Re: [ferret_users] Daily Averages from 6 hourly data

["William S. Kessler" <William.S.Kessler@xxxxxxxx>]

Hi Ashley -

One simple way is to successively average 4 points, then subsample:

! find all the 4-point forward averages
let adailyav = (a+a[l=@shf:1]+a[l=@shf:2]+a[l=@shf:3])/4
! subsample every 4th point
let adailyav_by4 = adailyav[l=1:`a[l=@ngd]`:4]

This assumes no missing values.

Note that @sbx:4 would not produce true daily averages because even  
arguments to @sbx are weighted 1/2 at the endpoints.

A possibly-undesirable feature of this method is that the time value  
of the points in adailyav_by4 will be that of the first point, but  
the daily average is in fact centered halfway between the second and  
third points. That is, if the original values are on  
0000,0600,1200,1800, then the above method makes a daily average  
centered at 0900, but the variable adailyav_by4 will have its time  
values at 0000.

Billy K

On Jun 20, 2007, at 5:15 AM, Ashley Watson wrote:

> Hi ferreters,
>           I am trying to get daily averages of a varaible which has  
> values from 01-oct-1998 to 01-jun-1999 with timestep of 6hours. As  
> it spreads
> over two years 1998 and 1999, I am getting problem in defining  
> output timestep. Any hint or suggestion how to calculate this daily  
> average.
>
>
> yes? show data
>      currently SET data sets:
>     1> ./latentheat.nc  (default)
>  name     title                             I         J            
> K          L
>  LTH      Latent_Heat             1:150     1:130     ...       1:973
>
>
> Thanks in Advance
> Ashley