排序数组元素（带数字的字符串），自然排序

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

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);

Answer 3

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 4

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 5

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)));

Answer 6

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 7

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

Answer 8

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