Multi-dimensional Fitting
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I have 3 parameters for my function. Let's say f(x,y,z)=(a*x+b)*exp(-y/c)*(z^2+d) (Or i have n parameters).
I constructed the custom function because I know the behavior of the function (that I desire).
I have many samples(around 5000). For example f(1000,10,2)= 35;
Is there a method to fit these samples into a shape,solid (for 3 parameters case) ?
Or is there a method that to find the coefficients (a,b,c,d in this case) for my custom function using all my samples?
The answer doesnt have to be a specific for 3 parameter case, i need actually a solution for n parameters case.
(I know Matlab has curve fitting and surface fitting tools but no more dimensions.)
Any support will help me, thanks.
采纳的回答
Walter Roberson
2019-12-25
1 个投票
You cannot use cftool it supports at most two independent variables and on dependent variable. You can, however, use your custom equation with fit() from the Curve Fitting Toolbox, or you could use nonlinear least squares https://www.mathworks.com/help/optim/nonlinear-least-squares-curve-fitting.html
fit() is often surprisingly efficient at what it does, and often generates coefficients that are close enough to optimal to be "good enough" for practical purposes.
However, in equations that have two or more major basins of attraction, fit() will typically go with the larger basin even when the smaller basin has a significantly better fit. Adding upper and lower bounds on the "reasonable" values of parameters can help a lot.
I would predict that the in your sample equation, the c value would tend to vary a lot. You do not have have an additive constant so the fitter would tend try to raise or lower the overall height by driving c large, but large values of c lead to small -y/c leading to near 1 value of exp() and relatively large changes in c have small effects, making it difficult to locate the best c value. This is a common problem for equations with exp() and a multiplicative coefficient inside.
5 个评论
Walter Roberson
2019-12-26
I wrote
%generate scattered points
x = rand(1,50)*10;
y = rand(1,50)*10-5;
z = rand(1,50)*2-1;
xyz = [x.', y.', z.'];
%create a fitting function
fun = @(C,xyz)(C(1)*xyz(:,1)+C(2)).*exp(-xyz(:,2)/500).*(1-C(3)./(C(4)+exp(-xyz(:,3)/C(5))))
%now generate points according to the function, adding a little
%noise to reflect uncertainty in measurements and round-off error
C = randn(1,5);
F = fun(C,xyz) + randn(50,1)*1e-10;
%now do some fitting
opt = optimoptions('lsqcurvefit', 'MaxFunctionEvaluations', 2000);
lsqcurvefit(fun,rand(1,5),xyz,F, [],[],opt)
Unfortunately the lsqcurvefit results were all over the place depending on what the initial value rand(1,5) turned out to be. Not reliable results at all.
The generated C that was used to populate the F values was
1.392207870887926 2.473900289376266 0.5039192426908656 -0.8224805538305699 0.2009818706463425
I then constructed
fvec = @(C) sum((fun(C,xyz) - F).^2);
and ran it through a custom optimizer that I wrote. The residue (sum of squared differences) with the known C values was 4.59715449612088e-19 -- any residue better than that would be "overfitting" -- fitting to the random noise.
After a fair bit of processing with my custom optimizer, the three best trial C values I found were
1.87327276773823e-18 @ [2.24518893244395 3.98961510448558 -0.461913715132666 -1.21583421679435 -0.200981870650839]
2.26305801333892e-18 @ [2.24518893250953 3.9896151040961 -0.461913715128887 -1.21583421679739 -0.200981870648803]
2.30133630608853e-18 @ [1.39220787088086 2.47390028942311 0.503919242681713 -0.822480553836453 0.20098187065357]
The third of those points has about 9 digit agreement with the actual values used to generate the points, so we can see that my fitting process was working pretty well. But notice that the two best points differ significantly.
