Precise Prediction Model Development

Question

suraj karki 2022-5-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724790-precise-prediction-model-development

评论： Star Strider 2022-5-22

I have four coefficient for prediction model, x1, x2, y1 and y2. with the help of these four cofficient I have develop quadratic 15 coefficient model.

Model: y = B0+B1*X1+B2*Y1+B3*X2+B4*Y2+B5*X1Y1+B6*X1X2+B7*X1Y2+B8*Y1X2+B9*Y1Y2+B10*X2Y2+B11*X1^2+B12*Y1^2+B13*X2^2+B14*Y2^2

However, I am going to precise this model

Precise model:

y = B0 + B6*X1X2 + B9*Y1Y2 + B11*X1^2 + B12*Y1^2 + B13*X2^2 + B14*Y2^2

My concern is how can I buid a funciton which could response for the Precise model instead of quadratic one.

respone will be appreciated. Thank you

3 个评论
显示 1更早的评论隐藏 1更早的评论

suraj karki 2022-5-22

编辑：suraj karki 2022-5-22

Hello Strider

I have used 'fitlm' function to fit the data and y will be the response of the precise model.

suraj karki 2022-5-22

with four independent variable x1, y1, x2 and y2 defined in tabular form I have used fitlm function with quadratic type.

Which have provided 15 coefficient model but I want to develop a model which is going to include 7 coefficient i.e

y = B0 + B6*X1X2 + B9*Y1Y2 + B11*X1^2 + B12*Y1^2 + B13*X2^2 + B14*Y2^2

coeff{i}= fitlm(coeff_table(i).table, 'quadratic','RobustOpts','on');

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2022-5-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724790-precise-prediction-model-development#answer_969335

在 MATLAB Online 中打开

What are you predicting?

How does ‘y’ fit in with this? Is it a separate vector?

With those concerns met, building the function and estimating ther parameters is straightforward —

xy = [X1(:) X2(:) Y1(:) Y2(:)]
% % b(1) = B0,  b(2) = B6,  b(3) = B9,  b(4) = B11,  b(5) = B12,  b(6) = B13,  b(7) = B14  
% yfcn = (b,xy) = b(1) + b(2).*xy(:,1).*xy(:,2) + b(3).*xy(:,3).*xy(:,4) + b(4).*xy(:,1).^2 + b(5).*xy(:,3).^2 + b(6).*xy(:,2).^2 + b(7).*xy(:,4).^2;     % Regression Function
DM = [yfcn = (b,xy) = [ones(size(xy(:,1))), xy(:,1).*xy(:,2), *xy(:,3).*xy(:,4), *xy(:,1).^2, *xy(:,3).^2, xy(:,2).^2, xy(:,4).^2];                     % Design Matrix
B = DM \ y(:)                                                                                                                                           % Calculate Coefficients

I wrote ‘yfcn’ for record-keeping purposes, although you can use it with nonlilnear parameter estimation problems if you wish. However, since this is a linear problem, the ‘DM’ design matrix and the parameter calculation for ‘B’ in the following line will be most efficient.

Be sure to check it for coding errors in ‘DM’ in case I missed something.

.

4 个评论
显示 2更早的评论隐藏 2更早的评论

Star Strider 2022-5-22

在 MATLAB Online 中打开

Actually, something like this —

T1 = array2table(randn(10,5),'VariableNames',{'y','X1','X2','Y1','Y2'})     % Create Data
T1 = 10×5 table
        y           X1           X2          Y1          Y2    
    _________    _________    ________    ________    _________

      0.49347     -0.39335      1.6959     -1.2509     -0.56473
      0.22331      0.31047      1.1232     0.47735      0.40803
      -0.6874    -0.069597      1.6491     -1.0242       1.0545
       -1.399      -2.1349      1.5589    -0.75855    -0.029821
      0.17651      -1.6427    -0.36282      0.1544       1.3891
    -0.045791     0.097463     -0.1876    -0.68584      0.25768
     -0.25065     -0.47585     0.13079    -0.59787     -0.73703
     -0.79733     -0.14262     0.54596    -0.46641      0.69454
      -1.3882      0.70808     0.82369    -0.19268     -0.16341
       1.3585       1.6211      1.2087      3.1796      -1.2615
y = T1.y;
X1 = T1.X1;
X2 = T1.X2;
Y1 = T1.Y1;
Y2 = T1.Y2;
% DM = [ones(size(X1)), X1.*X2, Y1.*Y2, X1.^2, Y1.^2, X2.^2, Y2.^2];
DM = [X1.*X2, Y1.*Y2, X1.^2, Y1.^2, X2.^2, Y2.^2];
mdl = fitlm(DM,y,'linear')
mdl = 
Linear regression model:
    y ~ 1 + x1 + x2 + x3 + x4 + x5 + x6

Estimated Coefficients:
                   Estimate       SE        tStat      pValue 
                   _________    _______    ________    _______

    (Intercept)     -0.68138    0.39737     -1.7147    0.18491
    x1              0.013709     0.4344    0.031559    0.97681
    x2               0.78314    0.43259      1.8103    0.16793
    x3              -0.17528    0.25733    -0.68114    0.54465
    x4               0.48237    0.21669      2.2261     0.1124
    x5             -0.069877    0.28059    -0.24904    0.81941
    x6               0.53053    0.51561      1.0289    0.37921


Number of observations: 10, Error degrees of freedom: 3
Root Mean Squared Error: 0.652
R-squared: 0.81,  Adjusted R-Squared: 0.429
F-statistic vs. constant model: 2.13, p-value = 0.286
B = mdl.Coefficients.Estimate;
fprintf('\n\tB0  = %10.4f\n\tB6  = %10.4f\n\tB9  = %10.4f\n\tB11 = %10.4f\n\tB12 = %10.4f\n\tB13 = %10.4f\n\tB14 = %10.4f\n',B)
	B0  =    -0.6814
	B6  =     0.0137
	B9  =     0.7831
	B11 =    -0.1753
	B12 =     0.4824
	B13 =    -0.0699
	B14 =     0.5305

The ‘B0’ (intercept) term is implicit in the model, so it does not need to be explicitly stated, as it would be using the mldivide function to do the regression.

This is how I would do it.

.

suraj karki 2022-5-22