Javascript: How to implement iterator over multidimensional array?

Question

I have 2 dim array, which looks like this:

var a = [[1,2,3],[4,5,6],[7,8,9]];

I want to write an iterator which will return one value a time when it's called.

iterator(); //returns 1
iterator(); //returns 2
iterator(); //returns 3

I tried such approach:

function iterator() {
    var a = [[1,2,3],[4,5,6],[7,8,9]];
    var i, j;
    return function() {
        for (var i = 0; i < a.length; i++) {
            var b = a[i];
            for (var j = 0; j < b.length; j++) {
                return a[i][j];
            }
        }
    }
};
var a = iterator();
a(); //1
a(); //1
a(); //1

It always returns me first element.

I can try this one:

function iterator() {
    var a = [[1,2,3],[4,5,6],[7,8,9]];
    var i = 0, j = 0;
    return function() {
        for (; i < a.length; i++) {
            var b = a[i];
            for (; j < b.length; j++) {
                return a[i][j];
            }
        }
    }
};

Also not works.

But if I try this one:

function test() {
    var a = [1,2,3,4,5], i = 0;
    return function() {
        while (i < a.length) {
            return a[i++];
        }
    }
}
var a = test();
a(); //1
a(); //2
a(); //3

It works fine.

What is the difference here? How to make for loop work?

One other visible problem for me is bounds. How should I stop when I reach array bounds?

you initialize i=0 , j=0 with the for loop each time you call the function.. — Hacketo
– Hacketo, Commented Oct 6, 2015 at 11:34
@Hacketo made an edit, check my second try for iterator. also won't work. — Viacheslav Kondratiuk
– Viacheslav Kondratiuk, Commented Oct 6, 2015 at 11:36
because you never increment i and j, it execute the return statement before — Hacketo
– Hacketo, Commented Oct 6, 2015 at 11:37

Hacketo · Accepted Answer · 2015-10-06 11:46:55Z

1

Instead of using an inner for-loop for the second dimension you can use a simple if to test the bounds of j

function iterator() {
    var a = [[1,2,3],[4,5,6],[7,8,9]];
    var i = 0, j = 0;
    return function() {
        for (; i < a.length; i++) {
            var b = a[i];
            if (j < b.length){
                return a[i][j++];
            }
            j = 0;
        }
        return undefined; // reached when there is no value left
    }
};

answered Oct 6, 2015 at 11:46

Hacketo

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2 Comments

Adrian Maire Over a year ago

Actually the loops are not necessary. You could just return a[i][j] or undefined.

Hacketo Over a year ago

@AdrianMaire the for-loop is only here to simplify the bounds check of i and i incrementation (save one line .) And I added return undefined because OP might want to return something else than undefined

Abhitalks · Accepted Answer · 2015-10-06 12:07:41Z

1

Why do you need a loop at all in this case. You are effectively flattening the array anyway. You could just increment the indices along with a bounds check:

function iterator() {
    var a = [[1,2,3],[4,5,6],[7,8,9]], i = 0, j = 0;
    return function() {
        if (j >= a[i].length) { j = 0; i++; }
        if (i >= a.length) { j = 0; i = 0; }
        snippet.log( a[i][j++] );
    }
};
var x = iterator();
x(); x(); x(); x(); x(); x(); x(); x(); x(); x(); 
x(); x(); x(); x(); x(); x(); x(); x(); x(); x();

<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

answered Oct 6, 2015 at 12:07

Abhitalks

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Comments

Adrian Maire · Accepted Answer · 2015-10-06 13:05:57Z

It is generally not a good idea to extend standard feature, but for a didactical example I will make an exception. For real use, I suggest to implement your own class.

General Idea

Array.prototype.beginIterator = function()
{
    var counter = 0;
    return function()
        { 
            if (counter<=this.length) return this[counter++]; 
            else return undefined;
        };
}

Then you could iterate like follow:

var a = [3,1,4,1,5];
var it = a.beginIterator();
for (var i=it(); i!=undefined; i=it())
{
    alert(i);
}

This at the moment work only with single-dimension array, but it could be applied with any logic to other arrays or object.

Multi-dimensional (Any) solution:

The following iterator allow any-dimension array in any combination:

Array.prototype.beginIterator = function()
{
    var counter = 0;
    var iterators = null;

    return function()
        { 
            val = undefined;
            if (iterators!=null)
            {
                val = iterators();
                if (val!==undefined) return val;
                else
                {
                    iterators = null;
                    counter++;
                }
            }

            while (counter <=this.length)
            {
                if (!(this[counter] instanceof Array)) return this[counter++];
                else
                {
                    iterators = this[counter++].beginIterator();
                    val = iterators();
                    if (val!==undefined) return val;
                }
            }
            return undefiend;
        };
}

Example of use:

var a = [3,[3,5,7],4,[1,[2,5,8]],5];
var it = a.beginIterator();
for (var i=it(); i!=undefined; i=it())
{
    alert(i);
}

trincot · Accepted Answer · 2024-12-17 09:47:55Z

0

Iterators implement a specific protocol. To make it work on a 2D array with a compliant iterator, you can make use of the iterator helper method flatMap which was introduced with ECMAScript 2025 (not to be confused with the array method bearing the same name):

const a = [[1,2,3],[4,5,6],[7,8,9]];
const it = a.values().flatMap(Object);

console.log(it.next().value); // 1
console.log(it.next().value); // 2
console.log(it.next().value); // 3
console.log(it.next().value); // 4

Object is passed as argument just to get an identity mapping. It could also be x => x. The goal of that callback is to return an iterable. Since the nested array is iterable, we can just return it as-is.

answered Dec 17, 2024 at 9:47

trincot

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