Interpolation of data that depends on two variables
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Hi guys,
I have the following set of data:
sigma=[0;0;0;0;0;1;1;1;1;1;2;2;2;2;2];
alpha=[0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
C=[10;11;12;13;14;15;16;17;18;19;20;21;22;23;24];
Now I want to know wat the value of C is if my calculate alpha is 2.5 and sigma is 1.5.
To solve this problem, I have tried to use the function interp2. But I get an error.
Can somebody help me with this?
Thanks in advance, Bas Siebers
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John D'Errico
am 21 Mai 2015
Bearbeitet: John D'Errico
am 21 Mai 2015
sigma=[0;0;0;0;0;1;1;1;1;1;2;2;2;2;2];
alpha=[0;1;2;3;4;0;1;2;3;4;0;1;2;3;4];
C=[10;11;12;13;14;15;16;17;18;19;20;21;22;23;24];
You have nicely gridded data already. Reshape will suffice to make it into a 2-d array. This is probably why you had an error, because you had the data strung out into vectors.
sigma = reshape(sigma,5,3);
alpha = reshape(alpha,5,3);
C = reshape(C,5,3);
So now we can plot C.
surf(sigma,alpha,C)

And interp2 will now work properly.
interp2(sigma,alpha,C,1.5,2.5)
ans =
20
It is important to understand that scatteredInterpolant is not needed, because your data is completely gridded already. That makes scatteredInterpolant less efficient than need be otherwise. As well, interp2 allows you to use a spline interpolant if you so desire, whereas scatterdInterpolant is limited to at most a linear interpolant. griddedInterpolant does allow the alternative (smoother) methods for interpolation.
In fact, interp2 looks to be something that MAY eventually be replaced by griddedInterpolant, at least they seem to be making hints along those lines in the help.
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Bas Siebers
am 21 Mai 2015
Andrei Bobrov
am 21 Mai 2015
Bearbeitet: Andrei Bobrov
am 21 Mai 2015
[y,x] = ndgrid(unique(alpha),unique(sigma));
v = reshape(C,size(x));
f = griddedInterpolant(x,y,v);
example of using:
>>f(1.5,2.5)
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