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From: Thomas 'PointedEars' Lahn <PointedEars@web.de>
Newsgroups: comp.lang.javascript
Subject: Re: Difference between two arrays
Date: Tue, 27 May 2014 17:21:21 +0200
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Andrew Poulos wrote:

> If I have two "simple" arrays and I need to create a third array of
> elements that are only in one of the arrays. I found this
> 
> Array.prototype.difference = function (a) {
>    return this.filter(function (i) {

This returns the elements of the calling array (“this”)…

>      return !(a.indexOf(i) > -1);

… that are *not* in the array referred to by “a” (“a”).  “i” is the value of 
the respective element of “this” in each call of the function.  (I would 
label it “e” for “element” instead, to avoid confusion with the index.)

If “i” is found in “a”, a.indexOf(i) returns the index of “i” in “a”.  The 
index of standard arrays is always greater than -1 as it starts with 0.

If “i” is not found in “a”, -1 is returned.  Boolean-inverting the 
expression “a.indexOf(i) > -1” thus evaluates to “true” if “a.indexOf(i) > 
-1” evaluates to “false”, that is, if “i” was not found in “a”.  Likewise, 
it evaluates to “false” if “a.indexOf(i) > -1” evaluates to “true”, that is, 
if “i” was found in “a”. 

>    });
> };
> 
> which I don't fully understand but the issue with it that I have is that
> I need to run it on both arrays to get all the differences. For example
> 
> var arrX = [1, 2, 4, 6, 8],
>      arrY = [4, 8, 9];
> 
> var arrRes1 = arrX.difference(arrY)); // 1,2,6
> var arrRes2 = arrY.difference(arrX)); // 9
> 
> var arrRes = arrRes1.concat(arrRes2); // 1,2,6,9

There may be elements in array #1 that are not in array #2; for example, 
array #1 contains 1, 2 and 6 here which are not in array #2.  Those are 
obtained if you run “difference()” on array #1 and pass array #2.

There may also be elements in array #2 that are not in array #1; for 
example, array #2 contains 9 here which is not in array #1.  Those are 
obtained if you run the method on array #2 and pass array #1.

Logic (set theory) dictates that the elements of the sets A an B that are 
either in A or in B (which *you* call the “differences” between A and B) can 
be obtained by the union {A ∖ B} ∪ {B ∖ A} [{A \ B} U {B \ A} from here]:

                 A  \ B

             ,-'''''..'''''.
            :/1////:4 :     :
          A :///2//: 8:  9  : B
            ://///6:  :     :
             '._____''____.'


                 B  \ A

             ,-'''''..'''''.
            : 1    :4 ://///:
          A :   2  : 8://9//: B
            :     6:  ://///:
             '._____''____.'


            {A \ B} U {B \ A}

             ,-'''''..'''''.
            :/1////:4 ://///:
          A :///2//: 8://9//: B
            ://///6:  ://///:
             '._____''____.'

HTH

-- 
PointedEars
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