60

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.

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9
  • Letter first, then number? Commented Mar 18, 2013 at 14:16
  • @BradChristie, yes sort by first two letters then the numbers, the others characters not required (but would be nice) Commented Mar 18, 2013 at 14:17
  • So if you had IL10 Hello & IL10 Bob, which comes first? Commented Mar 18, 2013 at 14:18
  • 1
    @RyanGates Except this isn't an alphabetical ordering and has nothing to do with elements in the DOM. Commented Mar 18, 2013 at 14:23
  • 1
    Does this answer your question? Javascript : natural sort of alphanumerical strings Commented Sep 5, 2021 at 10:21

8 Answers 8

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114

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();
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6 Comments

Seems to favor strings that start with alphabetical characters over numeric characters. Fixed by replacing $1 || 0 with typeof($1) == 'undefined' ? Infinity : $1
Could anyone help how this can be implemented for both ascending and descending sorting of array ?
Please note the .reverse will iterate the array again.
Can someone explain with some detial exactly what is happening with the regex replaces / Infinity? Then the while loop with shift mechanisms? I don't fully understand what is happening there
I just wanted to ask, for time-space complexity/Big O, this would be O(n) right?
|
22

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);

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2 Comments

Will it work if array contains undefined or null values? I guess it will just ignore it instead of placing them at the front or end depending on type of sort
@Umang, unfortunately localeCompare works only for strings. the parameter is converted to string.
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;  
});
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Comments

4

Pad numbers in string with leading zeros, then sort normally.

var naturalSort = function (a, b) {
    a = ('' + a).replace(/(\d+)/g, function (n) { return ('0000' + n).slice(-5) });
    b = ('' + b).replace(/(\d+)/g, function (n) { return ('0000' + n).slice(-5) });
    return a.localeCompare(b);
}

var naturalSortModern = function (a, b) {
    return ('' + a).localeCompare(('' + b), 'en', { numeric: true });
}

console.dir((["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"].sort(naturalSort)));

console.dir((["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"].sort(naturalSortModern)));

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Comments

3

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));

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Comments

2

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.

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Comments

1

Add one more alternative (why not):

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

// break out the three components in to an array
// "IL10 Bar" => ['IL', 10, 'Bar']
function getParts(i){
    i = i || '';
    var parts = i.match(/^([a-z]+)([0-9]+)(\s.*)$/i);
    if (parts){
        return [
            parts[1],
            parseInt(parts[2], 10),
            parts[3]
        ];
    }
    return []; // erroneous
}
ary.sort(function(a,b){
    // grab the parts
    var _a = getParts(a),
        _b = getParts(b);

    // trouble parsing (both fail = no shift, otherwise
    // move the troubles element to end of the array)
    if(_a.length == 0 && _b.length == 0) return 0;
    if(_a.length == 0) return -1;
    if(_b.length == 0) return 1;

    // Compare letter portion
    if (_a[0] < _b[0]) return -1;
    if (_a[0] > _b[0]) return 1;
    // letters are equal, continue...

    // compare number portion
    if (_a[1] < _b[1]) return -1;
    if (_a[1] > _b[1]) return 1;
    // numbers are equal, continue...

    // compare remaining string
    if (_a[2] < _b[2]) return -1;
    if (_a[2] > _b[2]) return 1;
    // strings are equal, continue...

    // exact match
    return 0;
});

jsfiddle example

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Comments

0

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

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1 Comment

that gives me the order ["IL0 Foo", "IL10 Baz", "IL3 Bob says hello", "PI0 Bar"]. IL3 should be the second element

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