Hey, seems like an easy thing to do but…

I have a list of elements(numbers), and i’m pulling unique ones from it - how can I take out a list of items that are NOT unique?

will it help if you swap the input to set difference…mean “list unique” goes into list2

Still returns Empty List, tried different lacing too, but Longest just breaks after first repetition…

@filip.kabelis I have a method for this. I had a similar situation. Please ensure you have clockwork package installed. Then follow this graph. I made a sample for you to understand.

Use the List.CountOccurences to get the count. then you can filter it. Use the In tab for elements that occur more than once and the Out for element that occurs only once!

Hope this solves the issue. Cheers!

List.UniqueItems returns the list as a unique set, meaning it removes duplicates and lists each item only once. It does not tell you which items only show up in the list once.

@nithesh23’s solution sounds like what you’re looking for except that you’d filter the uniqueElements list instead of the original (since the original doesn’t match the list length of count.)

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You were correct. i removed to avoid confusion in the thread.

Hi Nick,

My bad. Made the wrong connection for the bool filter. Take a look at this. The list that is expected is, values that repeat more than once should be excluded. The in side of the filtered list has every item that hasn’t occurred more than once.


@filip.kabelis Take a look at this and see if this resolves your issue.

Removing the first instance of every item should work


Hi @Vikram_Subbaiah , still the numbers that occur more than once come up in the final list. @filip.kabelis needs to confirm what he’s looking for. If it’s diff between the numbers, what you told is correct.

Thanks for all the responses, both @nithesh23 and @Vikram_Subbaiah replies were helpful. As for what i’m looking for… hehe i’m trying to create a script that matches two exact lenght points lists together by smallest distance, but avoiding at the same time matching two points to one same closest point (in this case point that is farther away has to look for another closest point)

So definetly this is one step to get closer to solve this :slight_smile: thanks again!

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Thanks filip. Please mark this topic as closed.

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