pentad time series

Arindam Chakraborty arch at IO.MET.FSU.EDU
Fri Jun 16 11:18:44 EDT 2006


hi Joythi,
        Since you have a free dimension (levels, z) in your control
file, you can do it easily. The idea is to rewrite the control file in
such a way that those 5 days of a pentad is represented by the the
z-dimension and all the 73 pentad time points is represented by the
t-dimension. This is possible because by default z varies faster than
t in grads. So, those two lines of your .ctl file should be:

zdef 5 linear 1 1 * actually 5-days of a pentad
tdef 73 linear 1jan1998 5dy * or whichever year that represents

Now you can average over the z dimension to get pentad mean and rest
is straightforward.

...
'set lat 0 30'
'set x 1'
'set t 1 73'
'd ave(ave(var,lon=80,lon=100),z=1,z=5)'
...



----------------------------
 ARINDAM CHAKRABORTY
 Department of Meteorology
 Florida State University
 Tallahassee, FL-32306, USA
 Tel: +001-850-6443524 (Off)
      +001-850-5758550 (Res)
 Fax: +001-850-6449642
____________________________

On Fri, 16 Jun 2006, N Jyothi wrote:

> Hello everyone
>           I have a data file, varying in longitude, latitude and time,
> with which I need to plot a latitude-time cross-section as:
>
> (1) first average the data between 80 to 100 degree longitude
> (2) then create a pentad time series data of the 365 days and plot from 0
> to 30 degree latitude (in the final dataset, the time will vary from 1 to
> 73 (365/5)).
>
> I was trying in the following manner, but it is getting wrong all the way.
> The dimension X varies from 1 to 512, Y from 1 to 256 and time from 1 to
> 365.
>
> 'open data.ctl'
> t=3
> 'set y 129 172'
> while(t<=10)
> 'define longave=(ave(precip,x=120,x=150))'
> 'd tloop(ave(longave,t-2,t+2))'
> 'set t 't+5
> endwhile
> ................
>
> The above does not provide any solution, as I need to have a variable in
> which the first time value will be average of first five time values
> (t=1,t=5), next time value will be average of the next five time values
> (t=6 to t=10) and final time value, which should be around 73, should have
> time averaged data from t=361 to t=365).
>
> I will be grateful for any help with this.
>
> Regards - Jyothi
>



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