Re: PSA script

Hi everyone,

To calculate of the PSA pattern is necessary to remove the zonal mean on
streamfunction data. Then I did:

  str = fstr->psi(tStrt:tLast,{levP},:,:)
  printVarSummary(str)

  zonal = dim_median_n(str,1)
  printVarSummary(zonal)

  dif = str-zonal
  printVarSummary(dif)

But str(time, lat, lon) and zonal(time,lon). Because of that, I cannot do
the difference (dif=str-zonal).

How can I do to transforming zonal(time,lon) in zonal(time,lat,lon) [where,
for all latitudes we have the same vector] ?

Renata

2013/5/22 Dennis Shea <shea@ucar.edu>

> No PSA script but ...
>
> Did you look at the climatology functions? They handle leap years
>
> http://www.ncl.ucar.edu/**Document/Functions/climo.shtml<http://www.ncl.ucar.edu/Document/Functions/climo.shtml>
>
> Specifically: clmDayTLL, clmDayTLLL, smthClmDayTLL
>
> Also, there are (as yet) undocumented functions in contributed.ncl
> That will determine the daily anomalies.
>
> calcDayAnomTLL , calcDayAnomTLLL
>
> Usage: xAnom = calcDayAnomTLL (x, yyyyddd, clmDay)
> xAnom = calcDayAnomTLLL(x, yyyyddd, clmDay)
>
> You can see the code via:
>
> %> less contributed.ncl
>
> search for the above function names.
>
>
>
>
> On 5/22/13 7:44 AM, Renata Tedeschi wrote:
>
>> Hi everyone.
>>
>> Does anyone have a script that calculates the PSA patterns? I'm trying to
>> do it but I'm having a lot of difficulty.
>>
>> The main difficulty is to calculate the daily climatology whereas daily
>> data have leap years.
>>
>> Renata
>>
>>
>>
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Received on Wed May 22 11:39:19 2013