NaN/Inf/Complex value warning using "fit"

Hi all,

I'm interpolating between points using "fit" and get the warning

Warning: NaN, Inf, or complex value detected in startpoint;
choosing random starting point instead. 
> In curvefit.attention.Warning/throw (line 30)
  In fit>iFit (line 299)
  In fit (line 108)
  In fit_test (line 12)

The things is, (I'm pretty sure) there is no NaN, Inf, or complex value involved. Also weird, the fit seems to be different every time I execute the code (and goes completely haywire from time to time). If the first row in the data is removed, the warning disappears.

I included a code snippet that replicates my problem:

clear
close

% data
data = [1,4.26993;2,3.316751;3,3.130543;4,2.785851;5,2.462480;6,2.220899;7,1.991443;8,1.713285;9,1.591862;10,1.426485;11,1.329915;12,1.212978;13,1.163305;14,1.099329;15,1.042767;16,0.9620282;17,0.9086622;18,0.8804317;19,0.8365277;20,0.8281681;21,0.7897473;196,0.08825890;197,0.08764152;198,0.08784707;199,0.08867170;200,0.09054118];

% if first row is removed, no warning appears
% data(1,:) = [];

% fit two-term exponential function to data
data_fit = fit(data(:,1),data(:,2),'exp2');

% plot fitted data and curve
figure 
plot(data_fit,data(:,1),data(:,2),'predfunc')

Any ideas? Thanks!

Answer 1

[Antoine de Comité]

I have the same problem and i'm pretty sure there is no NaN, Inf or complex values in my data. Did you find what was the problem?

Answer 2

[Stephan]

Hi,

if you do not specify any start points, this can happen, because the start Points are choosen by a heuristic normally. You can avoid this warning by adding a vector for the start point option:

data_fit = fit(data(:,1),data(:,2),'exp2', 'StartPoint',[0,0,0,0]);

In your case this example will work without a warning.

Best regards

Answer 3

[Jos (10584)]

Your model is ill-suited to this set of data, as you can verify by checking the output of data_fit. When you provide a proper starting point, the warning naturally disappears:

data_fit = fit(data(:,1),data(:,2),'exp2','Startpoint',[0 1 0 1]);

[Stephan]

Note that with [0 1 0 1] as start point you get this result:

using [0 0 0 0] will give:

Best regards

Stephan