Fit Curve to a user defined function

2022 10 27
1 个回答
回答已采纳
更新时间：2022 10 27
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0 个投票

Hello,
I have 2 defined variables, S and u, both with experimental data shown below:
S =
0.403000000000000
0.331000000000000
0.270400000000000
0.222200000000000
0.181600000000000
0.148900000000000
0.122100000000000
0.100500000000000
0.080800000000000
0.067000000000000
0.054000000000000
0.045200000000000
0.036300000000000
0.029500000000000
0.024800000000000
0.020100000000000
0.016400000000000
0.014200000000000
0.011900000000000
0.008800000000000
0.008000000000000
0.005100000000000
0.005100000000000
u =
0.957033515000000
0.903278685000000
0.815542061000000
0.732249907000000
0.656767397000000
0.578471565000000
0.506853196000000
0.440901480000000
0.379658753000000
0.324856464000000
0.276170883000000
0.233324443000000
0.196245474000000
0.164407417000000
0.137095397000000
0.113974220000000
0.094601760000000
0.078298766000000
0.064655532000000
0.053307707000000
0.043933989000000
0.036147733000000
0.029677689000000
I want to fit them to the Monod Equation using programmatic curve fitting, where the monod equation is given as u = (umax*S)/(Ks+S), where umax and Ks are constants.
However, I do not know how to define the Monod Equation as a function in MATLAB so that I can fit u and S to it, because I need to do so to obtain the 95% prediction intervals, instead I end up with the following error messages instead:
>>[fitresult,gof] = fit(S,u,'Monod');
Error using fittype>iCreateFromLibrary (line 414)
Library function Monod not found.
Error in fittype>iCreateFittype (line 345)
obj = iCreateFromLibrary( obj, varargin{:} );
Error in fittype (line 330)
obj = iCreateFittype( obj, varargin{:} );
Error in fit>iFit (line 165)
model = fittype( fittypeobj, 'numindep', size( xdatain, 2 ) );
Error in fit (line 116)
[fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ...
and
>> [fitresult,gof] = fit(S,u,'u = (umax*S)/(Ks+S)')
Error using fittype>iDeduceCoefficients (line 621)
The independent variable x does not appear in the equation expression.
Use x in the expression or indicate another variable as the independent variable.
Error in fittype>iCreateCustomFittype (line 477)
obj = iDeduceCoefficients(obj);
Error in fittype>iCreateFittype (line 353)
obj = iCreateCustomFittype( obj, varargin{:} );
Error in fittype (line 330)
obj = iCreateFittype( obj, varargin{:} );
Error in fit>iFit (line 165)
model = fittype( fittypeobj, 'numindep', size( xdatain, 2 ) );
Error in fit (line 116)
[fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ...
How would you all solve this problem?

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 采纳的回答

John D'Errico
John D'Errico 2022-10-27
编辑：John D'Errico 2022-10-27
Ran in:

1 个投票

Ran in:
You cannot assume that fit will know the name of every possible nonlinear model form. monod is apparently not one of the predefined types.
S = [0.403000000000000 0.331000000000000 0.270400000000000 0.222200000000000 0.181600000000000 0.148900000000000 0.122100000000000 0.100500000000000 0.080800000000000 0.067000000000000 0.054000000000000 0.045200000000000 0.036300000000000 0.029500000000000 0.024800000000000 0.020100000000000 0.016400000000000 0.014200000000000 0.011900000000000 0.008800000000000 0.008000000000000 0.005100000000000 0.005100000000000]';
u = [0.957033515000000 0.903278685000000 0.815542061000000 0.732249907000000 0.656767397000000 0.578471565000000 0.506853196000000 0.440901480000000 0.379658753000000 0.324856464000000 0.276170883000000 0.233324443000000 0.196245474000000 0.164407417000000 0.137095397000000 0.113974220000000 0.094601760000000 0.078298766000000 0.064655532000000 0.053307707000000 0.043933989000000 0.036147733000000 0.029677689000000]';
plot(S,u,'o')
mdl = fittype('(umax*S)/(Ks+S)','indep','S')
mdl =
General model: mdl(Ks,umax,S) = (umax*S)/(Ks+S)
fittedmdl = fit(S,u,mdl)
Warning: Start point not provided, choosing random start point.
fittedmdl =
General model: fittedmdl(S) = (umax*S)/(Ks+S) Coefficients (with 95% confidence bounds): Ks = 0.2603 (0.2524, 0.2683) umax = 1.593 (1.566, 1.621)
plot(fittedmdl,S,u)
The fit seems quite reasonable overall, even where the legend covers up the last data point. (Sorry about that. Blame legend, and I am too lazy now to force it to be better.)
Fit would have been happier if I had provided initial guesses, but even though it complained, it was ok at the end. This was not a difficult model to fit, and a random start is adequate.

4 个评论
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Gavin Leong
Gavin Leong 2022-10-27
This works perfectly! How would you plot the 95% prediction bands in this case?
John D'Errico
John D'Errico 2022-10-27
编辑：John D'Errico 2022-10-27
Ran in:
Ran in:
Make sure you understand the somewhat subtle difference. But then try it. My guess is you were asking for confint.
help confint
--- help for cfit/confint --- CONFINT Confidence intervals for the coefficients of a fit result object. CI = CONFINT(FITRESULT) returns 95% confidence intervals for the coefficient estimates. CI = CONFINT(FITRESULT,LEVEL) specifies the confidence level. LEVEL must be between 0 and 1, and has a default value of 0.95. Documentation for cfit/confint doc cfit/confint Other uses of confint sfit/confint
help predint
--- help for cfit/predint --- PREDINT Prediction intervals for a fit result object or new observations. CI = PREDINT(FITRESULT,X,LEVEL) returns prediction intervals for a new Y value at the specified X value. LEVEL is the confidence level and has a default value of 0.95. CI = PREDINT(FITRESULT,X,LEVEL,'INTOPT','SIMOPT') specifies the type of interval to compute. 'INTOPT' can be either 'observation' (the default) to compute bounds for a new observation, or 'functional' to compute bounds for the curve evaluated at X. 'SIMOPT' can be 'on' to compute simultaneous confidence bounds or 'off' to compute non-simultaneous bounds. If 'INTOPT' is 'functional', the bounds measure the uncertainty in estimating the curve. If 'INTOPT' is 'observation', the bounds are wider to represent the addition uncertainty in predicting a new Y value (the curve plus random noise). Suppose the confidence level is 95% and 'INTOPT' is 'functional'. If 'SIMOPT' is 'off' (the default), then given a single pre-determined X value you have 95% confidence that the true curve lies between the confidence bounds. If 'SIMOPT' is 'on', then you have 95% confidence that the entire curve (at all X values) lies between the bounds. Documentation for cfit/predint doc cfit/predint Other uses of predint sfit/predint
Gavin Leong
Gavin Leong 2022-10-27
Oh I was asking for Prediction Intervals, not the Confidence Intervals. Basically I want to know how to determine and plot those such intervals on a graph (I believe it is represented by a dotted curve as shown when you use the curve fitting toolbox.)
Gavin Leong
Gavin Leong 2022-10-27
Never mind, I have figured it out. Thank you so much for the help!
For reference I used the following code:
p11 = predint(fittedmdl,S,0.95,'observation','off')
plot(S,u, '.', 'markersize', 8), hold on, plot(fittedmdl), hold on, plot(S,p11,'m--'), xlim([0 0.5]), ylim([0 1.2])
title('Nonsimultaneous Prediction Bands','FontSize',9)
legend({'Data','Fitted curve', 'Prediction intervals'}, 'FontSize',8,'Location', 'southeast')

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