How to extrapolate an x value from a fitting curve?

Hello everyone,

I'm kind new to Matlab.

I have a set of data, from which I obtained a fitting curve by interpolation using Fitting Curve toolbox. This is the essential code I'm using:

%example values

xvalues = [10; 12; 22; 28]

yvalues = [1; 2; 3; 4]

% Fit model to data.

[fitresult, gof] = fit( xvalues, yvalues, 'linearinterp', 'Normalize', 'on' );

% Plot fit with data.

figure( 'Name', 'untitled fit 1' );

h = plot( fitresult, xvalues, yvalues );

Later, I would like to obtain the exact value from the fitting curve at a certain value on the y axis (that, obviously, is not contained in the yvalues vector). The problem is that the fit function does not return "fitresult" as a vector, so i don't know hot to do that. I know that it could be easily done graphically, but i need a code to automatize the process. What could I try?

I attach an image as graphical explanation:

EDIT

• I want to get an x value from a known y value;
• I can't use interp1 for what I know, because "fitresult" is a cfit and it is not a vector;
• I can't use x(fitresult==y_known_value) because it would return the error: "Operator '==' is not supported for operands of type 'cfit' ".
• Found a similiar problem here: Getting x for a given y value from a cfit object - MATLAB Answers - MATLAB Central (mathworks.com), I followed Steven Lord's method and I got this solved. The exact code is posted as answer to this topic.

Thank you all for help!

Answer 1

[David Hill]

feval(fitresult,xvalue)

[Bruno Galizia]

Thank you, but here we have the same problem as the answer from dpb:

as you can see from the image I posted, I know the y value and I want to get the corresponding x value. So, using feval would not work, 'cause xvalue is what I want to get, and it can't be my input.

Answer 2

[dpb]

xvalues = [10; 12; 22; 28];
yvalues = [1; 2; 3; 4];
% Fit model to data.
[fitresult, gof] = fit( xvalues, yvalues, 'linearinterp', 'Normalize', 'on' );
h = plot( fitresult, xvalues, yvalues );
hold on
X=25;
hP=plot(X,fitresult(X),'k*');
hF=gcf; hLg=findobj(hF,'type','Legend');
hLg.Location='northwest';

See fit documentation; the fit object evaluates the fit automagically for the points passed it...

[Bruno Galizia]

Thank you, but in your example you're giving a value from the x axis as it is known, to get the corresponding value on the y axis.

I have to do exactly the opposite: as you can see from the image I posted, I know the y value and I want to get the corresponding x value. So, writing fitresult(x_value) would not work.

The problem here, maybe, is that I want to get a value from the indipendet variable knowing the dependent one; and i can't use interp1 (I think) because I don't have a vector.

[dpb]

OK, I overlooked the original intent, sorry.

Since you're interpolating and not fitting a single expression, the resultant fit object is a piecewise polynomial and since it is linear and you're interpolating, then there's no reason to not just simply use interp1 on the original input data in the backwards lookup...

xvalues = [10; 12; 22; 28];
yvalues=[1:4].';
Y=3.5;
X=interp1(yvalues,xvalues,Y)

X = 25

Now, direct interpolation of your y-values vector with interp1 will mean it will have to be monotonic and unique; if that may not be so in the final dataset, then simply redo the fit t'other way 'round; I believe it will handle that case with the p-p instead (but I didn't test it).

[dpb]

xvalues = [10; 12; 22; 28; 12];
yvalues=[1:4 0].';
fitrev=fit(yvalues,xvalues,'linearinterp','Normalize', 'on' );
Y=3.5;
X=fitrev(Y)

X = 25

And, indeed, the pp serves the trick with non-monotonic data values -- although as @Star Strider notes, it would require deciding which section of the overall the lookup value is supposed to be in to get the result between 28;12 instead of that between 22;28 in the above.

Only you know what the real application is....

Answer 3

[Star Strider]

You can eliminate the monotonicity problem by first selecting the region of the desired value —

xvalues = [10; 12; 22; 28];
yvalues = [1; 2; 3; 4];

yq = 3.5;
idx = find(diff(sign(yvalues - yq)));
ixr = [-1 0 1]+idx;
xq = interp1(yvalues(ixr), xvalues(ixr), yq)

xq = 25

A more robust approach to defining ‘ixr’ (and the interpolation results), especially if it contains more than one value, is:

for k = 1:numel(idx)
    ixr = max(1,idx(k)-1) : min(numel(xvalues),idx(k)+1);
    xq(k) = interp1(yvalues(ixr), xvalues(ixr), yq)
end

xq = 25

This prevents ‘ixr’ from referencing indices beyond the index range of the vectors being interpolated.

.

Answer 4

[Bruno Galizia]

Thank you all for your answers.

Anyway, I got a working code by checking here:

Getting x for a given y value from a cfit object - MATLAB Answers - MATLAB Central (mathworks.com)

So, that's what I did:

xvalues = [10; 12; 22; 28]

yvalues = [1; 2; 3; 4]

y_known = 3.5950

% Fit model to data.

[fitresult, gof] = fit( xvalues, yvalues, 'linearinterp', 'Normalize', 'on' );

ftogetx = @(x) y_known - fitresult(x);

x_to_find = fzero(ftogetx, 0)

plot(fitresult, xvalues, yvalues)

hold on

plot(x_to_find, y_known, "k*")

[Star Strider]

Running that for the record —

xvalues = [10; 12; 22; 28]

xvalues = 4×1

10 12 22 28

yvalues = [1; 2; 3; 4]

yvalues = 4×1

1 2 3 4

y_known = 3.5950

y_known = 3.5950

% Fit model to data.
[fitresult, gof] = fit( xvalues, yvalues, 'linearinterp', 'Normalize', 'on' );
ftogetx = @(x) y_known - fitresult(x);
x_to_find = fzero(ftogetx, 0)

x_to_find = 25.5700

plot(fitresult, xvalues, yvalues)
hold on
plot(x_to_find, y_known, "k*")

.