# [ncl-talk] dividing by climatology

Jayant jayantkp2979 at gmail.com
Mon Feb 10 13:03:43 MST 2020

```Thank you Denis. I will discuss with my colleagues.
One more question:
in the example with nlat and nlon, "do loop" for nyrs or for each month on
top of lat-lon loop is not required?
best,

On Mon, Feb 10, 2020 at 12:53 PM Dennis Shea <shea at ucar.edu> wrote:

> 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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>
>
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