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How to get the k-th element in a jQuery set ... AS A JQUERY ELEMENT

So I encounter this situation often and I want to figure out a better way of going about it. I using the naming convention prefixing my jQuery elements with $ so that I can distinguish them from DOM elements. For example, the following:
$pointers = $('#pointer-box div.pointer');
The problem with that is, if I want to get the 3rd in the result as a jQuery, I have to do
$($pointers[2])
which looks awkward. Unless there's a better way of doing this?

You can use .eq() jQuery method
$pointers = $('#pointer-box div.pointer');
var element=$pointers.eq(2);
Given a jQuery object that represents a set of DOM elements, the .eq()
method constructs a new jQuery object from one element within that
set. The supplied index identifies the position of this element in the
set.

When in JS you need to identify element by array order?

Ok so I finally have a code example to show this!
if ($('#Snowsports-row')[0].classList.contains("hidden") == false) {
$('#snowsports-only').removeClass("hidden")
}
The code works ONLY as written above, i.e., if the [0] were moved to the second line and removed from the first line, or if it were present/absent in both lines, it would fail.
I understand the output difference...
$('#Snowsports-row')
=> [<div>...]
$('#Snowsports-row')[0]
=> <div>...
...but I'm not understanding under what circumstances you're OK to get an array of element(s) and in which you need to tease out the exact element.
THANKS FOR ALL ANSWERS! Very clearly helped me to figure out that the problem may have been confusing JS/jQuery methods. Final version:
if ($('#Snowsports-row').hasClass("hidden") == false) {
$('#snowsports-only').removeClass("hidden")
}

The .classList method is not widely supported (not in MSIE 9.0 for example) so it's not portable, although where it exists it's fast.
Since every ID in a document is supposed to be unique, and since calling removeClass for a class that isn't present is harmless, just replace your entire call with:
$('#Snowsports-row').removeClass('hidden')
Or better yet, if that class means what I think it does, use .hide() and let jQuery do its job for you, potentially animation the transition in the process.
Alternatively, if you actually wanted to stick with using DOM and classList, you should use the .remove() method that classList already supports:
document.getElementById('#Snowsports-row').classList.remove('hidden')
although there's a minor disadvantage in that this code will crash if that element isn't found (since .getElementById will return null) whereas jQuery silently ignores calls made on empty selectors.
As for the meta-question - you use [n] if you want to access the single DOM element at position n within the jQuery object, as you've done when you use .classList.
You use .eq(n) to obtain a jQuery object representing that DOM element, e.g. if you want to apply jQuery methods to that (single) element.
If there's only a single element, or you want the jQuery method to apply to every matching element, just call the method directly on the selector, as I've done above.

First off, by using jQuery for what it's good at, you can replace this:
if ($('#Snowsports-row')[0].classList.contains("hidden") == false) {
$('#snowsports-only').removeClass("hidden")
}
with this:
$('#Snowsports-row').removeClass("hidden");
Your first block of code does the following:
With $('#Snowsports-row'), make a jQuery object that contains all DOM elements that match the select '#Snowsports-row'.
Then reach into the jQuery object with [0] and get the first DOM object in that jQuery object.
Then, use a property/method on that DOM element to determine if a class exists on that DOM element with your .classList.contains("hidden") reference.
Then, if you find that class, remove it.
A jQuery object contains inside it an array of DOM elements. If you call a method on the jQuery object itself like:
$('.tableRows').html("hello");
Then, you are asking jQuery to operate on ALL the DOM elements inside the jQuery object. You must use jQuery methods, not DOM methods.
If, on the other hand, you want to use a method such as .classList.contains(), that is only a method on an actual DOM element. That isn't a jQuery method. So, you have to reach inside of the jQuery object to get a specific DOM element out of it. That's what the [0] does. It reaches into the jQuery object and gets the first DOM element from its internal data structure. Once you have that DOM element, you can then use any DOM element methods on that DOM object.
FYI, if you ever want to get just the first DOM element from a jQuery object, but want the result to be a jQuery object, not just a DOM element, instead of [0], you can use .eq(0) like ths:
$('#Snowsports-row').eq(0).removeClass("hidden");
Now, in this specific case, this is never necessary because $('#Snowsports-row') cannot ever contain more than one DOM element because internally jQuery will only return the first matching DOM element when you are searching for a ID value (since there's never supposed to be more than one matching element with the same ID).
Just keep in mind that DOM element and a jQuery object are completely different types of objects with different methods on them. What makes it slightly confusing is that a jQuery object contains an internal list of DOM elements. But, if the object you are operating on is a jQuery object, then you can only call jQuery methods on it. If you reach into the jQuery object and pull out a DOM element, then you can only call DOM methods on it.

