[ncl-talk] S1 skill score

[Dennis Shea]

No need to loop over longitude or time or level ...

>From example 6

dTdX = *new* <https://www.ncl.ucar.edu/Document/Functions/Built-in/new.shtml>
( (/ntim,klev,nlat,mlon/), *typeof*
<https://www.ncl.ucar.edu/Document/Functions/Built-in/typeof.shtml>(T),
T at _FillValue)

   do nl=0,nlat-1                      ; loop over each latitude
      dX = 6378388.**cos*
<https://www.ncl.ucar.edu/Document/Functions/Built-in/cos.shtml>(0.0174533*lat(nl))*dlon
 ; constant at this latitude
      dTdX(:,:,nl:nl,:) = *center_finite_diff_n* (T(:,:,nl:nl,:), dX ,
True,0,*3*)
   end do


On Tue, Aug 15, 2017 at 6:50 AM, Andrew Kren - NOAA Affiliate <
andrew.kren at noaa.gov> wrote:

> Dear ncl-talk,
>
> I have a question about calculating a metric. I am trying to calculate the
> S1 skill score (http://www.cawcr.gov.au/projects/verification/), which
> uses differences between forecast and observed adjacent points. I will be
> using centered finite differences for this and came across the example 6
> here (https://www.ncl.ucar.edu/Document/Functions/Built-in/
> center_finite_diff_n.shtml) which calculates dx at each latitude. My
> question would be, would I also loop over longitude to get dy or would this
> be included in the calculation already? I would calculate the S1 skill
> score over a global grid of 181x360 grid points in lat/lon.
>
> Thanks,
>
> --
> Andrew Kren, PhD
> Research Scientist I, Global Observing Systems Analysis (GOSA) Group
> NOAA ESRL Global Systems Division (Rm 3C515)
> 325 Broadway, Boulder, CO 80305
> (303) 497-5418
>
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