Nonlinear fit of segmented curve
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How would I go about getting a nonlinear least-squares fit of a segmented curve? In this case, I have a short, linear, lag period followed by a logistic growth phase (typical of bacterial growth in culture).
Thus, for x < T0, y = Y0; for x >= T0, y = Y0 + (Plateau-Y0)*(1 - exp(-K*(X-X0)).
I need least squares estimates for each of the parameters: T0, Y0, Plateau, and K
I've attempted to use a custom function in the curve fitting toolbox, but cannot figure out how to allow for the two curves.
Thanks!
采纳的回答
Teja Muppirala
2012-12-5
It is no problem to fit piecewise curves in MATLAB using the Curve Fitting Toolbox. You can deal with piecewise functions by multiplying each piece by its respective domain. For example:
rng(0); %Just fixing the random number generator for initial conditions
X = (0:0.01:10)';
% True Values
Y0_true = 3;
PLATEAU_true = 5;
K_true = 1;
X0_true = 4;
Y = [Y0_true] * (X <= X0_true) + [Y0_true + (PLATEAU_true-Y0_true)*(1 - exp(-K_true*(X-X0_true)))].* (X > X0_true);
Y = Y + 0.1*randn(size(Y));
plot(X,Y);
ftobj = fittype('[Y0] * (x <= X0) + [Y0 + (PLATEAU-Y0)*(1 - exp(-K*(x-X0)))].* (x > X0)');
cfobj = fit(X,Y,ftobj,'startpoint',rand(4,1))
hold on;
plot(X,cfobj(X),'r','linewidth',2);
5 个评论
Teja Muppirala
2012-12-5
The only situation where this approach might fail is if any of the piecewise functions can blow up to infinity, since 0*inf = nan, and that will leave you with a nan in your function. For example, if you were trying to do something like this:
y = 1/(x-3) for 0 <= x <= 2
y = x for 2 < x <= 4
In this case, I think you'll just have to write the piecewise function using if/else/end inside a separate file.
function y = f(x)
if x <= 2
y = 1/(x-3);
else
y = x;
end
It's slightly more work the the answer above, but you can still use such a user-defined file as the equation for a curve fitting operation.
Eric
2012-12-5
Mr M.
2015-9-17
I think this method is not working for me! The problem is that noncontinous parameters are not fitted, they remains unchanged, such as a in the following example:
X = [-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0]; Y = [0.1 -0.15 0.05 0.75 0.05 10.2 9.9 9.5 8.2 10.2 10.0 10.5 9.8 9.9 10.5 10.1];
equation = '(1+sign(x-a))*b';
hint = [0.5 15];
f = fit(X',Y',equation,'Start',hint);
plot(f,X,Y)
Jonathan Gößwein
2022-10-14
The problem are the startpoints, rand(4,1) does not work indeed, but with an appropriate selection the method works (e.g. the true values).
更多回答（2 个）
John Petersen
2012-12-4
Not sure how you would do that, but you could try using a sigmoid function which will get you close, relatively speaking. Something like, for example,
y2 = Y0 + (Plateau-Y0)./(1 + exp(-K*(X-X0)));
laoya
2013-5-14
0 个投票
Hi Teja Muppirala,
I am also interested in this topic. Now my problem is: if the express of curves are not expressed explicitly, but should be calculated by functions, how to use this function?
Thanks, Tang Laoya
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