Curve Fitting and Distribution Fitting

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Mj
Mj 2020-9-11
评论： Mj 2020-9-16
Hello. I have three sets of data and an equation based on it. How can I fit a curve on it and get the coefficients(c1,c2,c3,c4,c5,c6,c7)?and estimate root mean square errors (RMSE) and correlation coefficient (R2)
ye = [0.166393443
0.206557377
0.25204918
0.280737705
0.306967213
0.323770492
0.352868852
0.378278689
0.412704918
0.432786885
0.44057377
0.456147541
0.475819672];
Q = [0.040755669
0.067112106
0.10495312
0.132849355
0.160791371
0.179781479
0.214383688
0.246026104
0.290251909
0.316344405
0.326446015
0.346526208
0.371457751];
z=[0
0
0
0
0
0
0
0
0
0
0
0
0];
1.6<=c7<=3
Q= ((c1-(c2*(ye+z)))*(asin(ye)^c3)+c4*((ye+z)^c5)*ye+(z^c6))^c7;

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采纳的回答

Star Strider
Star Strider 2020-9-11
Try this:
yez = [ye, z];
Qf = @(c1,c2,c3,c4,c5,c6,c7,yez) ((c1-(c2.*(yez(:,1)+yez(:,2)))).*(asin(yez(:,1)).^c3)+c4.*((yez(:,1)+yez(:,2)).^c5).*yez(:,1)+(yez(:,2).^c6)).^c7;
Qfcn = @(cv,yez) Qf(cv(1),cv(2),cv(3),cv(4),cv(5),cv(6),cv(7),yez);
B = lsqcurvefit(Qfcn, [rand(6,1);2], yez, Q, [-Inf(1,6) 1.6], [Inf(1,6) 3])
The estimated parameters do not appear to be unique. This produces quite different results in different runs.
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Mj
Mj 2020-9-11
thanks alot for your response.I used less data to shorten my questions. Otherwise, 51 data from all three parameters are available.
Star Strider
Star Strider 2020-9-11
As always, my pleasure!
No worries! You can always save your data to a .mat file, then upload it here. Then, we have access to all of it. (There are size limitations for uploaded files, however even large data sets are compatible with MATLAB Answers.)

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更多回答（2 个）

John D'Errico
John D'Errico 2020-9-11
You have what, 13 data points? A model with many exponents in it? A model that is surely based on no real physical meaning of the terms in the model?
Expect numerical problems. You will need good starting values.
Worse, z == 0. z is IDENTICALLY zero. Yet you then expect to estimate the power z, z^c6? SIGH.
I can confindently predict that c6 = 17. Oh, wait, I remember, all parameters have the value 42, at least when any possible value will suffice.
The point is, you cannot intelligently estimate these parameters, because there is no unique solution. Any value of c6 will work.
That means your model reduces to
Q = ((c1-c2*ye)*(asin(ye)^c3)+c4*ye^(c5+1))^c7
It is still a mess of parameters put there just to get enough flexibility to fit your data.
So now, lets plot your data, something I should have done before anything else.
plot(ye,Q,'o-')
You are kidding me, right? A simple low order polynomial model is entirely adequate. Is there sufficient information content in those 13 data points to estimate the parameters tou wish to fit? From somewhere in the distance, you hear a deep, heartfelt sigh, even a groan. Is there any physical reason why you think this model is appropriate, that those parameters have any physical meaning in context?
If you prefer, a simple power curve will do quite well.
>> mdl = fittype('a + b*abs(ye-c)^d','indep','ye')
mdl =
General model:
mdl(a,b,c,d,ye) = a + b*abs(ye-c)^d
>> fittedmodel = fit(ye,Q,mdl,'start',[.01,.1,-.1,1.3])
fittedmodel =
General model:
fittedmodel(ye) = a + b*abs(ye-c)^d
Coefficients (with 95% confidence bounds):
a = 0.04029 (0.03285, 0.04772)
b = 1.508 (1.45, 1.566)
c = 0.1636 (0.1464, 0.1808)
d = 1.297 (1.227, 1.366)
>> plot(fittedmodel)
>> hold on
>> plot(ye,Q,'o-')
And that fits quite well, with a far simpler model.
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Mj
Mj 2020-9-11
hello.I used less data to shorten my questions. Otherwise, 51 data from all three parameters are available.
ye=[0.166393443
0.206557377
0.25204918
0.280737705
0.306967213
0.323770492
0.352868852
0.378278689
0.412704918
0.432786885
0.44057377
0.456147541
0.475819672
0.129098361
0.183196721
0.192213115
0.209016393
0.227868852
0.264754098
0.301639344
0.032786885
0.063114754
0.079098361
0.089344262
0.126229508
0.13647541
0.149180328
0.170491803
0.185245902
0.199590164
0.21147541
0.014691521
0.020238585
0.025862799
0.031509604
0.035940668
0.044923322
0.054482866
0.0717356
0.072919889
0.018024148
0.024373383
0.03174389
0.035398208
0.017063413
0.020637735
0.028203846
0.033120028
0.035555281
];
Q = [0.040755669
0.067112106
0.10495312
0.132849355
0.160791371
0.179781479
0.214383688
0.246026104
0.290251909
0.316344405
0.326446015
0.346526208
0.371457751
0.069843006
0.121450028
0.130852824
0.14882695
0.169537878
0.21088739
0.251709087
0.009151561
0.025042029
0.035476157
0.042806716
0.072740217
0.081892097
0.093653673
0.114235514
0.128963324
0.143531782
0.155708172
0.002992409
0.004835202
0.006977514
0.009371635
0.011403503
0.015891643
0.021148234
0.031683247
0.032449372
0.003571359
0.005565609
0.008182482
0.009580949
0.003294144
0.004359568
0.006889221
0.008701893
0.009642407
];
z=[0
0
0
0
0
0
0
0
0
0
0
0
0
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.33
0.33
0.33
0.33
0.33
0.33
0.33
0.33
0.33
0.33
0.33
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.66
0.66
0.66
0.66
0.66
0.66
0.66
0.66
0.66
];
Alex Sha
Alex Sha 2020-9-12
Hi, Hoda, for your all 51 data points, with:
1.6<=c7<=3
Q= ((c1-(c2*(ye+z)))*(asin(ye)^c3)+c4*((ye+z)^c5)*ye+(z^c6))^c7;
there are many local trap solutions. An unique global solution will be:
Root of Mean Square Error (RMSE): 0.00131207824399577
Sum of Squared Residual: 8.43559165999837E-5
Correlation Coef. (R): 0.999921364930492
R-Square: 0.999842736044457
Parameter Best Estimate
---------- -------------
c1 -0.128678458232134
c2 0.849061931670286
c3 -0.0162023653012152
c4 2.15335768995412
c5 -0.0699568937237459
c6 0.739734903785799
c7 1.6

