I think that you might be putting the cart before the horse. (I'm using that expression to give a hint at my age.)

If you are trying to obtain a prediction equation based on temperature (maybe predicted by elevation? spatial coordinates?, etc.), it would be helpful (although maybe not completely appropriate for this forum) to spell out the whole problem. Starting out with what to do with the predictor variable as you describe seems premature (again, at least for us who don't know the available data and objective).

And can you not define a function that just interpolates from the surrounding 4 grid points?

Finally, when faced with too much data: Sample! This is what the field of statistics is all about. That way you can get a manageable set of calculations without sacrificing the fineness of your spatial grid of data.

... or LinearModelFit if its linear...

Jaime,

You should probably post your data, otherwise it is impossible to say what is wrong. Maybe it needs a different function prototype to fit it. Maybe as Jim states below, the data needs to be sampled (with appropriate smoothing/interpolation). Maybe the data is so noisy it is problematic. Its hard to say in the abstract.

To get you started, I modified an example from the documentation. You can plot your prototype curve fit and your data and visually see what is going on to try to gain some insight as to what may be wrong.

model=aa Exp[-bb ((x-x0)^2+(y-y0)^2)];
data=MapThread[{#1[[1]],#1[[2]],1.2 Exp[-34((#1-.56).(#1-.56))]+#2}&,{RandomReal[1,{100,2}],RandomReal[{-.1,.1},100]}];
fit=NonlinearModelFit[data,model,{aa,bb,{x0, 0.5},{y0,0.6}},{x,y}];
Show[Plot3D[fit["BestFit"],{x,0,1},{y,0,1},PlotRange->All],ListPointPlot3D[data,PlotStyle->Directive[PointSize[Medium],Red]]]

Regards,

Neil

Setting one's goal with respect to $R^2$ is not recommended as $R^2$ is a measure of how much of the variability is explained but maybe there's a lot or a little to explain. (Sometimes an $R^2$ of 0.99 is not adequate.)

Using NLMXB["EstimatedVariance"]^0.5` (i.e., the root mean square error using your notation) gives an estimate of precision in terms of the units being predicted is more understandable and absolute.

Keep in mind that an interpolating polynomial will be of very high degree, which means it could wiggle considerably in regions where you would prefer it not do that, and also it will almost ceretainly give (huge) garbage if you go at all outside the rectangle (or other region) in which the points lie.

There are many ways to improve on this (radial basis methods, splines, local interpolating polynomials, convex combinations from neighboring values,...). But none of these comprise an "analytic" expression. Which leads me to suspect you are dealing with a limitation that will not be able to deliver sound results. I apologize for the pessimism. But I really doubt a huge interpolating polynomial will work well in the setting you describe.

Jaime,

Based on your last response, I seems that your question is one of creating the right structure and not trying to reduce the computation. If that is the case then you can restructure your data as follows:

In[1]:= xvect = {1, 2, 3, 4, 5};
yvect = {11, 22, 33, 44};
tmat = {{111, 122, 133, 144}, {211, 222, 233, 244}, {311, 322, 333, 
    344}, {411, 422, 433, 444}, {511, 522, 533, 544}};

In[4]:= ts = Flatten[tmat]

Out[4]= {111, 122, 133, 144, 211, 222, 233, 244, 311, 322, 333, 344, \
411, 422, 433, 444, 511, 522, 533, 544}

In[5]:= xys = Tuples[{xvect, yvect}]

Out[5]= {{1, 11}, {1, 22}, {1, 33}, {1, 44}, {2, 11}, {2, 22}, {2, 
  33}, {2, 44}, {3, 11}, {3, 22}, {3, 33}, {3, 44}, {4, 11}, {4, 
  22}, {4, 33}, {4, 44}, {5, 11}, {5, 22}, {5, 33}, {5, 44}}

In[6]:= data = MapThread[{#1, #2} &, {xys, ts}]

Out[6]= {{{1, 11}, 111}, {{1, 22}, 122}, {{1, 33}, 133}, {{1, 44}, 
  144}, {{2, 11}, 211}, {{2, 22}, 222}, {{2, 33}, 233}, {{2, 44}, 
  244}, {{3, 11}, 311}, {{3, 22}, 322}, {{3, 33}, 333}, {{3, 44}, 
  344}, {{4, 11}, 411}, {{4, 22}, 422}, {{4, 33}, 433}, {{4, 44}, 
  444}, {{5, 11}, 511}, {{5, 22}, 522}, {{5, 33}, 533}, {{5, 44}, 
  544}}

Is this what you wanted?

Regards,

Neil

Jaime,

do you really need an analytical expression for your data? I guess you would rather need ListInterpolation. Here is a minimal example:

data = RandomReal[{-1, 1}, {20, 30}];
xBorder = {0, 100};
yBorder = {-30, 50};
func = ListInterpolation[data, {xBorder, yBorder}];

Then you can use func like a "regular function", e.g. in:

Plot3D[func[x, y], {x, Sequence @@ xBorder}, {y, Sequence @@ yBorder}]

Does that help? Regards -- Henrik

... well, difficult! If your data are sufficiently "smooth", maybe you can greatly reduce the resolution.