Javascript : natural sort of alphanumerical strings

Question

This is now possible in modern browsers using localeCompare. By passing the numeric: true option, it will smartly recognize numbers. You can do case-insensitive using sensitivity: 'base'. Tested in Chrome, Firefox, and IE11.

Here's an example. It returns 1, meaning 10 goes after 2:

'10'.localeCompare('2', undefined, {numeric: true, sensitivity: 'base'})

For performance when sorting large numbers of strings, the article says:

When comparing large numbers of strings, such as in sorting large arrays, it is better to create an Intl.Collator object and use the function provided by its compare property. Docs link

var collator = new Intl.Collator(undefined, {numeric: true, sensitivity: 'base'});
var myArray = ['1_Document', '11_Document', '2_Document'];
console.log(myArray.sort(collator.compare));

So you need a natural sort ?

If so, than maybe this script by Brian Huisman based on David koelle's work would be what you need.

It seems like Brian Huisman's solution is now directly hosted on David Koelle's blog:

• Brian Huisman's javascript solutions
• David koelle's article on the subject

If you have a array of objects you can do like this:

myArrayObjects = myArrayObjects.sort(function(a, b) {
  return a.name.localeCompare(b.name, undefined, {
    numeric: true,
    sensitivity: 'base'
  });
});

var myArrayObjects = [{
    "id": 1,
    "name": "1 example"
  },
  {
    "id": 2,
    "name": "100 example"
  },
  {
    "id": 3,
    "name": "12 example"
  },
  {
    "id": 4,
    "name": "5 example"
  },

]

myArrayObjects = myArrayObjects.sort(function(a, b) {
  return a.name.localeCompare(b.name, undefined, {
    numeric: true,
    sensitivity: 'base'
  });
});
console.log(myArrayObjects);

To compare values you can use a comparing method-

function naturalSorter(as, bs){
    var a, b, a1, b1, i= 0, n, L,
    rx=/(\.\d+)|(\d+(\.\d+)?)|([^\d.]+)|(\.\D+)|(\.$)/g;
    if(as=== bs) return 0;
    a= as.toLowerCase().match(rx);
    b= bs.toLowerCase().match(rx);
    L= a.length;
    while(i<L){
        if(!b[i]) return 1;
        a1= a[i],
        b1= b[i++];
        if(a1!== b1){
            n= a1-b1;
            if(!isNaN(n)) return n;
            return a1>b1? 1:-1;
        }
    }
    return b[i]? -1:0;
}

But for speed in sorting an array, rig the array before sorting, so you only have to do lower case conversions and the regular expression once instead of in every step through the sort.

function naturalSort(ar, index){
    var L= ar.length, i, who, next, 
    isi= typeof index== 'number', 
    rx=  /(\.\d+)|(\d+(\.\d+)?)|([^\d.]+)|(\.(\D+|$))/g;
    function nSort(aa, bb){
        var a= aa[0], b= bb[0], a1, b1, i= 0, n, L= a.length;
        while(i<L){
            if(!b[i]) return 1;
            a1= a[i];
            b1= b[i++];
            if(a1!== b1){
                n= a1-b1;
                if(!isNaN(n)) return n;
                return a1>b1? 1: -1;
            }
        }
        return b[i]!= undefined? -1: 0;
    }
    for(i= 0; i<L; i++){
        who= ar[i];
        next= isi? ar[i][index] || '': who;
        ar[i]= [String(next).toLowerCase().match(rx), who];
    }
    ar.sort(nSort);
    for(i= 0; i<L; i++){
        ar[i]= ar[i][1];
    }
}