How can one compare string and numeric values (respecting negative values, with null always last)?

Question

I'm trying to sort an array of values that can be a mixture of numeric or string values (e.g. [10,"20",null,"1","bar","-2",-3,null,5,"foo"]). How can I sort this array such that

• null values are always placed last (regardless of sorting order, see jsFiddle)
• negative numbers are sorted correctly (i.e. they are less than positive numbers and sort correctly amongst themselves)

? I made a jsFiddle with detailed numeric and string examples (using localeCompare and the numeric option), but will paste the numeric version of my sorting algorithm below as a starting point.

// Sorting order
var order = "asc"; // Try switching between "asc" and "dsc"

// Dummy arrays
var numericArr = [10,20,null,1,-2,-3,null,5];

// Sort arrays
$(".output1").append(numericArr.toString());
numericArr.sort(sortByDataNumeric);
$(".output2").append(numericArr.toString());

// Numeric sorting function
function sortByDataNumeric(a, b, _order) {

    // Replace internal parameters if not used
    if (_order == null) _order = order;

    // If values are null, place them at the end
    var dflt = (_order == "asc" ? Number.MAX_VALUE : -Number.MAX_VALUE);

    // Numeric values
    var aVal = (a == null ? dflt : a);
    var bVal = (b == null ? dflt : b);
    return _order == "asc" ? (aVal - bVal) : (bVal - aVal);
}

The problem with my string sorting algorithm (see jsFiddle) is that I can't find a way to always place null values last and negative values aren't correctly sorted within themselves (e.g. -3 should be less than -2)

Edit

To answer the comments, I expect [10,"20",null,"1","bar","-2",-3,null,5,"foo"] to sort to [-3,"-2","1",5,10,"20","bar","foo",null,null]

Answer 1

You should first check to see if either value is null and return the opposite value.

---

On a side note:

For your default _order value, you should check if the parameter is undefined instead of comparing its value to null. If you try to compare something that is undefined directly you will get a reference error:

(undefinedVar == null) // ReferenceError: undefinedVar is not defined

Instead, you should check if the variable is undefined:

(typeof undefinedVar == "undefined") // true

Also, it's probably a better idea to wrap your compare function in a closure instead of relying on a global order variable.

Sometime like:

[].sort(function(a, b){ return sort(a, b, order)})

This way you can sort at a per-instance level.

---

http://jsfiddle.net/gxFGN/10/

JavaScript

function sort(a, b, asc) {
    var result;

    /* Default ascending order */
    if (typeof asc == "undefined") asc = true;

    if (a === null) return 1;
    if (b === null) return -1;
    if (a === null && b === null) return 0;

    result = a - b;

    if (isNaN(result)) {
        return (asc) ? a.toString().localeCompare(b) : b.toString().localeCompare(a);
    }
    else {
        return (asc) ? result : -result;
    }
}

Answer 2

function sortByDataString(a, b) {
    if (a === null) {
        return 1;
    }
    if (b === null) {
        return -1;
    }
    if (isNumber(a) && isNumber(b)) {
        if (parseInt(a,10) === parseInt(b,10)) {
            return 0;
        }
        return parseInt(a,10) > parseInt(b,10) ? 1 : -1;
    }
    if (isNumber(a)) {
        return -1;
    }
    if (isNumber(b)) {
        return 1;
    }
    if (a === b) {
        return 0;
    }
    return a > b ? 1 : -1;
}

fiddle here: http://jsfiddle.net/gxFGN/6/

I left out the order parameter, but you could always reverse the array at the end if needed.