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Java - Distinct List of Objects

Asked 2009-06-19T20:17:21.053
37

I have a list/collection of objects that may or may not have the same property values. What's the easiest way to get a distinct list of the objects with equal properties? Is one collection type best suited for this purpose? For example, in C# I could do something like the following with LINQ.

var recipients = (from recipient in recipientList
                 select recipient).Distinct();

My initial thought was to use lambdaj (link text), but it doesn't appear to support this.

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The ordinary way of doing this would be to convert to a Set, then back to a List. But you can get fancy with Functional Java. If you liked Lamdaj, you'll love FJ.

recipients = recipients
             .sort(recipientOrd)
             .group(recipientOrd.equal())
             .map(List.<Recipient>head_());

You'll need to have defined an ordering for recipients, recipientOrd. Something like:

Ord<Recipient> recipientOrd = ord(new F2<Recipient, Recipient, Ordering>() {
  public Ordering f(Recipient r1, Recipient r2) {
    return stringOrd.compare(r1.getEmailAddress(), r2.getEmailAddress());
  }
});

Works even if you don't have control of equals() and hashCode() on the Recipient class.

answered 2009-06-19T21:10:45.050

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