Calculating overall average of multidimensional array in JavaScript

Question

How would I go about calculating the average of all the values in a multidimensional array? I've written a function to calculate the average from a 1-dimensional array, but I'm not sure what the best method is when there are more than 1-dimensions.

For example, let's say we have the following:

var A = Array(3);
for (i=0; i<A.length; i++) {
    A[i] = new Array(2);
    for (j=0; j<A[i].length; j++) {
        A[i][j] = i+j;
    }
}

Therefore, A is a 2-dimensional array, or 3x2 matrix:

A = 0   1
    1   2
    2   3

So I'd like to find the average of all the values, which in this case would equal 1.5. I imagine I need to create a new 1-dimensional array of all the values, which I could then feed into my averaging function. However, I'm not sure of the easiest way to do this when the array is highly-dimensional (e.g. 5x3x6x9).

Thanks!

EDIT

Thanks everyone! I've used your advice and flattened the array using code I found in one of the attached links which uses the reduce function. My averaging function is now like this:

function average(x) {

    // Flatten multi-dimensional array
    while (x[0] instanceof Array) {
        x = x.reduce( function(a, b) { return a.concat(b); } );
    }

    // Calculate average
    return x.reduce( function(a, b) { return a + b; } )/x.length;
}

You have to define the expected logic. How would you implement this feature in a mathematical assignment? — Rob W
– Rob W, Commented Nov 3, 2011 at 17:00

Benjie · Accepted Answer · 2011-11-03 17:09:18Z

1

You can use this code to flatten the multi-dimensional array:

function flatten(array){
    var flat = [];
    for (var i = 0, l = array.length; i < l; i++){
        var type = Object.prototype.toString.call(array[i]).split(' ').pop().split(']').shift().toLowerCase();
        if (type) { flat = flat.concat(/^(array|collection|arguments|object)$/.test(type) ? flatten(array[i]) : array[i]); }
    }
    return flat;
}

and then just sum and divide:

var total = 0;
for (var i = 0, l = flattenedArray.length; i<l; i++) {
  total += flattenedArray[i];
}
var average = total/flattenedArray.length;

edited Nov 3, 2011 at 17:09

answered Nov 3, 2011 at 16:59

Benjie

7,9765 gold badges31 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

T.J. Crowder Over a year ago

External links are a great adjunct to an answer, but always copy the relevant code into the answer as well. External resources can move, get deleted, get modified in a way that makes them irrelevant, etc. Stack Overflow is meant not just for answering the OP's question right now, but for others with the same sort of question in the future.

Benjie Over a year ago

Okay @T.J.Crowder, will do :)

T.J. Crowder · Accepted Answer · 2011-11-03 17:09:47Z

1

I don't see any particular reason you need to create a new array. Just loop through the ones you have:

var i, j, sub, total, count, avg;

total = count = 0;
for (i = 0; i < A.length; ++i) {
    sub = A[i];
    count += sub.length;
    for (j = 0; j < sub.length; ++j) {
        total += sub[j];
    }
}
avg = count === 0 ? NaN : total / count;

Note that the above assumes the arrays are not sparse (they aren't in your example).

answered Nov 3, 2011 at 17:09

T.J. Crowder

1.1m201 gold badges2k silver badges2k bronze badges

Comments

kennebec · Accepted Answer · 2011-11-03 17:44:26Z

1

// The simplest method is to flatten the array

Array.prototype.flatten= function(){
var A= [];
this.forEach(function(itm){
    if(!itm || !itm.flatten)A.push(itm);
    else{
        A= A.concat(itm.flatten());
    }
});
return A;
}
// shim for older browsers (without array forEach)
Array.prototype.forEach= [].forEach || function(fun, scope){
    var T= this, L= T.length, i= 0;
    if(typeof fun== 'function'){
        while(i< L){
            if(i in T){
                fun.call(scope, T[i], i, T);
            }
            ++i;
        }
    }
    return T;
}

var a=[[1,2,3,[1,2,3]],[4,5,6,[7,8,[1,2,3,[1,2]]]],11,[1,[2,[2,4,[5]]]]];
a.flatten().join('\n')

/*  returned value: (String)
1
2
3
1
2
3
4
5
6
7
8
1
2
3
1
2
11
1
2
2
4
5
*/

edited Nov 3, 2011 at 17:44

answered Nov 3, 2011 at 17:12

kennebec

105k32 gold badges109 silver badges127 bronze badges

Comments

Jordão · Accepted Answer · 2011-11-03 18:05:57Z

1

You can flatten your multidimensional-array with this function:

function flatten(arr) {
    var acc = [];
    var f = function(arr) {
        for (var i = 0; i < arr.length; ++i) {
            if (arr[i] instanceof Array) {
                f(arr[i]);
            }
            else {
                acc.push(arr[i]);
            }
        }
    };
    f(arr);
    return acc;
}

And then you can calculate the average with your function.

edited Nov 3, 2011 at 18:05

answered Nov 3, 2011 at 17:10

Jordão

56.8k13 gold badges118 silver badges147 bronze badges

3 Comments

T.J. Crowder Over a year ago

x instanceof Array is not reliable in an environment where multiple windows (such as iframes) may be used. The most reliable way to find out whether you're really dealing with an array is Object.prototype.toString.call(x) === "[object Array]".

Jordão Over a year ago

@T.J.Crowder: good advice if multiple windows are used. Thanks!

frogstarr78 Over a year ago

You can also use Array.isArray(). Although I'm not sure about browser support

Collectives™ on Stack Overflow

Calculating overall average of multidimensional array in JavaScript

4 Answers 4

2 Comments

Comments

Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related