[ncl-talk] dividing by climatology

[Dennis Shea]

re: "... not sure whether this means normalization?"

You can "normalize by any non-zero quantity:
eg: Max value, Mean, standard deviation,....
You should consult with a colleague/advisor for which is appropriate for
you.
---
The following is a simple method of normalizing

 nmos=12
 nyrs=20
 ntim= nyrs*nmos
 monClim = random_uniform(10,100,nmos)   ; [*]
 monData = random_uniform(10,100,ntim)    ; [*]
 do nmo=0,nmos-1
    monData(nmo::nmos) = monData(nmo::nmos)/ monClim(nmo)
 end do

 nlat = 72
 mlon = 144

 monClim* :=* random_uniform(10,100,(/nmos,nlat,mlon/))  ;[*][*][*]
 monData *:=* random_uniform(10,100,(/ntim,nlat,mlon/))
 do nl=0,nlat-1
   do ml=0,mlon-1
     monData(nmo::nmos,nl,ml) = monData(nmo::nmos,nl,ml)/ monClim(nmo,nl,ml)
   do
 end do

On Mon, Feb 10, 2020 at 6:16 AM Jayant via ncl-talk <ncl-talk at ucar.edu>
wrote:

> Hi,
> I have monthly fire data for 20 years. I have computed climatology. I want
> to calculate ratio of each month and its respective climatology to obtain
> weights. Is there a function that does this calculation? I am not sure
> whether this means normalization?
> Cheers,
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