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copied!<p>As far as I'm aware, you can't do this with a single predicate. However, you can do it with a key-path operator and a predicate in two separate calls.</p> <p>So, you have an array of arrays of terms. You want a single array of terms where its name contains 'A' (for example). Here's what you do:</p> <pre><code>NSArray * allTerms = ...; //your array of arrays of terms NSArray * collapsedTerms = [allTerms valueForKeyPath:@"@unionOfArrays.self"]; NSPredicate * filter = [NSPredicate predicateWithFormat:@"name CONTAINS[cd] %@", @"A"]; NSArray * filteredTerms = [collapsedTerms filteredArrayUsingPredicate:filter]; </code></pre> <p>The magic here is the <code>valueForKeyPath:</code> bit. The <a href="http://developer.apple.com/library/ios/documentation/Cocoa/Conceptual/KeyValueCoding/Concepts/ArrayOperators.html#//apple_ref/doc/uid/20002176-178611" rel="noreferrer"><code>@unionOfArrays</code> operator</a> takes an array of arrays of objects, and returns an array of objects. Here's what it's doing: The <code>NSArray</code> is going to break the keypath up by <code>.</code>, giving it <code>@unionOfArrays</code> and <code>self</code>. It's going to recurse and invoke <code>valueForKeyPath:</code> on itself using the second bit (self), and it's going to get back an array of arrays. It's then going to mash all those arrays together into a single array, ignoring duplicates. (If you want to remove duplicates, use <code>@distinctUnionOfArrays</code> instead)</p> <p>Once we have that array of objects, we can then filter it normally using a simple <code>name CONTAINS 'blah'</code> predicate.</p> <p>(I thought I might be able to get this to work using a <code>SUBQUERY</code>, but that's a strange beast and nothing was immediately obvious to me.)</p> <p>For the super intrepid, you could do this with a single method call using KickingBear's collection extensions (<a href="http://kickingbear.com/blog/archives/9" rel="noreferrer">available here</a>), but I would strongly advise against using hacks like that in a production setting. :)</p>

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