Re: variogram
- To: mathgroup at smc.vnet.net
- Subject: [mg42164] Re: variogram
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Sat, 21 Jun 2003 02:49:39 -0400 (EDT)
- Organization: The University of Western Australia
- References: <bcjulj$hsd$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <bcjulj$hsd$1 at smc.vnet.net>, civnrn at hotmail.com (Rees)
wrote:
> Given the following dataset:
>
> x y T
> 2.9 0.876 0.138
> 2.5 0.188 0.214
> 4.7 2.716 2.119
> 4.2 2.717 2.685
> 4.2 3.739 0.031
> 2.1 1.534 1.534
> 2.4 2.078 0.267
> 5.8 3.324 1.670
>
> I want to formulate the variogram: ie the square difference
> 1/2[T(x,y)-T(x,y)']**2 for all measurement pairs. For the above 8
> measurements there are (n(n-1))/2 = 28 such pairs.
>
>
> so for T=0.138, the first step is to compute:
>
> 0.138-0.214 then 0.138-2.119 then 0.138-2.685....then 0.138-1.670
>
> the next step is:
> 0.214-2.119 then 0.214-2.685 etc [BUT NOT 0.214-0.138]
Others have given full solutions to your problem. I note that you do not
want to compute the full matrix, just the upper or lower triangular
part. If you did want to compute the full (symmetric) matrix then Outer
provides an elegant solution.
t={0.138,0.214,2.119,2.685,0.031,1.534,0.267,1.67}
1/2 Outer[Subtract, t, t]^2
Cheers,
Paul
--
Paul Abbott Phone: +61 8 9380 2734
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