How do you determine the number of defined elements in an array?

Question

If I have an array in Javascript that has been filled sporadically, how do I determine the number of non-nothing elements in the array?

for example:

var zipCodes = [];
zipCodes[10006] = 'New York, NY';
zipCodes[90210] = 'Los Angeles, CA';

zipCodes.length returns 90,211 here and I really don't want to loop through 90,209 nothings to find out I have two valid elements. Using a for-in loop means dealing with anything that might bubble up the prototype chain and trying to figure out if it's part of this array.

Is there any way, given the array defined above, I can extract "2" as the number of defined elements?

Answer 1

You would need an object with key-value pairs (kind of like an associative array / hashtable) instead of an array:

var zipCodes = {};
zipCodes[10006] = 'New York, NY';
zipCodes[90210] = 'Los Angeles, CA';


for(var zipCode in zipCodes) {
    if(zipCodes.hasOwnProperty(zipCode)) {//makes sure prototypes aren't taken into account

    }
}

edit: or you can store objects in an array like this:

var zipCodes = [];
zipCodes.push({10006:'New York, NY'});
zipCodes.push({90210: 'Los Angeles, CA'});

//zipCodes.length = 2

Answer 2

You could also (depending on a large variety of current unknowns) populate a second array that holds indexes (indices?) of populated entries in the first array.

So you could have a method that zooms over the entire array once, populating the second array:

var indexesNext = 0;
var indexes = new Array();
for(i = 0 ; i < zipCodes.length ; i++) 
     if(zipCodes[i] != null) indexes[indexesNext++] = i;

Or you could use add and delete accessor methods on the zipCodes array to update the indexes array if you're expecting it to change often.

Or you could use the Javascript sort() method on the array, and iterate through it until you hit an empty cell. You could use this to optimise the previous techniques as well (YMMV on this built-in method's efficiency, but the new Javascript engines in FFX 3.1 and Chrome will certainly do quite well, I'm sure.)