排序数字和字母元素的数组（自然排序）

Question

Suppose I have an array

var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"abc","huds"];

and I try sorting it, I get something like ...

[1, 1, 10, 2, 2, 3, 5, 55, 7, 75, 8, "abc", "ahsldk", "huds"]

notice 10 is before 2, how can I have something more like

[1,1,2,2,3,5 ..., "abc", "ahs...",...]

Answer 1

Short and sweet, per the original question:

var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"abc","huds"];
arr.sort(function(a,b){
  var a1=typeof a, b1=typeof b;
  return a1<b1 ? -1 : a1>b1 ? 1 : a<b ? -1 : a>b ? 1 : 0;
});
// [1, 1, 2, 2, 3, 5, 7, 8, 10, 55, 75, "abc", "ahsldk", "huds"]

(Sort first by type, then by value.)

---

A more-full-featured natural sort:

 var items = ['a1c', 'a01', 'a1', 'a13', 'a1a', 'a1b', 'a3b1', 'a1b0', 'a1b3', 'a1b1', 'dogs', 'cats', 'hogs', 'a2', '2', '20', 1, 13, 1.1, 1.13, '1.2', 'a']; console.log(naturalSort(items)) function naturalSort(ary, fullNumbers) { var re = fullNumbers ? /[\d\.\-]+|\D+/g : /\d+|\D+/g; // Perform a Schwartzian transform, breaking each entry into pieces first for (var i=ary.length;i--;) ary[i] = [ary[i]].concat((ary[i]+"").match(re).map(function(s){ return isNaN(s) ? [s,false,s] : [s*1,true,s]; })); // Perform a cascading sort down the pieces ary.sort(function(a,b){ var al = a.length, bl=b.length, e=al>bl?al:bl; for (var i=1;i<e;++i) { // Sort "a" before "a1" if (i>=al) return -1; else if (i>=bl) return 1; else if (a[i][0]!==b[i][0]) return (a[i][1]&&b[i][1]) ? // Are we comparing numbers? (a[i][0]-b[i][0]) : // Then diff them. (a[i][2]<b[i][2]) ? -1 : 1; // Otherwise, lexicographic sort } return 0; }); // Restore the original values into the array for (var i=ary.length;i--;) ary[i] = ary[i][0]; return ary; }

With naturalSort , pass true as the second parameter if you want "1.13" to sort before "1.2".

Answer 2

From http://snipplr.com/view/36012/javascript-natural-sort/ by mrhoo:

Array.prototype.naturalSort= function(){
    var a, b, a1, b1, rx=/(\d+)|(\D+)/g, rd=/\d+/;
    return this.sort(function(as, bs){
        a= String(as).toLowerCase().match(rx);
        b= String(bs).toLowerCase().match(rx);
        while(a.length && b.length){
            a1= a.shift();
            b1= b.shift();
            if(rd.test(a1) || rd.test(b1)){
                if(!rd.test(a1)) return 1;
                if(!rd.test(b1)) return -1;
                if(a1!= b1) return a1-b1;
            }
            else if(a1!= b1) return a1> b1? 1: -1;
        }
        return a.length- b.length;
    });
}

Or, from Alphanum: Javascript Natural Sorting Algorithm by Brian Huisman :

Array.prototype.alphanumSort = function(caseInsensitive) {
  for (var z = 0, t; t = this[z]; z++) {
    this[z] = [];
    var x = 0, y = -1, n = 0, i, j;

    while (i = (j = t.charAt(x++)).charCodeAt(0)) {
      var m = (i == 46 || (i >=48 && i <= 57));
      if (m !== n) {
        this[z][++y] = "";
        n = m;
      }
      this[z][y] += j;
    }
  }

  this.sort(function(a, b) {
    for (var x = 0, aa, bb; (aa = a[x]) && (bb = b[x]); x++) {
      if (caseInsensitive) {
        aa = aa.toLowerCase();
        bb = bb.toLowerCase();
      }
      if (aa !== bb) {
        var c = Number(aa), d = Number(bb);
        if (c == aa && d == bb) {
          return c - d;
        } else return (aa > bb) ? 1 : -1;
      }
    }
    return a.length - b.length;
  });

  for (var z = 0; z < this.length; z++)
    this[z] = this[z].join("");
}

Answer 3

You could do this in one line using String.prototype.localCompare() and get the result you are looking for. Note that the numeric collation option is enabled.

