Most efficient way to convert an HTMLCollection to an Array

Most efficient way to convert an HTMLCollection to an Array

2 min read
javascript
arrays
object

Is there a more efficient way to convert an HTMLCollection to an Array, other than iterating through the contents of said collection and manually pushing each item into an array?

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3 posted solutions
1K

harpo wrote:

var arr = Array.prototype.slice.call( htmlCollection )

will have the same effect using "native" code.

Edit

Since this gets a lot of views, note (per @oriol's comment) that the following more concise expression is effectively equivalent:

var arr = [].slice.call(htmlCollection);

But note per @JussiR's comment, that unlike the "verbose" form, it does create an empty, unused, and indeed unusable array instance in the process. What compilers do about this is outside the programmer's ken.

Edit

Since ECMAScript 2015 (ES 6) there is also Array.from:

var arr = Array.from(htmlCollection);

Edit

ECMAScript 2015 also provides the spread operator, which is functionally equivalent to Array.from (although note that Array.from supports a mapping function as the second argument).

var arr = [...htmlCollection];

I've confirmed that both of the above work on NodeList.

A performance comparison for the mentioned methods: http://jsben.ch/h2IFA

122

mido wrote:

not sure if this is the most efficient, but a concise ES6 syntax might be:

let arry = [...htmlCollection] 

Edit: Another one, from Chris_F comment:

let arry = Array.from(htmlCollection)
25

Codesmith wrote:

I saw a more concise method of getting Array.prototype methods in general that works just as well. Converting an HTMLCollection object into an Array object is demonstrated below:

[].slice.call( yourHTMLCollectionObject );

And, as mentioned in the comments, for old browsers such as IE7 and earlier, you simply have to use a compatibility function, like:

function toArray(x) {
    for(var i = 0, a = []; i < x.length; i++)
        a.push(x[i]);

    return a
}

I know this is an old question, but I felt the accepted answer was a little incomplete; so I thought I'd throw this out there FWIW.

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