JavaScript sorting on array in array based on calculation

Question

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:

I have an array of objects named objects.
Each object has an array of vectors named vectors.
Each vector has three coordinates: x,y,z.
Each object also has an array called faces.
Faces is an array of smaller array of IDs for the vectors. Each small array is a cluster of vectors.

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.

Can you include value of vectors as well as the expected output? — p.s.w.g
– p.s.w.g, Commented Feb 12, 2014 at 21:27

Bergi · Accepted Answer · 2014-02-12 21:46:23Z

1

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;
})

answered Feb 12, 2014 at 21:46

Bergi

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1 Comment

Niddro Over a year ago

It looks very obvious when you say it and it worked beautifully. Thank you!

user3299259 · Accepted Answer · 2014-02-12 21:30:27Z

0

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

answered Feb 12, 2014 at 21:30

user3299259

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