Check array if there is a duplicate value

[hiNI]

Hello,

I know that this should be simple, but I am not converging to a simple solution.

I would like to just generate a TRUE value if there is a duplicate value in an array.

This is what I have thus far (but doesn't work):

Can you offer some guidance as to a simple elegant solution to return TRUE if there is a duplicate?

Thanks for your help and time!

[davy.k]

The following code should work. This is going to run O(n^2) so It'll get really slow for larger arrays, but I'm not sure you can do much better with unordered arrays.

[maxnoder1995 · Accepted solution]

[davy.k]

Didn't even know there were sets in LabVIEW, so thanks! This is a better solutions than mine for sure.

[Yamaeda]

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@davy.k wrote:

Didn't even know there were sets in LabVIEW, so thanks! This is a better solutions than mine for sure.

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It was added in 2020 or something. 

[altenbach]

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@hiNI wrote:

This is what I have thus far (but doesn't work):

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This code makes absolutely no sense!

• It does not return anything! (There is no indicator)
• You search the entire array for an element that is guaranteed to exist, so the search output can never be <0.

Yes, Maps and Sets were introduce in LabVIEW 2019, have a look at my presentation for details. They offer a good solution.

Even without the use of a set, this will probably OK for typical scenarios:

[davy.k]

Without sets this is possibly a better solution with a runtime of O(n*log(n)), though it might lead to the array being copied.

[altenbach]

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@davy.k wrote:

Without sets this is possibly a better solution with a runtime of O(n*log(n)), though it might lead to the array being copied.

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Yes, sorting first can be an advantage for very large arrays and very few duplicates, but why would you need two instances of index array and that +1??? It is sufficient to resize to two outputs and wire [i] to the top index and the second one will automatically be [i]+1.

You can also delete the first element, autoindex over the rest, and compare current to previous. Here are two possibilities: Version B is your example, but cleaned up a bit. 😄

[altenbach]

Be aware that anything that relies on an equal comparison will not be able to tell if there are duplicate NaNs in the array.

If that is a concern, the "set" or my earlier version will still work correctly!

[davy.k]

Could be a relevant edge case, though checking floating point numbers for equality is code smell and usually a bad idea IMO. I assumed that we're just using floating point in these examples because it's the default.