Re: [ferret_users] calculating spatial root mean square error

[Peter Szabo <szabpet83@xxxxxxxxx>]

Ah, sorry,
i thought you wanted spatially-averaged RMSE, not a variance through twelve months.
Then you need to change something like this. But variance of the climatological months are not that common to have :)
...
yes? let diff_var=diff[l=@var]
...
yes? fill diff_stdev !have x and y axes

Cheers, P

On Fri, Oct 10, 2008 at 10:05 AM, Peter Szabo <szabpet83@xxxxxxxxx> wrote:

Hi Satish,
I dont know why you need to change grids, but here is what might be useful for your problem:

yes? let diff=x1-x2
yes? let diff_var=diff[x=@var,y=@var] !it creates the variance in x-y direction:
first computes the mean in one x/one y direction, then takes the sum of the (data - mean) squared, then divides the sum by n
yes? let diff_stdev=diff_var^0.5  !standard deviation (~RMSE) is the square root of variance
yes? plot diff_stdev !this only has L axis now.


Cheers, Peter

On Thu, Oct 9, 2008 at 6:51 PM, Satish Bastola <bastola@xxxxxxxxxxxxxxxxxxx> wrote:

Dear ferret user

I have two climatological data (12 months). I was trying to estimate the  root means square error (spatially) between two climatological data set. I would be very much thankful for your help and suggestion.

to calculate the spatial root mean square error  between two data set (one obs and next estimtaed)i tried the following way..
the two climatological data set x1.nc and x2.nc
use climatological axis
use x1.nc
use x2.nc
!grid defination
define axis/x=1:5:1
define axis/y=2:20:1
define grid/x=xrfe/y=yrfe/t=month_reg grfe
!manupulation
let diff=x1[gx=grfe,gy=grfe,gt=grfe]-x2[gx=grfe, gy=grfe,gt=grfe]
let sdiff=diff[gx=grfe, gy=grfe,gt=grfe]^2
let ave=sdiff[ts=month_reg@ave]
let ave1=ave^0.5


!!
truly
satish bastola