CODEWITHSUNDEEP
javascript
Niddro
Niddro

Reputation: 1725

JavaScript sorting on array in array based on calculation

I have an array of small arrays. The small arrays are IDs to objects that belong together.

faces = [[0,1,2,3],[4,5,6,7],[0,3,7,4],[1,2,5,6],[0,1,5,4],[2,3,7,6]];

I want to sort the array faces based on a calculation of the values inside the smaller arrays. Each value represent a z-value.

The Setup:

What I want to do is to sort the array faces based on the average value of the z-value that are represented inside the small arrays.

I'm basing my try on this:

myArray.sort(function(a,b){
return b-a;
})

This is what I have so far in terms of sorting the objects[i].faces array:

objects[i].faces.sort(function(a,b){
    var value1=0;
    var value2=0;
    for (var j = 0; j<objects[i].faces[a].length; j++) {
        value1+=objects[i].vectors[faces[a][j]].z;
    }
    for (var j = 0; j<objects[i].faces[b].length; j++) {
        value2+=objects[i].vectors[faces[b][j]].z;
    }
return value1/objects[i].faces[a].length-value2/objects[i].faces[b].length;
})

It's currently complaining that he can't read the property "length" of undefined.

Upvotes: 0

Views: 78

Answers (2)

Bergi
Bergi

Reputation: 664766

a and b are the to-be-compared items, not their indices. Use

var vectors = objects[i].vectors;
objects[i].faces.sort(function(a,b){
    var value1 = 0,
        value2 = 0;
    for (var j=0; j<a.length; j++)
        value1 += vectors[a[j]].z;
    for (var j=0; j<b.length; j++)
        value2 += vectors[b[j]].z;
    return value1/a.length - value2/b.length;
})

Upvotes: 1

user3299259
user3299259

Reputation:

if you say faces.sort (function (a,b).... then a already is an item of faces so why call faces[a]... its a.length

Upvotes: 0

Related Questions