I have an array of arrays in javascript set up within an object that I'm storing with local storage. It's high scores for a Phonegap game in the format {'0':[[score,time],[score,time]} where 0 is the category. I'm able to see the scores using a[1][0]. I want to sort with the high scores first.

var c={'0':[[100,11],[150,12]};
c.sort(function(){
    x=a[0];
    y=b[0];
    return y-x;
}

However, b[0] always gives an undefined error.

I'm new to Javascript and making this is my first major test as a learning experience. Though I've looked at a number of examples on stackoverflow still can't figured this one out.

I have an array of arrays in javascript set up within an object that I'm storing with local storage. It's high scores for a Phonegap game in the format {'0':[[score,time],[score,time]} where 0 is the category. I'm able to see the scores using a[1][0]. I want to sort with the high scores first.

var c={'0':[[100,11],[150,12]};
c.sort(function(){
    x=a[0];
    y=b[0];
    return y-x;
}

However, b[0] always gives an undefined error.

I'm new to Javascript and making this is my first major test as a learning experience. Though I've looked at a number of examples on stackoverflow still can't figured this one out.

• javascript
• arrays
• object
• undefined
• sorting

• Where/how are a and b defined? – Jeff Commented Feb 16, 2012 at 17:40
• The examples you've given are syntactically incorrect: this is missing a square bracket: {'0':[[score,time],[score,time]} – Matt Fenwick Commented Feb 16, 2012 at 17:40
• @Jeff a and b are the default .sort parameters. So a[0] should be the first item in the array? – Sketchee Commented Feb 16, 2012 at 18:01
• @Matt thanks! I was re-typing for the post and missed that – Sketchee Commented Feb 16, 2012 at 18:03

3 Answers 3

You need to declare the parameters to the parator function.

c.sort(function(){

should be

c.sort(function(a, b){

and you need to call sort on an array as "am not i am" points out so

var c={'0':[[100,11],[150,12]]};
c[0].sort(function(a, b){
    var x=a[0];
    var y=b[0];
    return y-x;
});

leaves c in the following state:

{
  "0": [
    [150, 12],
    [100, 11]
  ]
}

function largestOfFour(arr) {

    for(var x = 0; x < arr.length; x++){
        arr[x] = arr[x].sort(function(a,b){
            return b - a;
        });
    }

    return arr;
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

var myArray = [ [11, 5, 6, 3, 10], [22,55,33,66,11,00], [32, 35, 37, 39], [1000, 1001, 1002, 1003, 999]];

_.eachRight(myArray, function(value) {
  var descending = _.sortBy(value).reverse();
  console.log(descending);
});

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