2nd Order Polynomial Coefficient Solving
显示 更早的评论
I have been having very good success with guidance from the community with using Curve Fitter. I now would like to produce coefficients with an added variable.
I have the Data for R,S,P,T.
T = a + b*P + c*S + d*S^2 + e*(S*P) + f*(S^2*P)
Trying to find coefficients a,b,c,d,e,f.
The coefficients would a median seperated by 'R' breakpoints [550 725 950] , so realistically I would have a 6x3 matrix of coefficients.
At the moment linear regression will work, but the affect of P on T is linear, and the affect of R and S on T is non-linear in reality.
I have a 63000 x 4 matrix for data.
R S P T
716 28.5000000000000 291.727272727300 184.407961051320
721 28.5000000000000 291.625000000000 187.140145995908
721 28.5000000000000 291.625000000000 187.220504376631
722.5 28.5000000000000 291.625000000000 187.140145995908
722.5 28.5000000000000 291.625000000000 187.140145995908
I tried this along with a couple other methods to no avail.
Ignore the column indices, that was another iteration from the above data.
TRQ1 = @(c,VAR) c(1).*VAR(:3) + c(2) + c(3).*VAR(:,2) + c(4).*VAR(:,4) + c(5).*VAR(:,5) + c(6).*VAR(:,6);
F_TRQ= linsolve(VAR(:,[1 3] ) , TRQ, TRQ1)
采纳的回答
I don't see R in your regression equation
T = a + b*P + c*S + d*S^2 + e*(S*P) + f*(s^2*P)
And I assume that "s" means "S".
M = [ones(63000,1),P,S,S.^2,S.*P,S.^2.*P];
y = T;
x = M\y;
a = x(1)
b = x(2)
c = x(3)
d = x(4)
e = x(5)
f = x(6)
11 个评论
Eric
2024-8-29
Torsten
2024-8-29
Thus problem solved ?
If you have to put constraints on the coefficients, use "lsqlin" to solve the overdetermined linear system M*x = y.
E.g.
M = [ones(63000,1),P,S,S.^2,S.*P,S.^2.*P];
y = T;
lb = [...]; % lower bounds on the coefficients
ub = [...]; % upper bounds on the coefficients
x = lsqlin(M,y,[],[],[],[],lb,ub);
a = x(1)
b = x(2)
c = x(3)
d = x(4)
e = x(5)
f = x(6)
Eric
2024-8-29
Eric
2024-8-30
Torsten
2024-8-30
From what I understand, you want something similar to this ?
T1 = T(R<=550);
P1 = P(R<=550);
S1 = S(R<=550);
n1 = numel(T1);
T2 = T(R>550 & R<=725);
P2 = P(R>550 & R<=725);
S2 = S(R>550 & R<=725);
n2 = numel(T2);
T3 = T(R>725);
P3 = P(R>725);
S3 = S(R>725);
n3 = numel(T3);
M = [[ones(n1,1);zeros(n2,1);zeros(n3,1)],[zeros(n1,1);ones(n2,1);zeros(n3,1)],...
[zeros(n1,1);zeros(n2,1);ones(n3,1)],[P1;P2;P3],[S1;S2,S3],[S1.^2;S2.^2;S3.^2],...
[S1.*P1;S2.*P2;S3.*P3],[S1.^2.*P1;S2.^2*P2;S3.^2*P3]];
y = [T1;T2;T3];
x = M\y;
a1 = x(1)
a2 = x(2)
a3 = x(3)
b = x(4)
c = x(5)
d = x(6)
e = x(7)
f = x(8)
Eric
2024-8-31
Walter Roberson
2024-8-31
M = [[ones(n1,1);zeros(n2,1);zeros(n3,1)],[zeros(n1,1);ones(n2,1);zeros(n3,1)],...
[zeros(n1,1);zeros(n2,1);ones(n3,1)],[P1;P2;P3],[S1;S2,S3],[S1.^2;S2.^2;S3.^2],...
[S1.*P1;S2.*P2;S3.*P3],[S1.^2.*P1;S2.^2*P2;S3.^2*P3]];
I think that sould be [S1;S2;S3] instead of [S1;S2,S3]
Torsten
2024-8-31
Thank you.
You get 6 coefficients. The first coefficient depends on R. The regression equation solved above is
T = x(1) + x(4)*P + x(5)*S + x(6)*S^2 + x(7)*(S*P) + x(8)*(S^2*P) if R <= 550
T = x(2) + x(4)*P + x(5)*S + x(6)*S^2 + x(7)*(S*P) + x(8)*(S^2*P) if R > 550 & R <=725
T = x(3) + x(4)*P + x(5)*S + x(6)*S^2 + x(7)*(S*P) + x(8)*(S^2*P) if R > 725
Eric
2024-9-1
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Fit Postprocessing 的更多信息
Translated by 