First of all, ids must be unique, so if you have more than one #Snowsports-only elements you can experience problems.
In your question, you are mixing jQuery code with pure Javascript code.
This:
if ($('#Snowsports-row')[0].classList.contains("hidden") {
...
}
Means that you get the first instance of #Snowsports-row (remember that is better if there is only one element with this id), but you get the DOM object (pure javascript) with the jQuery selector. You can do the same thing in jQuery like this:
$('#Snowsports-row').hasClass("hidden")
See more:
https://api.jquery.com/hasclass/
https://developer.mozilla.org/es/docs/Web/API/Element/classList

Sure, because you are operating over a list. Now, you're kind of mistaking the jQuery/javascript code. If you would like to use the same line twice you can basically drop jQuery altogether and write something like this:
var el = document.getElementById('Snowsports-row');
if (el.classList.contains('hidden')){
el.classList.remove('hidden');
}

In the first line you're selecting one specific DOM element, whereas in the second line you are selecting ALL elements in the DOM that fit that selector and removing the "hidden" class from all of them. Basically checking whether the element has a class can only be performed over an element (that's why you need to select the index, specifying a given element), but jQuery allows you to remove the class of every element inside a list (hence your second line)

Use jQuery's .eq() function. So:
var el = $('#Snowsports-row').eq(0);
if (el.hasClass("hidden")) {
$(el.removeClass("hidden")
}
There's also no harm in calling removeClass on an element that might not have that class... so:
$('#Snowsports-row').eq(0).removeClass('hidden');

array.eq() vs. array[] in Javascript and Jquery

When accessing an array, when is it appropriate to use the .eq() function?
For example, I have...
slides.eq(slidesLength-1).css("z-index", (slidesLength-1));
and later I have...
for(i=0; i<slidesLength-1; i++) {
$(slides[i]).css("left", "-100%");
}
In the first piece of code, the slideshow stops functioning if I don's use the .eq() function. However, the second piece seems to function whether I use the .eq() function or not. Why is this?

slides is not an array. It's a jQuery object. The .eq() method returns you the element at the specified index as a jQuery object.
While jQuery objects may not be arrays, they can pretend to be by having a length property as well as properties corresponding to the indexes. (Since they are not arrays, you can't call methods like .pop(), .forEach(), etc. on them.)
When you do slides[i], you are actually getting the DOM element, not a jQuery object. The $() function turns the DOM element into a jQuery object.
So, when you do slides.eq(1), internally jQuery is doing $(slides[i]).
P.S. Objects, like jQuery objects, that pretend to be arrays are called "array-like objects". If you console.log(slides), it may look like an array. This is just your console trying to make things convenient for you. (See this question for more info: Creating array-like objects in JavaScript)

.eq() is a jQuery method which returns a jQuery object, while accessing by index returns plain DOM element. You should use eq() when you want to use jQuery methods (css() in this case) on the returned selection.
The reason $(slides[i]) works is because you're constructing a jQuery object by passing the plain element to $() constructor.

Your slides variable is not an Array, but a jQuery object.
.eq() returns a jQuery object, eventually empty if index is out of bounds, and a negative index is counted from the end.
.get() returns a DOM Element, or undefined if index is out of bounds, and a negative index is counted from the end.
[] returns a DOM Element, or throw an Error if index is out of bounds.
...
Additionally, jQuery methods let you interact with a set of elements as it was alone. So you if you do:
slides.css("left", "-100%");
It is applied on every matched elements contained in the jQuery object. It is unnecessary to loop over them.
...
Also the preferred way to loop over matched elements is using the each() method:
slides.each(function (i, el) {
var $el = $(el);
});
...
Also it is an established convention to prefix jQuery variables with a $ sign; it let you to easily differentiate DOM elements from jQuery objects. But that's only a matter of taste.

What do [] brackets after a Jquery selector mean?