If we compare the ideal,
/ y \ / 391871678034571 x 2785364105346679 \ / 2269450513607405 \
-exp| - --- | | ----------------- + ---------------- | | -------------------------------------------------------------------- - 1 |
\ 500 / \ 281474976710656 1125899906842624 / | / / 18014398509481984 z \ 1852061557875417 \ |
| | exp| - ------------------- | - ---------------- | 4503599627370496 |
\ \ \ 3620567511004373 / 2251799813685248 / /
to the best point found in my search,
/ y \ / 5055716019765459 x 8983814548956485 \ / 1040137217674399 \
exp| - --- | | ------------------ + ---------------- | | ------------------------------------------------------------------ + 1 |
\ 500 / \ 2251799813685248 2251799813685248 / | / / 18014398509481984 z \ 2737815262849665 \ |
| | exp| ------------------- | - ---------------- | 2251799813685248 |
\ \ \ 3620567511085367 / 2251799813685248 / /
we observe a overall sign difference, but that is partly negated by the different sign of the last term: -(termA-1) expands to (1-termA) which compares to (1+termB) , We also observe a much more serious sign difference on the z coefficient of the exp(). Substituting in the first of my random points,
xyz = [7.0223663266789 2.55914120907961 -0.411394746307526]
then the formula at the known generation values would predict
/ 576267369779971 \ / 2269450513607405 \
exp| - ------------------ | | ------------------------------------------------------------------------------ - 1 | 1941168252945175763659330182021
\ 112589990684262400 / | / 1852061557875417 \ |
| 4503599627370496 | exp(7411028904691008/3620567511004373) - ---------------- | |
\ \ 2251799813685248 / /
- --------------------------------------------------------------------------------------------------------------------------------------------------
158456325028528675187087900672
and the formula at the best point I found would predict
/ 576267369779971 \ / 1040137217674399 \
exp| - ------------------ | | ----------------------------------------------------------------- + 1 | 25043900806793194606380187720381
\ 112589990684262400 / | / / 7411028904691008 \ 2737815262849665 \ |
| 2251799813685248 | exp| - ---------------- | - ---------------- | |
\ \ \ 3620567511085367 / 2251799813685248 / /
--------------------------------------------------------------------------------------------------------------------------------------
1267650600228229401496703205376
For the known generation values, the terms to be multiplied come out as
[ 0.99489479367074606620566129200363, -0.92719579975957747573178766456934, -12.250493961636970947965676803273]
and for the best point found in my search, they come out as
[ 0.99489479367074606620566129200363, 0.57494016275805730759843536485652, 19.756154260712107376891803209107]
which multiply out to nearly exactly the same thing.
This is the sort of problem I referred to before, that when you have exp() involved, it is pretty common for the minimizers to have difficulty resolving between two different configurations.
fmincon with a seed point of [1 1 -1 -1 -1] and 10000 iterations, does a respectable job of getting near-ish to the best point found in the search, and with a seed point of [1 1 1 -1 1] does a respectable job of getting near-ish to the known generating points.
For that matter, lsqcurvefit() does a very good job of getting quite close to those values given those respective seed points -- enough so that despite my earlier exploration, I would recommend lsqcurvefit if you can provide a reasonable starting point -- but it can do a quite bad job if your starting point is in the wrong area.
... Though as I demonstrated above, the coefficients that best fit the points might turn out to be quite different than expected, because of the mathematics of the equation to be fit.
Walter Roberson
2019-12-27
In general, if you have equations with multiple terms multiplied together, and you do not constrain all but one of the terms to be a specific sign, then you are at risk of there being multiple solutions.
(a*x+b) is not constrained as to sign: (-a)*x+(-b) is -(a*x+b) so if one of the other terms can be made negative you will have multiple solutions.
Can (1-c/(d+exp(-z/e))) be made negative through a transformation of values? My checks suggest that there is no direct negative, but that Yes, you can flip signs leading to a range in the negative of the other range; although it might not be the direct negative, it could potentially be balanced in a least-squared sense by changes to a and b.
Muhammet Dabak
2019-12-27
Walter Roberson
2019-12-27
If the first term cannot be have a negative coefficient then put in bounds on the values.
Muhammet Dabak
2019-12-27
更多回答（2 个）
Muhammet Dabak
2019-12-25
编辑：Walter Roberson
2019-12-26
1 个评论
Walter Roberson
2019-12-26
You are right, fit() cannot be used for 3 or more independent variables. You will need to use a nonlinear least squares such as https://www.mathworks.com/help/optim/ug/lsqcurvefit.html .
Unfortunately with the random data I generated, lsqcurvefit did not do a good job. I am experimenting further.
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