evaluate the interpolant+correcting script+equation from interpolated data

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milad babaei
milad babaei 2011-7-23
Hi all,
i am trying to evaluate interpolant by this code,would somebody plz help me to correct that??
clear
clc;
BHv= [1:1:8];
phiv= [0:10:30];
h = [ 1 1.02 1.11 1.21 1.3 1.4 1.59 1.78; ...
1 1.11 1.35 1.62 1.95 2.33 3.34 4.77; ...
1.01 1.39 2.12 3.29 5.17 8.29 22 61; ...
1.13 2.5 6.36 17.5 50 150 1400 14800];
mesh(BHv,phiv,h)
[phim BHm] = meshgrid(phiv,BHv);
hh=interp2(BHv,phiv,h,phim,BHm,'spline');
mesh(phim,BHm,hh);
hold on
[fm,Bm] = meshgrid(phiv,BHv);
plot3(fm,Bm,h+.1,'0');
hold off
xlabel('x-axis,(B/H)')
ylabel('y-axis,phi')
zlabel('depth factor,hc')
title('depth factor(hc) measurements,mandel & salencon 1969')
and is there any way get the equation and coefficients of the interpolated data????

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回答（2 个）

John D'Errico
John D'Errico 2011-7-24
Interp2 is a spline. There is no simple set of "coefficients" you can extract and use or look at. At best, you can extract a set of coefficients for each small piece of the result. In this case, there will be dozens of such sets of coefficients. It would be far less illuminating than just looking at the original data points, and those coefficients will probably take more space to write down.
As far as getting the "equation", this is perhaps the most common question asked about tools like this, and the answer is always - don't bother.
If you really need an equation, then you must pose a mathematical form for that surface. YOU must do this first. Given a mathematical form, a model, you can then use tools to estimate the coefficients of the model. Look at the curve fitting toolbox, or many other tools. But not interp2. Interp2 will give you only an interpolant, a reasonably good one in general, but nothing more.
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milad babaei
milad babaei 2011-7-24
thank u so much for helping me.ive already tried that fitting data with weighted polynomial and other toolboxes like sftool but because of large range of h values 1 to 14800 didnt possible to get a good fit,and i hoped to do this with spline tools,and still looking for an idea to transform those data to equation,using that in a another script.
milad babaei
milad babaei 2011-7-24
the general format of my 32set of data(on a table) is (phi,BHv,h),which phi values changes from 0 to 30,BHv values are in this limit [1:8] and h values are above.i would like to break data in a specified zone and then fit them to get better result in fitting,i mean for example: when BHv are between 1 and 6 and phi values are in[0,10].making equation for h values,but still large diffirences of h values make a problem.....

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the cyclist
the cyclist 2011-7-24
I don't think there is a function that will output this info. I think you will have to use "edit interp2", which will lead you into "edit spline", to get into the guts of the calculation. I see that inside the spline() function (line 66 of R2011a) there is the comment "the interpolant is a parabola", that is probably getting the coefficients you need.
I realize that this is not a very satisfying answer, so I hope someone has a better one for you!
  2 个评论
the cyclist
the cyclist 2011-7-24
Milad: Please read John's answer before digging into the code I mention!
John: I totally agree with the points you made in your answer, about the almost-always-useless nature of extracting the coefficients of each piecewise polynomial. (I had assumed that Milad knew he would be getting a gazillion coefficients, but now I think he might believe it is one over-arching equation.)

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