Sort Array Elements (string with numbers), natural

Question

I have an array like;

["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"]

And need to sort it so it appears like;

["IL0 Foo", "IL3 Bob says hello", "IL10 Baz", "PI0 Bar"]

I have tried a sort function;

function compare(a,b) {
  if (a < b)
     return -1;
  if (a > b)
    return 1;
  return 0;
}

but this gives the order

["IL0 Foo", "IL10 Baz", "IL3 Bob says hello", "PI0 Bar"]

I have tried to think of a regex that will work but can't get my head around it.
If it helps the format will always be 2 letters, x amount of numbers, then any number of characters.

Answer 1

This is called "natural sort" and can be implemented in JS like this:

function naturalCompare(a, b) {
    var ax = [], bx = [];

    a.replace(/(\d+)|(\D+)/g, function(_, $1, $2) { ax.push([$1 || Infinity, $2 || ""]) });
    b.replace(/(\d+)|(\D+)/g, function(_, $1, $2) { bx.push([$1 || Infinity, $2 || ""]) });
    
    while(ax.length && bx.length) {
        var an = ax.shift();
        var bn = bx.shift();
        var nn = (an[0] - bn[0]) || an[1].localeCompare(bn[1]);
        if(nn) return nn;
    }

    return ax.length - bx.length;
}

/////////////////////////

test = [
    "img12.png",
    "img10.png",
    "img2.png",
    "img1.png",
    "img101.png",
    "img101a.png",
    "abc10.jpg",
    "abc10",
    "abc2.jpg",
    "20.jpg",
    "20",
    "abc",
    "abc2",
    ""
];

test.sort(naturalCompare)
document.write("<pre>" + JSON.stringify(test,0,3));

To sort in reverse order, just swap the arguments:

test.sort(function(a, b) { return naturalCompare(b, a) })

or simply

test = test.sort(naturalCompare).reverse();

Answer 2

I liked georg's solution a lot, but I needed underscores ("_") to sort before numbers. Here's how I modified his code:

var chunkRgx = /(_+)|([0-9]+)|([^0-9_]+)/g;
function naturalCompare(a, b) {
    var ax = [], bx = [];
    
    a.replace(chunkRgx, function(_, $1, $2, $3) {
        ax.push([$1 || "0", $2 || Infinity, $3 || ""])
    });
    b.replace(chunkRgx, function(_, $1, $2, $3) {
        bx.push([$1 || "0", $2 || Infinity, $3 || ""])
    });
    
    while(ax.length && bx.length) {
        var an = ax.shift();
        var bn = bx.shift();
        var nn = an[0].localeCompare(bn[0]) || 
                 (an[1] - bn[1]) || 
                 an[2].localeCompare(bn[2]);
        if(nn) return nn;
    }
    
    return ax.length - bx.length;
}

/////////////////////////

test = [
    "img12.png",
    "img10.png",
    "img2.png",
    "img1.png",
    "img101.png",
    "img101a.png",
    "abc10.jpg",
    "abc10",
    "abc2.jpg",
    "20.jpg",
    "20",
    "abc",
    "abc2",
    "_abc",
    "_ab_c",
    "_ab__c",
    "_abc_d",
    "ab_",
    "abc_",
    "_ab_cd",
    ""
];

test.sort(naturalCompare)
document.write("<pre>" + JSON.stringify(test,0,3));

Answer 3

You could do a regex like this to get non-numeric and numeric parts of the string:

var s = "foo124bar23";
s.match(/[^\d]+|\d+/g)

returns: ["foo", "124" , "bar" , "23"]

Then in your compare function you can iterate through the parts of the two strings comparing them part-by-part. The first non-matching part determines the result of the overall comparison. For each part, check if the part starts with a digit and if so parse it as a number before doing the comparison.

Answer 4

Not pretty, but check the first two char codes. If all equal parse and compare the numbers:

var arr = ["IL0 Foo", "IL10 Baz", "IL3 Bob says hello", "PI0 Bar"];
arr.sort(function (a1, b1) {
    var a = parseInt(a1.match(/\d+/g)[0], 10),
        b = parseInt(b1.match(/\d+/g)[0], 10),
        letterA = a1.charCodeAt(0),
        letterB = b1.charCodeAt(0),
        letterA1 = a1.charCodeAt(1),
        letterB1 = b1.charCodeAt(1);
    if (letterA > letterB) {
        return 1;
    } else if (letterB > letterA) {
        return -1;
    } else {
        if (letterA1 > letterB1) {
            return 1;
        } else if (letterB1 > letterA1) {
            return -1;
        }
        if (a < b) return -1;
        if (a > b) return 1;
        return 0;
    }
});

Example

Answer 5

var re = /([a-z]+)(\d+)(.+)/i;
var arr = ["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"];
var order = arr.sort( function(a,b){
    var ma = a.match(re),
        mb = b.match(re),
        a_str = ma[1],
        b_str = mb[1],
        a_num = parseInt(ma[2],10),
        b_num = parseInt(mb[2],10),
        a_rem = ma[3],
        b_rem = mb[3];
    return a_str > b_str ? 1 : a_str < b_str ? -1 : a_num > b_num ? 1 : a_num < b_num ? -1 : a_rem > b_rem;  
});

Answer 6

You could use String#localeCompare with options

sensitivity

Which differences in the strings should lead to non-zero result values. Possible values are:

• "base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
• "accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A.
• "case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A.
• "variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A.

The default is "variant" for usage "sort"; it's locale dependent for usage "search".

numeric

Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.

var array = ["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"];

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

console.log(array);