Correction of code for Curve fitting

Question

WILLIAM BAYA MWARO 2018-6-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/406598-correction-of-code-for-curve-fitting

编辑： Walter Roberson 2018-6-21

d=...
[1   0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0.000037   0.999963
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0.000074   0.999926
 0   1
 0   1
 0   1
 0   1
 0   1
 0   1
 0.000105   0.999895
 0   1
 0.000074   0.999926
 0   1
 0.000167   0.999833
 0.000011   0.999989
 0.000137   0.999863
 0.000163   0.999837
 0   1
 0.000052   0.999948
 0.000159   0.999841
 0.000295   0.999705
 0.001635   0.998365
 0.012518   0.987482
 0.029448   0.970552
 0.022927   0.977073
 0.014557   0.985443
 0.008347   0.991653
 0.005198   0.994802
 0.003137   0.996863
 0.002566   0.997434
 0.004579   0.995421
 0.009629   0.990371
 0.0566   0.9434
 0.11717   0.88283
 0.136539   0.863461
 0.077873   0.922127
 0.097373   0.902627
 0.213279   0.786721
 0.439796   0.560204
 0.679317   0.320683
 0.79856   0.20144
 0.835889   0.164111
 0.854733   0.145267
 0.870323   0.129677
 0.883056   0.116944
 0.89085   0.10915
 0.896808   0.103192
 0.901832   0.098168
 0.906941   0.093059
 0.912774   0.087226
 0.916352   0.083648
 0.923203   0.076797
 0.926324   0.073676
 0.930451   0.069549
 0.933807   0.066193
 0.935963   0.064037
 0.938749   0.061251
 0.941213   0.058787
 0.944072   0.055928
 0.946399   0.053601
 0.948934   0.051066
 0.950979   0.049021
 0.952639   0.047361
 0.954867   0.045133
 0.955925   0.044075
 0.959433   0.040567
 0.95997   0.04003
 0.962362   0.037638
 0.963256   0.036744
 0.965582   0.034418
 0.965949   0.034051
 0.967992   0.032008
 0.969444   0.030556
 0.970288   0.029712
 0.971594   0.028406
 0.972755   0.027245
 0.974647   0.025353
 0.975369   0.024631
 0.977052   0.022948
 0.977606   0.022394
 0.9784   0.0216
 0.979068   0.020932
 0.980051   0.019949
 0.981965   0.018035
 0.981666   0.018334
 0.981889   0.018111
 0.983013   0.016987
 0.983737   0.016263
 0.984486   0.015514
 0.984662   0.015338
 0.985374   0.014626
 0.986094   0.013906
 0.986944   0.013056
 0.988056   0.011944
 0.987725   0.012275
 0.989148   0.010852
 0.988741   0.011259
 0.989182   0.010818
 0.990009   0.009991
 0.990414   0.009586
 0.991178   0.008822
 0.99144   0.00856
 0.992042   0.007958
 0.992104   0.007896
 0.992555   0.007445
 0.992916   0.007084
 0.993574   0.006426
 0.993736   0.006264
 0.994157   0.005843
 0.994434   0.005566
 0.994576   0.005424
 0.994984   0.005016
 0.995202   0.004798
 0.99526   0.00474
 0.995948   0.004052
 0.995785   0.004215
 0.996173   0.003827
 0.996602   0.003398
 0.996508   0.003492
 0.996771   0.003229
 0.997203   0.002797
 0.997377   0.002623
 0.997366   0.002634
 0.997502   0.002498
 0.997718   0.002282
 0.997964   0.002036
 0.997837   0.002163
 0.998121   0.001879
 0.998246   0.001754
 0.99846   0.00154
 0.998757   0.001243
 0.998778   0.001222
 0.998876   0.001124
 0.999046   0.000954
 0.999119   0.000881
 0.999218   0.000782
 0.999086   0.000914
 0.999439   0.000561
 0.999629   0.000371
 0.999496   0.000504
 0.999669   0.000331
 0.999708   0.000292
 1   0
 0.999843   0.000157
 0.999882   0.000118
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
 1   0
];
x=d(:,1);
% y1=d(:,2);
y2=d(:,3);
% plot(x,y,'b*')
constant=lsqcurvefit(@L5P7,[1.0035631;0.000015740085;90;-15;0.15],x,y2);
A=constant(1);
B=constant(2);
C=constant(3);
D=constant(4);
E=constant(5);
xfit=0:1:200;
yfit=L5P7(constant,xfit);
figure
plot(x,y2,'g*')
hold on
plot(xfit,yfit,'b-','linewidth',2)
% grid on
disp(A)
disp(B)
disp(C)
disp(D)
disp(E)

