如何获得所有孩子的每个父母的position（）

Question

Hi I need to get position of all element that have attribute "data-branch" per parent - .components

<div class="component">
  <ul class="component--list">
    <li data-branch="1"><p>item<p></li>
    <li><p>item<p></li>
    <li>
      <p>item<p>
      <ul class="component--list">
        <li><p>item<p></li>
        <li data-branch="2"><p>item<p></li>
      </ul>
    </li>
    <li><p>item<p></li>
    <li><p>item<p></li>
  </ul>
</div>

I can't use offset() because I have another elements on page with changing hight like expander etc. I must get position only per .components

$('[data-branch="'+ branch.attribute +'"]').each(function(){
                var position  = $(this).position();
                console.log("Top: " + position.top + " Left: " + position.left);
            });

Result:

Top: 594.515625 Left: 90 
Top: 155.625 Left: 90 

Top: 155.625 Left: 90 because position() get position relative to .component--list not to .component :/

Answer 1

What about jQuery UI position ?

$('[data-branch="2"]').position({of: '.component'})

in resut jquery ui change position data-branch="2" an change position relative to .component :/ but i need to get coords.. How to use position UI to get top and left?