I'm using this code to play a preloaded mp3 file.
var shuffle = $("#shuffle")[0];
shuffle.play();
Shuffle is my id. I got the code off the net but I CANNOT figure out what the [0] after the jquery selector does. The sound does not play if I remove it.What does it do?
thanks

jQuery is an array-like object that contains all of your matched elements. Often times, jQuery will by default apply its changes to the first element in the collection:
$("li").css("display"); // display val of first element, not all elements.
Even though many li elements could have been found, the jQuery object tells us about the first implicitly. We could explicitly instruct it to do so by using the $.get method:
$("li").get(0); // Returns first DOM element
$("li")[0]; // Also returns first DOM element
We could check the nodeName to verify this:
$("li").get(0).nodeName; // LI
$("li")[0].nodeName; // LI
If we look under the covers, we can see how $.get() is implemented:
get: function(num) {
return num == null
? this.toArray()
: ( num < 0
? this[ this.length + num ]
: this[ num ] );
}
From this we can see that when no argument is provided, the entire collection of element is converted to an array, and then returned. When an argument is provided, for instance 2, we return the element as index 2. If -2 is provided, this is added to the length (suppose the length is 5, 5+(-2) is 3) and the resulting number is used as the index.
So with regards to your particular example:
var shuffle = $("#shuffle")[0];
shuffle.play();
jQuery is used to get any element that has the id value of shuffle. This returns the jQuery array-like object. But your play() method doesn't exist on the jQuery object, it exists on the #shuffle object. As such, you need to get the first element in the collection.
You could use $.get(0), however as we just saw, this would just be adding one more step. Internally, jQuery would perform the same code you're performing above, [0].

In the direct context of your question, $("#shuffle") is a selector of an id, which returns a jQuery object (not an array per-se, but it has an array-like structure), then the [0] part is actually returning a native DOMElement object of the element with id shuffle instead of the jQuery object returned by calling $('#shuffle') (without the []).
Essentially the same as doing document.getElementById('shuffle')
EDIT (as Matt pointed out)
this will then allow you to do your .play() call to start your audio stream.

The brackets after the $('#shuffle') get the first element of that selector provided.
$('div.test')[0];
<div class="test"></div> <-- this one would get returned
<div class="test"></div>
<div class="test"></div>

It returns the native javascript object containing the first element in the matched set of elements.

JQuery returns an array. [0] takes the first item in the array.

The nth element of the returned array. The same as in regular javascript or php and a good part of the programming languages which support arrays.

Its means the chronological order of object to be process always used on array [0],[1],[2]... you can check Here

Jquery returns a list using id selector

I'm having trouble with jquery and selectors using the following code:
<div id="test"></div>
console.log($('#test'));
This always returns a list like [<div id=​"test">​</div>​] instead of the single element.
This results on always having to write $('#test')[0] for every operations instead of only $('#test'). Any idea on why?
Regards

Jquery will not return the HtmlElement, it returns a jQuery object.
A jQuery object contains a collection
of Document Object Model (DOM)
elements that have been created from
an HTML string or selected from a
document. Since jQuery methods often
use CSS selectors to match elements
from a document, the set of elements
in a jQuery object is often called a
set of "matched elements" or "selected
elements"
The jQuery object itself behaves much
like an array; it has a length
property and the elements in the
object can be accessed by their
numeric indices [0] to [length-1].
Note that a jQuery object is not
actually a Javascript Array object, so
it does not have all the methods of a
true Array object such as join().
http://api.jquery.com/Types/#jQuery
This is an example of the Composite Design Pattern
The Composite Pattern describes a group of objects that can be treated in the same way a single instance of an object can. Implementing this pattern allows you to treat both individual objects and compositions in a uniform manner. In jQuery, when we're accessing or performing actions on a single DOM element or a group of DOM elements, we can treat both in a uniform manner. http://www.addyosmani.com/resources/essentialjsdesignpatterns/book/#designpatternsjquery

jQuery encapsulates any objects found for a number of reasons. For one, it ensures that no matter what, null will never be returned for example. Rather, the elements found are inserted into a list. The fact that even though you're searching for a single element, the mechanism always remains the same and therefore the results will be placed into an array of one element rather than the element itself.

In jQuery it is better to think of selectors as matching multiple items, and your solution would be best if you use the each syntax to iterate through the matches items...
$('#test').each(function() {
console.log($(this));
});

As ID is not unique, jQuery looks for every element with such ID. So it's always returns a list, because threre is no guarantee that the element is exactly one