*************************

function

**************************

function cf=L5P7(constant,x)
cf = constant(1)+((constant(2)-constant(1))./((1+((x./constant(3)).^constant(4))).^constant(5)));
end

***********************************************************************************************

I have written the above code to fit a the 5Parameter Logistic equation (L5P). However as can be seen in the graph, the fit is not inline with the data as expected. If someone could adjust or modify my code so as to fit the data accurately and let me have the modified data. Kindly not that the plot is for the x-values (the 1st column and one of the two y-column i.e. column 2 and 3. one is a reverse of the other.

And by the way, I posted a question on this platform about installation of Matlab, student version. Am glad to report that I was able to install it successfully thanks to the advice I got from friends on this platform.

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John D'Errico 2018-6-21

What you don't seem to appreciate is this model simply does not fit your data. You cannot make it do so, by trying slightly other sets of parameters, by letting the optimizer try harder. This is now the second time you have asked the question.

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Answer 1

Walter Roberson 2018-6-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/406598-correction-of-code-for-curve-fitting#answer_325582

编辑：Walter Roberson 2018-6-21

在 MATLAB Online 中打开

If you increase the maximum number of function evaluations and switch to marquardt, you can do slightly better, reaching parameters

0.992753402229052 0.0165694036747047 50.6403017630613 -29.0747052519667 176.250137513041

which has pretty much no visible effect on the display.

There is a catch-basin near that set of parameters: with plotting you can see that there is a favored point quite near those parameters, with residue about 0.128.

However, if you do a global search, you reach parameters close to

0.015026059670688 0.992253891841139 59.2243915203201 100.091982688792 0.208908924049189

which is a catch basin that has slightly better results than the other one, with residue about 0.113.

Notice that all of the parameters are quite different than the other ones, except for C. See the recent discussion at https://www.mathworks.com/matlabcentral/answers/405549-custom-equation-in-curve-fitting#comment_578562 about why that kind of big difference can happen in curve fitting.

To my eye, the curve with these adjusted parameters is steeper and looks like it should be less of a fit on the rights side. However, with those repeated exact zeros at the end of your data, I can accept that plausibly it comes out marginally better on residue.

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Walter Roberson 2018-6-21

在 MATLAB Online 中打开

It turns out that the set of parameters you were getting, the ones with D near -29.07: it turns out that if you let F increase quite a bit, that there are slightly better solutions for that grouping, with C decreasing a little, D decreasing just a small bit, but E increasing greatly. For example near

0.992729299839471779, 0.0165568048796285068, 23.2109699489501367, -28.9055223508293366, 1058962010575.32788

This is consistent with the analysis I linked to for fitting a different curve: that fitting a curve that has a wide range of values can end up being a situation of driving one component to effectively zero contribution (or to non-zero but having to balance a highly non-linear change.)

All of the residues I have found in the extended search with D near -29 are still not quite as good as the residues for the parameter set I posted with D near 100, still over 0.128 (so far) whereas the 100-ish was about residue 0.113 .

The first parameters above are not a bad fit by any means, but you should not have any confidence in some of the coefficients. A and B change only a little, but there is variation in the other parameters and F in particular is pretty much nonsense because it looks plausible that it can increase arbitrarily large.

The second set of parameters, which concentrate the fitting on the other components, seem a little more stable, but I would have to check them more to be sure.

Walter Roberson 2018-6-21

在 MATLAB Online 中打开

x0 = [0.0150260600194573 0.992253891130622 59.2243915462699 100.091986983342 0.208908919228146] .';
opts = optimoptions('lsqcurvefit','Algorithm','levenberg-marquardt','MaxFunctionEvaluations', 5000);
lb = []; ub = [];
constant = lsqcurvefit(@L5P7, x0, x, y2, lb, ub, opts);

Walter Roberson 2018-6-21

编辑：Walter Roberson 2018-6-21

My tests show that near that point, improvements in residue are due to numeric round-off instead of due to theoretical improvements in position. So the theoretical parameters might be near there rather than exactly there, but the above is pretty close to the best you can do numerically... at least with the order that the Symbolic Toolbox happened to do the calculations in when I set it up.

... And what that should tell you is that your model is visibly not the right one for your data.

Note that to produce a continuous curve that goes exactly through all those repeated 0 and 1, you would probably need to go for a polynomial of order 199... but that would fail since it would involve values on the order of 200^199 which is about 10^458.

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