Why would feval return NaN in MATLAB

Question

I have a bunch of points in 2D for which I know the value, and I'd like to fit a cubic spline through them to interpolate some other data around using MATLAB.

My code looks like:

fitobject = fit(x,y,'cubicinterp');
yy=feval(fitobject,xx)

with the following inputs:

Coordinates

x = [...
   313     3;
   313     5;
   313     7;
   315     3;
   315     5;
   317     3;
   319     5];

Values

y = [...
   28.0779;
   28.0186;
   11.6220;
   16.7640;
   23.7139;
  -14.7882;
  -20.4626];

Interpolation points

xx = [...
   313     3;
   313     4;
   313     5;
   313     6;
   313     7;
   313     8;
   313     9;
   314     3;
   314     5;
   314     7;
   315     3;
   315     4;
   315     5;
   315     6;
   315     7;
   316     3;
   316     5;
   317     3;
   317     4;
   317     5;
   318     3;
   319     5;
   319     6;
   319     7;
   320     5];

In my output vector yy, I get several NaN values. To me, the input data look clean (they are all finite values and there's no NaN). I don't get what would cause feval to return NaN when fitting data. Why couldn't it give the best possible fit, even if it is bad? Is there an error in my approach?

I browsed a bit and it seems that the same question had been asked a bunch of times on mathworks forums, but no one gave a clear answer.

Thanks in advance for your help.

Answer 1

It's because an interpolation cannot be used as an extrapolation:

  %xx(:,1)    xx(:,2)  yy

  313.0000    3.0000   28.0779
  313.0000    4.0000   29.5074
  313.0000    5.0000   28.0186
  313.0000    6.0000   22.3233
  313.0000    7.0000   11.6220
  313.0000    8.0000       NaN   % xx exceeds bounds of original x interval
  313.0000    9.0000       NaN   % xx exceeds bounds of original x interval
  314.0000    3.0000   24.1239
  314.0000    5.0000   27.5130
  314.0000    7.0000       NaN   % xx exceeds bounds of original x interval
  315.0000    3.0000   16.7640
  315.0000    4.0000   21.7028
  315.0000    5.0000   23.7139
  315.0000    6.0000   11.2710
  315.0000    7.0000       NaN   % xx exceeds bounds of original x interval
  316.0000    3.0000    1.4641   
  316.0000    5.0000   13.9662
  317.0000    3.0000  -14.7882
  317.0000    4.0000   -5.4876
  317.0000    5.0000    2.7781
  318.0000    3.0000       NaN   % xx exceeds bounds of original x interval
  319.0000    5.0000  -20.4626
  319.0000    6.0000       NaN   % xx exceeds bounds of original x interval
  319.0000    7.0000       NaN   % xx exceeds bounds of original x interval
  320.0000    5.0000       NaN   % xx exceeds bounds of original x interval

In other words, you're trying to get data beyond the boundaries of your original surface data (extrapolation), which is usually already quite dangerous, and fit does not even allow you to do it.

Answer 2

It looks like the points which come up as NaN lie outside of the interpolation. You can plot it to take a look.

The code I used to play with this is as follows: (Note that I set the NaNs to -25 just so that they could be plotted)

x = [313     3;
    313     5;
    313     7;
    315     3;
    315     5;
    317     3;
    319     5];
y = [
    28.0779
    28.0186
    11.6220
    16.7640
    23.7139
    -14.7882
    -20.4626];

fitobject = fit(x,y,'cubicinterp');

xx = [
    313     3
    313     4
    313     5
    313     6
    313     7
    313     8
    313     9
    314     3
    314     5
    314     7
    315     3
    315     4
    315     5
    315     6
    315     7
    316     3
    316     5
    317     3
    317     4
    317     5
    318     3
    319     5
    319     6
    319     7
    320     5];


yy = fitobject(xx);
badindices = isnan(yy);
yy(badindices) = -25;

plot(fitobject, xx, yy, 'Exclude', badindices)

By the way, notice that I'm not using feval, but a direct call to fitobject