how to sort strings in javascript numerically

Question

I would like to sort an array of strings (in javascript) such that groups of digits within the strings are compared as integers not strings. I am not worried about signed or fl

Answer 1

Another variant is to use an instance of Intl.Collator with numeric option:

var array = ["a100","a20","a3","a3b","a3b100","a3b20","a3b3","!!","~~","9","10","9.5"];
var collator = new Intl.Collator([], {numeric: true});
array.sort((a, b) => collator.compare(a, b));
console.log(array);

Answer 2

I think this does what you want

function sortArray(arr) {
    var tempArr = [], n;
    for (var i in arr) {
        tempArr[i] = arr[i].match(/([^0-9]+)|([0-9]+)/g);
        for (var j in tempArr[i]) {
            if( ! isNaN(n = parseInt(tempArr[i][j])) ){
                tempArr[i][j] = n;
            }
        }
    }
    tempArr.sort(function (x, y) {
        for (var i in x) {
            if (y.length < i || x[i] < y[i]) {
                return -1; // x is longer
            }
            if (x[i] > y[i]) {
                return 1;
            }
        }
        return 0;
    });
    for (var i in tempArr) {
        arr[i] = tempArr[i].join('');
    }
    return arr;
}
alert(
    sortArray(["a1b3", "a10b11", "a10b2", "a9b2"]).join(",")
);

Answer 3

Assuming what you want to do is just do a numeric sort by the digits in each array entry (ignoring the non-digits), you can use this:

function sortByDigits(array) {
    var re = /\D/g;

    array.sort(function(a, b) {
        return(parseInt(a.replace(re, ""), 10) - parseInt(b.replace(re, ""), 10));
    });
    return(array);
}

It uses a custom sort function that removes the digits and converts to a number each time it's asked to do a comparison. You can see it work here: http://jsfiddle.net/jfriend00/t87m2/.

If this isn't what you want, then please clarify as your question is not very clear on how the sort should actually work.

Answer 4

Use this compare function for sorting ..

function compareLists(a,b){
    var alist = a.split(/(\d+)/), // split text on change from anything to digit and digit to anything
        blist = b.split(/(\d+)/); // split text on change from anything to digit and digit to anything

    alist.slice(-1) == '' ? alist.pop() : null; // remove the last element if empty
    blist.slice(-1) == '' ? blist.pop() : null; // remove the last element if empty

    for (var i = 0, len = alist.length; i < len;i++){
        if (alist[i] != blist[i]){ // find the first non-equal part
           if (alist[i].match(/\d/)) // if numeric
           {
              return +alist[i] - +blist[i]; // compare as number
           } else {
              return alist[i].localeCompare(blist[i]); // compare as string
           }
        }
    }

    return true;
}

Syntax

var data = ["a1b3","a10b11","b10b2","a9b2","a1b20","a1c4"];
data.sort( compareLists );
alert(data);

demo at http://jsfiddle.net/h9Rqr/7/

Answer 5

Sorting occurs from left to right unless you create a custom algorithm. Letters or digits are compared digits first then letters.

However, what you want to accomplish as per your own example (a1, a9, a10) WON'T EVER HAPPEN. That would require you knowing the data before hand and spliting the string in every possible way before applying the sorting.

One final alternative would be:

a) break each and every string from left to right whenever is a change from letter to digit and vice versa; & b) then start the sorting on those groups from RIGHT-TO-LEFT. That will be a very demanding algorithm. Can be done!

Finally, if you are the GENERATOR of the original "text", you should consider NORMALIZING the output where a1 a9 a10 could be outputed as a01 a09 a10. This way you could have full cotnrol of the final version of the algorithm.

Good luck!

Answer 6

Here's a more complete solution that sorts according to both letters and numbers in the strings

function sort(list) {
    var i, l, mi, ml, x;
    // copy the original array
    list = list.slice(0);

    // split the strings, converting numeric (integer) parts to integers
    // and leaving letters as strings
    for( i = 0, l = list.length; i < l; i++ ) {
        list[i] = list[i].match(/(\d+|[a-z]+)/g);
        for( mi = 0, ml = list[i].length; mi < ml ; mi++ ) {
            x = parseInt(list[i][mi], 10);
            list[i][mi] = !!x || x === 0 ? x : list[i][mi];
        }
    }

    // sort deeply, without comparing integers as strings
    list = list.sort(function(a, b) {
        var i = 0, l = a.length, res = 0;
        while( res === 0 && i < l) {
            if( a[i] !== b[i] ) {
                res = a[i] < b[i] ? -1 : 1;
                break;
            }

            // If you want to ignore the letters, and only sort by numbers
            // use this instead:
            // 
            // if( typeof a[i] === "number" && a[i] !== b[i] ) {
            //     res = a[i] < b[i] ? -1 : 1;
            //     break;
            // }

            i++;
        }
        return res;
    });

    // glue it together again
    for( i = 0, l = list.length; i < l; i++ ) {
        list[i] = list[i].join("");
    }
    return list;
}