Wolfgang Schwanghart

Hi Adam, right!

  d.val = gammaexp
  d.dist = h

I know it's a little pedestrian...

Regards, Wolfgang

Hi Jorge. I have just uploaded a new version with a bug removed. Please take this one as soon as it is available (probably by 25 Nov 2011). Then, in order to obtain variograms for different directions, just calculate the anisotropic variogram. E.g. for obtaining variograms in 0 and 90° direction, you can do following:

d = variogram([x y],z,...
'nrbins',50,'anisotropy',true,...
'thetastep',30);

plot(d.distance,...
d.val(:,find(d.theta==0 | d.theta==pi/2)));

You may choose tighter angular bins. In this case it will be 0+-15 and 90+-15.

HTH, Wolfgang

Hi Aditi,

yes, this is possible, but you have to make some changes to the function.

Note that the general variogram parameters are in vector b.

% b(1) range
% b(2) sill
% b(3) nugget

So let's say you fit your horizontal data such that
b(1) = 1;
b(2) = 5;
b(3) = []; % no nugget

Then, before fitting your next model, edit to the following lines (302-313)

% generate upper and lower bounds when fminsearchbnd is used
switch lower(params.solver)
    case {'fminsearchbnd'};
        % lower bounds
        % lb = zeros(size(b0));
        lb = [0 5];
        % upper bounds
        if nugget;
            ub = [inf max(gammaexp) max(gammaexp)]; %
        else
            % ub = [inf max(gammaexp)];
            ub = [inf 5];
        end
end

This should work. Make the obvious changes if you employ a nugget variance.

Is this a common problem and should it be implemented in variogramfit? If yes, let me know.

Best regards,
Wolfgang

Hi Adam, right!

  d.val = gammaexp
  d.dist = h

I know it's a little pedestrian...

Regards, Wolfgang

Please one stupid function,if I am doing right.
For this function I use output from function variogram d(val)=gammaexp and d(dist)=h?
Thank you a lot for those two functions!