Is there a quick way to wrap jQuery around every element in an array?
<p>Can this be done in a more concise way?</p>

<pre><code>var src = $('div');
var dest = [];

for ( var i = 0; i &lt; src.length; i++)
    dest[i] = $(src[i]);
</code></pre>

<p>I can only think of this, which is still pretty verbose:</p>

<pre><code>var dest = [];
$('div').each(function() { dest.push($(this)); });
</code></pre>

<p>Does jQuery offer anything better for this case? I can't find it.</p>

<hr>

<p>To address some of the recurring questions in the comments:</p>

<blockquote>
  <p><code>src[i]</code> is already a jQuery object, calling <code>jQuery(src[i])</code> doesn't do anything.</p>
</blockquote>

<p>No, it's a plain DOM node, without any jQuery wrappings at all.</p>

<blockquote>
  <p>Just out of curiosity, why do it at all?</p>
</blockquote>

<p>Because afterwards I'll need to do quite a lot of mucking around with each element individually. For example, I need to go over all of them to find the tallest one, then set the rest of them to the same height. After that I need to get all their widths and perform some layouting based on that (so each element gets its x &amp; y coordinates based on the widths of other elements). Etc.</p>

<p>All these manipulations are done more easily if I can use the shorthand functions provided by jQuery, but that means that I need to wrap each element individually (<code>$('div')</code> only returned a wrapper around <strong>all</strong> of them). And since I need to visit every element multiple times, I'd like to increase performance by wrapping each of them only once, not on every visit.</p>
