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Re: Mean = NaN if NaN present [message #91778 is a reply to message #91766]

Tue, 25 August 2015 07:08

Jeremy Bailin
Messages: 618
Registered: April 2008

Senior Member

On Monday, August 24, 2015 at 6:56:04 PM UTC-5, LMH wrote:
> Hi,
>
> I (obviously) using data with NaNs filling in for bad data. I would like to take the mean of a subset of the data and have it fail (return NaN) if any NaNs are included in the subarray. If I use
>
> a = mean(subarray, /NaN)
>
> this only eliminates the NaNs from the computation, meaning that a mean would be returned even if there was only one good value in the subarray. Is there any way to do this besides incorporating an IF statement before the computation, such as
>
> if (total(finite(subarray)) eq n_elements(subarray)) then a = mean(subarray) else a = !Values.F_NAN
>
> which is not only convoluted but may be a nuisance to implement when indices are used to define the subarray?
>
> Thanks,
>
> Larry

Don't you just want MEAN without the /NAN flag?

-Jeremy.

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[Message index]

		Mean = NaN if NaN present By: laura.hike on Mon, 24 August 2015 16:56
		Mean = NaN if NaN present By: Helder Marchetto on Tue, 25 August 2015 01:11
		Re: Mean = NaN if NaN present By: Jeremy Bailin on Tue, 25 August 2015 07:08
		Re: Mean = NaN if NaN present By: laura.hike on Tue, 25 August 2015 14:45

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