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Mean = NaN if NaN present [message #91766]

Mon, 24 August 2015 16:56

laura.hike
Messages: 87
Registered: September 2013

Member

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

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Mean = NaN if NaN present [message #91770 is a reply to message #91766]

Tue, 25 August 2015 01:11

Helder Marchetto
Messages: 520
Registered: November 2011

Senior Member

How about defining a function to do that?

function mn, subarray
if (total(finite(subarray)) eq n_elements(subarray)) then return, mean(subarray) else return, !Values.F_NAN
end

Cheers,
Helder

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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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Re: Mean = NaN if NaN present [message #91781 is a reply to message #91778]

Tue, 25 August 2015 14:45

laura.hike
Messages: 87
Registered: September 2013

Member

On Tuesday, August 25, 2015 at 7:08:41 AM UTC-7, Jeremy Bailin wrote:

>
> Don't you just want MEAN without the /NAN flag?
>
> -Jeremy.

Doh! I've been controlling for NaNs for too long! Thanks, Jeremy.

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