Using Javascript to sort an array of numeric arrays

Question

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, comparing 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 comparison operator for comparing 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 comparison 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?

Answer 1

As I said in my comment, 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 compFunc = 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 common length of the two arrays. If the common length is exactly the same, then it sorts on the basis of array length. Here's a working fiddle.

Answer 2

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

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

Answer 3

When you do X.sort(), Javascript is comparing 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 compare 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 comes 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;
});

Answer 4

You need to sort and compare between the 2 arrays: http://jsfiddle.net/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);

Answer 5

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.