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Curve fitting with multiple variables?

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Liril Silvi
Liril Silvi 2024-4-22
评论： Sam Chak 2024-4-23
Ran in:
I have three different curves for S=2, S=1 and S=0.5 as shown in Figure below. The Book1.xlsx is attached.
I want to use curve fitting to find out one common equation such as, HF = f(S,T).
I used the following comman line to get the equation which I got from previous post (link). But I am getting following error:
Error using nlinfit (line 213)
Error evaluating model function '@(b,x)exp(b(1)).*x(:,1).^b(2)+b(3).*x(:,2)
What should I do? Is there any other better method as well?
T1 = readtable('Book1.xlsx', 'VariableNamingRule','preserve')
T1 = 10x4 table
T S2 S1 S0.5 __ __________ __________ __________ 10 2.2639e+05 1.63e+05 44925 12 3.2139e+05 2.3122e+05 1.3253e+05 14 4.0539e+05 2.9211e+05 2.0969e+05 16 4.7838e+05 3.4568e+05 2.764e+05 18 5.4038e+05 3.9192e+05 3.3265e+05 20 5.9137e+05 4.3084e+05 3.7846e+05 22 6.3136e+05 4.6243e+05 4.1383e+05 24 6.6034e+05 4.867e+05 4.3874e+05 26 6.7833e+05 5.0363e+05 4.532e+05 28 6.8531e+05 5.1325e+05 4.5722e+05
VN = T1.Properties.VariableNames
VN = 1x4 cell array
{'T'} {'S2'} {'S1'} {'S0.5'}
Viv = regexp([VN{:}], '(\d*\.\d*)|\d*', 'match')
Viv = 1x3 cell array
{'2'} {'1'} {'0.5'}
Viv = str2double(Viv)
Viv = 1x3
2.0000 1.0000 0.5000
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Viym = T1{:,2:4};
Viyv = Viym(:); % 'Vi' Vector
Vivm = repmat(Viv, size(T1,1), 1);
Vivv = Vivm(:); % 'Viv' Vector
y = Viyv;
SoCv = repmat(T1.T, size(Viym,2), 1) % 'SoC' Vector
Error using . (line 229)
Unrecognized table variable name 'deltaT'.
x = [SoCv Vivv]; % Concatenate Vectors
objfcn = @(b,x) exp(b(1)).*x(:,1).^b(2) + b(3).*x(:,2);
mdl = fitnlm(x, y, objfcn, rand(3,1))
  3 个评论
显示 1更早的评论
Torsten
Torsten 2024-4-22
编辑：Torsten 2024-4-22
@Liril Silvi
x must be the matrix
x = [10 0.5;12 0.5;...;28 0.5;10 1;12 1;...;28 1;10 2;12 2;...;28 2]
and y must be the matrix
y = [S0.5;S1;S2]
in your code.
And why do you use exp(b(1)) in your model function ? Why not simply b(1) ?
Liril Silvi
Liril Silvi 2024-4-23
编辑：Torsten 2024-4-23
@Matt J
deltaT should be T (I edited this one in the original question). Now, you will get the error as I said.

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采纳的回答

Sam Chak
Sam Chak 2024-4-23
Ran in:
Hi @Liril Silvi,
See if the polynomial surface function is acceptable
tb = readtable('Book2.xlsx', 'VariableNamingRule','preserve');
S = tb{1:30,2};
T = tb{1:30,1};
HF = tb{1:30,3};
[sf, gof] = fit([S, T], HF, 'poly24')
sf =
Linear model Poly24: sf(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p21*x^2*y + p12*x*y^2 + p03*y^3 + p22*x^2*y^2 + p13*x*y^3 + p04*y^4 Coefficients (with 95% confidence bounds): p00 = -1.028e+06 (-1.028e+06, -1.028e+06) p10 = 1.173e+06 (1.173e+06, 1.173e+06) p01 = 1.108e+05 (1.108e+05, 1.108e+05) p20 = -4.329e+05 (-4.329e+05, -4.329e+05) p11 = -9.663e+04 (-9.663e+04, -9.662e+04) p02 = -2110 (-2110, -2110) p21 = 4.004e+04 (4.004e+04, 4.004e+04) p12 = 2021 (2021, 2021) p03 = -5.616e-12 (-5.367e-11, 4.243e-11) p22 = -826.9 (-826.9, -826.9) p13 = -1.061e-12 (-5.705e-12, 3.584e-12) p04 = 8.182e-14 (-5.452e-13, 7.089e-13)
gof = struct with fields:
sse: 2.0286e-17 rsquare: 1 dfe: 18 adjrsquare: 1 rmse: 1.0616e-09
plot(sf, [S, T], HF), xlabel('S'), ylabel('T'), zlabel('HF')
  2 个评论
Liril Silvi
Liril Silvi 2024-4-23
This polynomial equation has excellent fit. But, I was more into a simple and concise equation rather than a long one with lots of coefficients. Something like as you said about paraboloid function.
Sam Chak
Sam Chak 2024-4-23
Ran in:
Hi @Liril Silvi
The Hyperbolic Paraboloid has this polynomial form when expanded. Thus, you can select 'poly22'.
syms a [1 4]
syms x y
%% Hyperbolic Paraboloid
hp = a1*(x - a3)^2 + a2*(y - a4)^2;
expand(p)
ans = 
tb = readtable('Book2.xlsx', 'VariableNamingRule','preserve');
S = tb{1:30,2};
T = tb{1:30,1};
HF = tb{1:30,3};
[sf, gof] = fit([S, T], HF, 'poly22')
sf =
Linear model Poly22: sf(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 Coefficients (with 95% confidence bounds): p00 = -5.246e+05 (-6.033e+05, -4.459e+05) p10 = 8.896e+04 (2.158e+04, 1.563e+05) p01 = 6.545e+04 (5.808e+04, 7.282e+04) p20 = 2105 (-2.134e+04, 2.555e+04) p11 = 2343 (827.7, 3858) p02 = -1199 (-1386, -1012)
gof = struct with fields:
sse: 4.9789e+09 rsquare: 0.9936 dfe: 24 adjrsquare: 0.9922 rmse: 1.4403e+04
plot(sf, [S, T], HF), xlabel('S'), ylabel('T'), zlabel('HF')

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