 var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"abc","huds"]; arr.sort((a,b) => ("" + a).localeCompare(b, undefined, {numeric: true})); console.log(arr); // [1, 1, 2, 2, 3, 5, 7, 8, 10, 55, 75, "abc", "ahsldk", "huds"]

Maybe add some logic to handle nulls.

Please be aware that this will only work with integer numbers . Floating point numbers will not be sorted the way you would hope.

Answer 4

// Most natural sorts are for sorting strings, so file2 is sorted before file10 .

If you are mixing in actual numbers you need to sort them to the front of the array, because negative numbers and digits separated by hyphens are a pain to interpret. Strings with leading zeroes need to be careful, so part002 will sort before part010 .

var natSort=function(as, bs) {
    var a, b, a1, b1,
    rx=  /(\d+)|(\D+)/g, rd= /\d/, rz=/^0/;
    if(typeof as=='number' || typeof bs=='number'){
        if(isNaN(as))return 1;
        if(isNaN(bs))return -1;
        return as-bs;
    }
    a= String(as).toLowerCase();
    b= String(bs).toLowerCase();
    if(a=== b) return 0;
    if(!(rd.test(a) && rd.test(b))) return a> b? 1: -1;
    a= a.match(rx);
    b= b.match(rx);
    while(a.length && b.length){
        a1= a.shift();
        b1= b.shift();
        if(a1!== b1){
            if(rd.test(a1) && rd.test(b1)){
                return a1.replace(rz,'.0')- b1.replace(rz,'.0');
            }
            else return a1> b1? 1: -1;
        }
    }
    return a.length - b.length;
}

array.sort(natSort)

Answer 5

This is a refined.

var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"56","abc","huds"];
    arr.sort(
                function (a,b){
                    if ( isNaN(a)&&isNaN(b)) return a<b?-1:a==b?0:1;//both are string
                    else if (isNaN(a)) return 1;//only a is a string
                    else if (isNaN(b)) return -1;//only b is a string
                    else return a-b;//both are num
                }
    );

result: 1|1|2|2|3|5|7|8|10|55|56|75|abc|ahsldk|huds|

Answer 6

If you have only alphabetical and integer items, you can stick with simple code:

var arr = [1,5,"ahsldk",10,55,3,2,7,8,1,2,75,"abc","huds"];
arr.sort(function(a, b)
{
    if (a == b)
        return 0;

    var n1 = parseInt(a, 10);
    var n2 = parseInt(b, 10);
    if (isNaN(n1) && isNaN(n2)) {
        //both alphabetical
        return (a > b) ? 1 : 0;
    }
    else if (!isNaN(n1) && !isNaN(n2)) {
        //both integers
        return (n1 > n2) ? 1 : 0;
    }
    else if (isNaN(n1) && !isNaN(n2)) {
        //a alphabetical and b is integer
        return 1;
    }

    //a integer and b is alphabetical
    return 0;
});

Working example: http://jsfiddle.net/25X2e/

Answer 7

If you can always assume numbers and strings of unmixed alphas, I would just divide and conquer. slice out numbers into a new array using typeof. Sort both independently and then just join the two arrays.

Answer 8

I knew the following way also which might sort the array alphanumerically order.

 const arr = [1, 5, "ahsldk", 10, 55, 3, 2, 7, 8, 1, 2, 75, "abc", "huds"]; arr.sort((a, b) => a - b || a.toString().localeCompare(b.toString())); console.log(arr)

Answer 9

var sorted =  ['as', '21sasa0', 'bssll'].sort((a,b) => a.replace(/[0-9]/g,"").localeCompare(b.replace(/[0-9]/g,"")));