set` function without having to get everything else...
Blixt
@Blixt I believe you can download it all, and just include just the set.js file
Samuel Carrijo
+2  A: 

Just thinking... for the sake of a challenge ;-) would this work... (for basic arrays of strings, numbers, etc.) no nested arrays

function diffArrays(arr1, arr2, returnUnion){
  var ret = [];
  var test = {};
  var bigArray, smallArray, key;
  if(arr1.length >= arr2.length){
    bigArray = arr1;
    smallArray = arr2;
  } else {
    bigArray = arr2;
    smallArray = arr1;
  }
  for(var i=0;i<bigArray.length;i++){
    key = bigArray[i];
    test[key] = true;
  }
  if(!returnUnion){
    //diffing
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = null;
      }
    }
  } else {
    //union
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = true;
      }
    }
  }
  for(var i in test){
    ret.push(i);
  }
  return ret;
}

array1 = "test1", "test2","test3", "test4", "test7"
array2 = "test1", "test2","test3","test4", "test5", "test6"
diffArray = diffArrays(array1, array2);
//returns ["test5","test6","test7"]

diffArray = diffArrays(array1, array2, true);
//returns ["test1", "test2","test3","test4", "test5", "test6","test7"]

Note the sorting will likely not be as noted above... but if desired, call .sort() on the array to sort it.

scunliffe
A: 

I assume you are comparing normal array. If no, you need to change for loop to for .. in loop.

function arr_diff(a1, a2)
{
  var a=[], diff=[];
  for(var i=0;i<a1.length;i++)
    a[a1[i]]=true;
  for(var i=0;i<a2.length;i++)
    if(a[a2[i]]) delete a[a2[i]];
    else a[a2[i]]=true;
  for(var k in a)
    diff.push(k);
  return diff;
}
Thinker
This may work but it does three loops to accomplish what can be done in one line of code using the filter method of Array.
Joshaven Potter
+1  A: 

How about this:

Array.prototype.contains = function(needle){
  for (var i=0; i<this.length; i++)
    if (this[i] == needle) return true;

  return false;
} 

Array.prototype.diff = function(compare) {
    return this.filter(function(elem) {return !compare.contains(elem);})
}

var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));

So this way you can do array1.diff(array2) to get their difference (Horrible time complexity for the algorithm though - O(array1.length x array2.length) I believe)

Cat
Using the filter option is a great idea... however, you don't need to create a contains method for Array. I converted your idea into a one liner... Thanks for the inspiration!
Joshaven Potter
A: 

I wanted a similar function which took in an old array and a new array and gave me an array of added items and an array of removed items, and I wanted it to be efficient (so no .contains!).

You can play with my proposed solution here: http://jsbin.com/osewu3/12.

Can anyone see any problems/improvements to that algorithm? Thanks!

Code listing:

function diff(o, n) {
  // deal with empty lists
  if (o == undefined) o = [];
  if (n == undefined) n = [];

  // sort both arrays (or this won't work)
  o.sort(); n.sort();

  // don't compare if either list is empty
  if (o.length == 0 || n.length == 0) return {added: n, removed: o};

  // declare temporary variables
  var op = 0; var np = 0;
  var a = []; var r = [];

  // compare arrays and add to add or remove lists
  while (op < o.length && np < n.length) {
      if (o[op] < n[np]) {
          // push to diff?
          r.push(o[op]);
          op++;
      }
      else if (o[op] > n[np]) {
          // push to diff?
          a.push(n[np]);
          np++;
      }
      else {
          op++;np++;
      }
  }

  // add remaining items
  if( np < n.length )
    a = a.concat(n.slice(np, n.length));
  if( op < o.length )
    r = r.concat(o.slice(op, o.length));

  return {added: a, removed: r}; 
}
Ian Grainger
+3  A: 
Array.prototype.diff = function(a) {
    return this.filter(function(i) {return !(a.indexOf(i) > -1);});
};

////////////////////  
// Examples  
////////////////////

[1,2,3,4,5,6].diff( [3,4,5] );  
// => [1, 2, 6]

["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);  
// => ["test5", "test6"]
Joshaven Potter
How about converting `!(a.indexOf(i) > -1)` to `a.indexOf(i) < 0`, just for the sake of simplicity?
Nikita Rybak
Also, you may find useful [editing tutorial](http://stackoverflow.com/editing-help)
Nikita Rybak