Fitting a data with the best fit

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TTA
TTA 2022-10-20
编辑： Alex Sha 2022-10-22
Ran in:
Hi guys,
Please, I need a best fitting method for the data below. It is also attached in a text file.
I have tried tried few fitting methods but they are not giving good result.
M = [-40 242.118600000000
-39 242.717400000000
-38 239.597400000000
-37 236.149700000000
-36 241.291400000000
-35 243.789200000000
-34 244.742700000000
-33 237.290600000000
-32 239.856600000000
-31 236.112800000000
-30 233.345900000000
-29 235.154400000000
-28 233.210300000000
-27 235.488400000000
-26 230.774800000000
-25 237.738900000000
-24 237.066600000000
-23 242.997300000000
-22 241.418500000000
-21 243.022000000000
-20 251.080400000000
-19 255.720000000000
-18 257.327500000000
-17 255.298700000000
-16 257.262300000000
-15 259.756300000000
-14 260.172100000000
-13 261.738600000000
-12 262.423100000000
-11 269.295300000000
-10 266.707900000000
-9 263.961600000000
-8 263.291300000000
-7 267.552400000000
-6 266.540300000000
-5 272.394900000000
-4 266.560100000000
-3 265.945800000000
-2 269.381300000000
-1 269.702600000000
0 276.367300000000
1 277.915600000000
2 274.844200000000
3 274.439700000000
4 274.654000000000
5 277.024300000000
6 277.782500000000
7 278.889500000000
8 280.760300000000
9 281.655000000000
10 279.313700000000
11 278.863900000000
12 277.457200000000
13 275.902800000000
14 274.092300000000
15 273.431400000000
16 263.025100000000
17 255.005200000000
18 254.097200000000
19 255.055100000000
20 243.951200000000
21 236.658300000000
22 244.101800000000
23 240.492600000000
24 245.866900000000
25 239.299800000000
26 233.284500000000
27 234.925100000000
28 246.933500000000
29 231.857000000000
30 238.856600000000
31 234.920200000000
32 232.561800000000
33 242.456800000000
34 237.296900000000
35 235.602100000000
36 246.238300000000
37 240.722800000000
38 256.414100000000
39 243.881000000000];
plot(M(:,1),M(:,2))
Thanks
  2 个评论
Matt J
Matt J 2022-10-20
What curve model would you be trying to fit to?
dpb
dpb 2022-10-20
"Fit" for what purpose? There's nothing going to model that closely. What detail are you willing to sacrifice for a model of some sort?
One of the smoothing or interpolating splines or somesuch is about all I can imagine doing with the above.

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采纳的回答

David Hill
David Hill 2022-10-20
Ran in:
M = [-40 242.118600000000
-39 242.717400000000
-38 239.597400000000
-37 236.149700000000
-36 241.291400000000
-35 243.789200000000
-34 244.742700000000
-33 237.290600000000
-32 239.856600000000
-31 236.112800000000
-30 233.345900000000
-29 235.154400000000
-28 233.210300000000
-27 235.488400000000
-26 230.774800000000
-25 237.738900000000
-24 237.066600000000
-23 242.997300000000
-22 241.418500000000
-21 243.022000000000
-20 251.080400000000
-19 255.720000000000
-18 257.327500000000
-17 255.298700000000
-16 257.262300000000
-15 259.756300000000
-14 260.172100000000
-13 261.738600000000
-12 262.423100000000
-11 269.295300000000
-10 266.707900000000
-9 263.961600000000
-8 263.291300000000
-7 267.552400000000
-6 266.540300000000
-5 272.394900000000
-4 266.560100000000
-3 265.945800000000
-2 269.381300000000
-1 269.702600000000
0 276.367300000000
1 277.915600000000
2 274.844200000000
3 274.439700000000
4 274.654000000000
5 277.024300000000
6 277.782500000000
7 278.889500000000
8 280.760300000000
9 281.655000000000
10 279.313700000000
11 278.863900000000
12 277.457200000000
13 275.902800000000
14 274.092300000000
15 273.431400000000
16 263.025100000000
17 255.005200000000
18 254.097200000000
19 255.055100000000
20 243.951200000000
21 236.658300000000
22 244.101800000000
23 240.492600000000
24 245.866900000000
25 239.299800000000
26 233.284500000000
27 234.925100000000
28 246.933500000000
29 231.857000000000
30 238.856600000000
31 234.920200000000
32 232.561800000000
33 242.456800000000
34 237.296900000000
35 235.602100000000
36 246.238300000000
37 240.722800000000
38 256.414100000000
39 243.881000000000];
p=polyfit(M(:,1),M(:,2),5);%change to desired degree
plot(M(:,1),polyval(p,M(:,1)))
hold on;
plot(M(:,1),M(:,2))

更多回答（2 个）

Torsten
Torsten 2022-10-20
编辑：Torsten 2022-10-20
Try "fit" with fitType
gaussEqn = 'a*exp(-((x-b)/c)^2)+d'
https://de.mathworks.com/help/curvefit/fit.html#d124e31956
John D'Errico
John D'Errico 2022-10-20
As others have said, there is no magical way to know what is the "best" fitting model for such a problem. If you lack any model based on physical principles, then there is little magic you can do.
These are typically problems where a spine model of some sort may be best, since a spline makes very little in the way of assumptions, except that the curve be relatively smooth.
For example, using my SLM toolbox, there is relatively no information to provide, beyond a rough degree of fit in terms of how many knots to use, so I would do something like this:
slm = slmengine(x,y,'knots',7,'plot','on')
In the end, the result seems reasonable, but it depends on how much you want to chase what may well be noise.
  1 个评论
Alex Sha
Alex Sha 2022-10-22
编辑：Alex Sha 2022-10-22
If want to have an explicit function expression，refer to one below:
y = p4*(1/(1+exp(p1*x+p8)))+p5*(1/(1+exp(p2*x+p9)))+p6*(1/(1+exp(p3*x+p10)))+p11+p7*x;
Sum Squared Error (SSE): 931.576981042263
Root of Mean Square Error (RMSE): 3.41243494634378
Correlation Coef. (R): 0.976919416755453
R-Square: 0.954371546833815
Parameter Best Estimate Std. Deviation Confidence Bounds[95%]
--------- ------------- -------------- --------------------------------
p1 2.17071637227502 2.00375241483378 [-1.82666032070769, 6.16809306525773]
p2 -0.866846695995501 0.502259597641453 [-1.86882717750392, 0.135133785512921]
p3 0.357015983521599 0.0432728726246512 [0.270688964680532, 0.443343002362667]
p4 -13.1100496826724 2.55813476865172 [-18.2133889106205, -8.00671045472424]
p5 -16.4803026879968 2.88984006155108 [-22.2453758691813, -10.7152295068122]
p6 64.8893226472405 4.37934503986771 [56.1527683387838, 73.6258769556972]
p7 0.948998342347286 0.0993732061413023 [0.750754220371189, 1.14724246432338]
p8 44.370594281755 40.924919037562 [-37.2723853158642, 126.013573879374]
p9 -27.8000726945408 16.0846620989817 [-59.8880956024535, 4.2879502133719]
p10 -6.25791814028238 0.774805804682342 [-7.80361342793189, -4.71222285263287]
p11 225.224588689396 3.53354087170702 [218.17536752829, 232.273809850502]

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