In Javascript, if I have an array of arrays, like the following:

X = [ [1,2,3,4],
      [1,1,2,3],
      [1,1,3],
      [1,4],
      [2,1,2],
      [2,2]
    ]

Javascript sorts my array, paring first entry first, then second, and so on, so that X.sort() returns the following:

[ [1,1,2,3],
  [1,1,3],
  [1,2,3,4],
  [1,4],
  [2,1,2],
  [2,2]
]

Which is what I want. The problem is that the parison operator for paring the elements in the arrays is lexicographical, so [10,2] < [2,2], and, for example,

[[10,2],[1,1,3],[2,2]].sort() -> [[1,1,3],[10,2],[2,2]]

I need it to sort numerically, so that I get a sorted array of [[1,1,3],[2,2],[10,2]].

I tried using a parison function of function(a,b){return (a-b) }, which would work for sorting an array of numbers, but this fails to properly sort my array, which makes sense (I think) because [10,2] - [1,1,3] yields NaN

How do I go about sorting an array of numeric arrays?

In Javascript, if I have an array of arrays, like the following:

X = [ [1,2,3,4],
      [1,1,2,3],
      [1,1,3],
      [1,4],
      [2,1,2],
      [2,2]
    ]

Javascript sorts my array, paring first entry first, then second, and so on, so that X.sort() returns the following:

[ [1,1,2,3],
  [1,1,3],
  [1,2,3,4],
  [1,4],
  [2,1,2],
  [2,2]
]

Which is what I want. The problem is that the parison operator for paring the elements in the arrays is lexicographical, so [10,2] < [2,2], and, for example,

[[10,2],[1,1,3],[2,2]].sort() -> [[1,1,3],[10,2],[2,2]]

I need it to sort numerically, so that I get a sorted array of [[1,1,3],[2,2],[10,2]].

I tried using a parison function of function(a,b){return (a-b) }, which would work for sorting an array of numbers, but this fails to properly sort my array, which makes sense (I think) because [10,2] - [1,1,3] yields NaN

How do I go about sorting an array of numeric arrays?

• javascript
• sorting
• numbers

• 5 If you change the pare function to account for the fact that the arguments passed are arrays, it should work. jsfiddle/SYHr2 – Rikonator Commented Mar 28, 2013 at 15:35
• @Rikonator Great solution, you might want to post that as an answer, it seems to work – Ian Commented Mar 28, 2013 at 15:39
• @Rikonator Fantastic! Gives me the sorting capability I need with a slim function that doesn't overwrite the natural .sort method. If you write it up as an answer I'll accept it. – ckersch Commented Mar 28, 2013 at 15:48

5 Answers 5

As I said in my ment, the sort function needs to account for the fact that it's receiving arrays as arguments and not plain values. So you need to handle them accordingly.

I suggest this;

var pFunc = function (a, b) {
    var len = a.length > b.length ? b.length : a.length;

    for(var i=0; i<len; ++i) {
        if(a[i] - b[i] !== 0)
            return a[i] - b[i];
    }

    return (a.length - b.length);
};

It first tries to look for differences in the mon length of the two arrays. If the mon length is exactly the same, then it sorts on the basis of array length. Here's a working fiddle.

What you want is to run a natural sort. For your pare function, replace it with the script mentioned in this article

http://my.opera./GreyWyvern/blog/show.dml/1671288

When you do X.sort(), Javascript is paring your individual arrays as strings. It's basically doing a.toString().localeCompare(b.toString()). This is not what you want.

_{a.toString() is usually the same as a.join(',')}

What I would do is pare each element in the arrays by using a for loop.

Something like this:

X.sort(function(a,b){
    // Start off assuming values are equal
    var ret = 0;

    // Loop through a
    for(var a_i = 0, a_length = a.length; a_i < a_length; a_i++){
        // If b is shorter than a, it es first
        if(typeof b[a_i] === 'undefined'){
            ret = 1;
            break;
        }
        // if the element in a and b are *not* the same, then we can sort
        else if(a[a_i] !== b[a_i]){
            ret = a[a_i] - b[a_i];
            break;
        }
    }

    return ret;
});

You need to sort and pare between the 2 arrays: http://jsfiddle/pXzB6/

var arr = [[10,2],[1,1,3],[2,2]];

arr.sort(function(a,b){
    for(var i=0;i<a.length;i++){
       var item_a = a[i];
       for(var j=0;j<b.length;b++){   
           var item_b = b[j];   
           if(item_a == item_b){
                 continue;
           }
           else{
               return item_a > item_b;
           }
       }
    }

    if(a.length == b.length){
       return 0;
    }
    else{
      return a.length > b.length;
    }
});

console.log(arr);

var points = [40, 100, 1, 5, 25, 10];

points.sort(function(a, b){return a-b});

then the result is : 1,5,10,25,40,100

This is simplest way i think, it's worked.