4

Say I have an array like this: 

let arr = ["1.2.5", "1", "10", "2.0.4", "3.3.3.3"];

What would be the best way to sort this and get result like this: 
let arr = ["1", "1.2.5", "2.0.4", "3.3.3.3", "10"];

First I thought of converting each item in the array into 'float' may work but then multiple decimals won't give expected results.

I can also go for a for loop and doing stuff like item.split(".") and then check one by one, but I do not think this is the best way.

Any suggestions, please?

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5
  • is 3.3.3.3 a number? no right? Commented Jul 17, 2017 at 6:44
  • yeah i know but what i don't know what else i call it to explain what result i expect. However i do hope you get my point here. Commented Jul 17, 2017 at 6:45
  • 2.0.4 is not a number just like others, and hence it can't be sorted like numericals. Commented Jul 17, 2017 at 6:48
  • ok, what i want is not 'sort like numbers' but just sorting. Please let me know if i can rewrite my question in a better for explanation. Commented Jul 17, 2017 at 6:50
  • You can try sorting by first number. arr.sort(function(a,b){ return a.split('.')[0] - b.split('.')[0]; }); Commented Jul 17, 2017 at 6:52

5 Answers 5

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7

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 = ["1.2.5", "1", "10", "2.0.4", "3.3.3.3"];

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

console.log(array);

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4 
function compare(a, b) {
    var aSplit = a.split(".");
    var bSplit = b.split(".");


    var length = Math.min(aSplit.length, bSplit.length);
    for (var i = 0; i < length; ++i) {
        if (parseInt(aSplit[i]) < parseInt(bSplit[i])) {
            return -1;
        } else if (parseInt(aSplit[i]) > parseInt(bSplit[i])) {
            return 1;
        }
    }

    if (aSplit.length < bSplit.length) {
        return -1;
    } else if (aSplit.length > bSplit.length) {
        return 1;
    }

    return 0;
}

You can use it like: arr.sort((a, b) => compare(a, b));

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1

Here is my solution (uses ES6 syntax)

const arr = ["1.2.5", "1", "10", "2.0.4", "3.3.3.3"];

const result = arr
.map((n) => n.split('.').map((c) => c.padStart(10, '0')).join('.'))
.sort()
.map((n) => n.split('.').map((c) => +c).join('.'))

console.log(result);

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3 Comments

can you please explain this a little bit
the 1st map() adds leading zeros, i.e it does the following conversion "1.2.3" => "0000000001.0000000002.0000000003" and after that sorting works correctly. The 2nd map() just removes leading zeros.
well that's a nice way to do it, thanks for explaining. I didn't knew about padStart
0

var arr = ["1.2.5", "1", "10", "2.0.4", "3.3.3.3"];

var arr = ["1.2.5", "1", "10", "2.0.4", "3.3.3.3"];
console.log(arr.sort(function(a,b){

  var arr1 = a.split('.');
  var arr2 = b.split('.');
  
  var maxlen = Math.max(arr1.length,arr2.length);
  
  var i;

  for(i = 0;i<maxlen;i++){
  		var i1 = parseInt(arr1[i] || 0)
      var i2 = parseInt(arr2[i] || 0)
			if(i1 < i2){
      			return -1;
      }
      else if(i1 > i2){
      			return 1;
      }
  }
  
  return 0;

}));

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Comments

-1
  • sort 1.0a notation correct
  • use native localeCompare to sort 1.090 notation

function log(label,val){
  document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}

const sortVersions = (
  x,
  v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/) 
                             ? v(b) < v(a) ? 1 : -1 
                             : a.localeCompare(b, 0, {numeric: true}))

let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1","1.0a"];
log(' input : ',v);
log('sorted: ',sortVersions(v));
log('no dups:',[...new Set(sortVersions(v))]);

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