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  • $\begingroup$ there is another option: Values@AssociationThread[myNewList -> myList] $\endgroup$ Commented Jan 19, 2016 at 14:13
  • $\begingroup$ @garej, That one takes the cake timewise. For a larger list, it's hard to see how it chooses which of the elements in myList to keep for the various duplicates, but I think it does it correctly. All the other methods choose the first corresponding entry, making it easy to check visually on a small list that it did the right thing. I can add it here, or you can add it to your answer, on a large list it's the fastest method. $\endgroup$ Commented Jan 19, 2016 at 14:35
  • $\begingroup$ @JasonB it always keeps the last one per key, isn't it? It's like <|1 -> old, 1->new|>. $\endgroup$ Commented Jan 19, 2016 at 15:10
  • $\begingroup$ @Kuba, that's it, the old one is replaced by the new one. $\endgroup$ Commented Jan 19, 2016 at 15:11
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    $\begingroup$ If order is important you can Reverse both the old and new lists, perform the operation, then Reverse the result, and you'll have the original order. $\endgroup$ Commented Jan 20, 2016 at 0:41