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Home/ Questions/Q 3606958
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T21:18:54+00:00 2026-05-18T21:18:54+00:00

Given some jQuery object set, and also a subset of that set, is there

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Given some jQuery object set, and also a subset of that set, is there a method that will return the inverse subset?

If the answer is “No“, is there any way to avoid making a second selection to get the inverse subset?

Explanatory example:

<ul>
  <li class="subset"></li>
  <li class="subset"></li>
  <li class="inverse"></li>
  <li class="inverse"></li>
</ul>

First I want to do something to all the <li>, then something to only .subset, and finally something else to only .inverse:

$('li').css('background-color','blue')
  .filter('.subset')
    .css('color','black')
  .inverse()  // <-- White Whale?!?
    .css('color','white');

I know this could be easily done with .end().filter('.inverse'), but suppose the selector were actually big and nasty and running it twice would be a big performance hit. What then?

Haven’t found anything like this in the API docs, but I’m new to jQuery and may have overlooked something obvious (also, the existence of .andSelf() means this would not be unreasonable to expect…).

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  1. Editorial Team
    Editorial Team
    2026-05-18T21:18:55+00:00Added an answer on May 18, 2026 at 9:18 pm

    There’s no built-in function for this, the reverse filter is what you’re after…the alternative isn’t chainable. The alternative would be .not(), like this:

    var all = $('li').css('background-color','blue');
    var sub = all.filter('.subset').css('color','black');
    all.not(sub).css('color','white');
    

    …but again it’s not pretty, I’d stick with the .end().filter() approach. However, if performance is paramount, use the .not() approach above, it is more efficient (since no selector runs for the inverse set).


    For your specific example (though not the general question), it’d be pretty quick to do this:

    $('li').css('background-color','blue').css('color','white')
           .filter('.subset').css('color','black');
    
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