Finding the optimal solution for a data with two variables

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AAAAAA 2023-6-11

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https://ww2.mathworks.cn/matlabcentral/answers/1981369-finding-the-optimal-solution-for-a-data-with-two-variables

评论： Alan Stevens 2023-6-12

I have a data set for two variables (two columns) in which each variable has thousands of solutions (rows). I also have the actual solution.

I want to find the best solution (row) with the lowest error considering both variables (not only one variable).

As a simple example containing only 10 solutions (rows), I have the following:

% actual solution:
X1_actual = 0.4722;
X2_actual = 4.4;
% predicted data:[X1_predicted  X2_predicted]
Predicted_data = [0.4742    4.4557
                  0.4739    4.4553
                  0.4732    4.4549
                  0.4730    4.4545
                  0.4725    4.4540
                  0.4723    4.4536
                  0.4715    4.4532
                  0.4714    4.4528
                  0.4713    4.4505
                  0.4701    4.4501]

Where the first and second columns represent the predicted values of X1 and X2, respectively.

The problem is that the minimum errors for both variables are not at the same row. As can be seen in this example, the minimum error for X1 is in the 6th row while for X2 it is in the last row.

Is there a scientific method to find the optimal row that considers the minimum errors of both variables?

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AAAAAA 2023-6-11

OK. Thanks What are the other methods so that I search for these methods and compare the results with this method?

Torsten 2023-6-11

编辑：Torsten 2023-6-11

As I wrote: all norms on IR^2 can be used:

https://uk.mathworks.com/help/matlab/ref/norm.html

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Answer 1

Alan Stevens 2023-6-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1981369-finding-the-optimal-solution-for-a-data-with-two-variables#answer_1253764

在 MATLAB Online 中打开

Here's a possible approach (I've used relative errors as the values of the two columns are an order of magnitude different):

% actual solution:
X1_actual = 0.4722;
X2_actual = 4.4;
% predicted data:[X1_predicted  X2_predicted]
Predicted_data = [0.4742    4.4557
                  0.4739    4.4553
                  0.4732    4.4549
                  0.4730    4.4545
                  0.4725    4.4540
                  0.4723    4.4536
                  0.4715    4.4532
                  0.4714    4.4528
                  0.4713    4.4505
                  0.4701    4.4501];
  Relative_error = [Predicted_data(:,1)/X1_actual, ...
                    Predicted_data(:,2)/X2_actual] ...
                   - ones(10,2);
             
  combined_error = sqrt(Relative_error(:,1).^2 + Relative_error(:,2).^2);
  
  minerror = min(combined_error) 
minerror = 0.0116
  minerror_row = find(combined_error == minerror)
minerror_row = 9

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AAAAAA 2023-6-12

Great This worked very well. But is the scaling you did very compulsory? How if we use the data as it is without doing any scaling? Will this affect the results?

Alan Stevens 2023-6-12

If the errors of X2 are much larger than those of X1 they will dominate the combined contribution and you might find that the result is the same as if you only used X2!

However, you could try it and see!

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