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Mj
Mj 2020-9-12
thanks alot. Can you please send me the code you wrote?
  2 个评论
Alex Sha
Alex Sha 2020-9-13
Hi, Hoda, your problem looks like a general curve fitting issue with one constrained condition, however, it seems to be a prety hard job for existing fitting function in Matlab, such as lsqcurvefit, fmincon, cftool, etc., those functions are all influenced heavy by the guessing initial start-values. Although there is a Global Optimization toolbox in Matlab, its performance is far away as expected. The results shown you above are obtained by 1stOpt, one of software package with global optimization function, easy for using without required to guess initial start values, the code are as below:
Parameter c(6),1.6<=c7<=3;
Variable ye,z,q;
Function Q= ((c1-(c2*(ye+z)))*(arcsin(ye)^c3)+c4*((ye+z)^c5)*ye+(z^c6))^c7;
Data;
ye=[0.166393443,0.206557377,0.25204918,0.280737705,0.306967213,0.323770492,0.352868852,0.378278689,0.412704918,0.432786885,0.44057377,0.456147541,0.475819672,0.129098361,0.183196721,0.192213115,0.209016393,0.227868852,0.264754098,0.301639344,0.032786885,0.063114754,0.079098361,0.089344262,0.126229508,0.13647541,0.149180328,0.170491803,0.185245902,0.199590164,0.21147541,0.014691521,0.020238585,0.025862799,0.031509604,0.035940668,0.044923322,0.054482866,0.0717356,0.072919889,0.018024148,0.024373383,0.03174389,0.035398208,0.017063413,0.020637735,0.028203846,0.033120028,0.035555281];
z=[0,0,0,0,0,0,0,0,0,0,0,0,0,0.25,0.25,0.25,0.25,0.25,0.25,0.25,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.66,0.66,0.66,0.66,0.66,0.66,0.66,0.66,0.66];
q=[0.040755669,0.067112106,0.10495312,0.132849355,0.160791371,0.179781479,0.214383688,0.246026104,0.290251909,0.316344405,0.326446015,0.346526208,0.371457751,0.069843006,0.121450028,0.130852824,0.14882695,0.169537878,0.21088739,0.251709087,0.009151561,0.025042029,0.035476157,0.042806716,0.072740217,0.081892097,0.093653673,0.114235514,0.128963324,0.143531782,0.155708172,0.002992409,0.004835202,0.006977514,0.009371635,0.011403503,0.015891643,0.021148234,0.031683247,0.032449372,0.003571359,0.005565609,0.008182482,0.009580949,0.003294144,0.004359568,0.006889221,0.008701893,0.009642407];
Hope there will be some similar function in Matlab in the future.
Mj
Mj 2020-9-16
hi, Alex Sha, thank you for your responce.

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