How to perform linear regression between respective columns of two matrix/table?

Question

Ranit 2023-1-20

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1897660-how-to-perform-linear-regression-between-respective-columns-of-two-matrix-table

评论： Star Strider 2023-1-23

采纳的回答： Star Strider

I have two matix/ table. For example,

Table 1:

Table 2:

I want to fit linear regression between respective columns of the two tables. For example, column 1 in Table 1 are y values. Column 1 in Table 2 are x values. I want to get slope and intercept of linear regression between these two columns. Then the same for column 2, 3 and so on. At the end I should have n different slope and intercept pair (suppose each table has total n columns) collected in a matrix or table.

P.S.:

Edit: I would also really like to know any approach without using a for loop.

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Answer 1

Star Strider 2023-1-20

1
链接

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https://ww2.mathworks.cn/matlabcentral/answers/1897660-how-to-perform-linear-regression-between-respective-columns-of-two-matrix-table#answer_1152845

在 MATLAB Online 中打开

One approach —

A1 = randn(5,10)                                    % Y-Values
A1 = 5×10
    0.0299    1.5214    0.3536   -1.1142   -0.1150   -0.5494    0.2548   -1.4189    1.0701    0.1450
   -0.8383    1.7322   -0.2779   -1.0632    0.8443    0.0851    0.4722   -0.1375   -0.6431    0.4131
   -0.8640   -1.6304    2.2374   -0.8050    1.1452   -0.2773   -0.4974   -1.6779    0.9647    0.7920
   -1.8258   -1.1512    1.4659   -1.1022    0.8539   -0.0612   -0.3418   -0.0167    0.8839   -0.3638
    0.3277   -0.3655    0.2634    0.7859   -2.1518   -1.2548    0.2227   -0.1939   -0.4167   -1.5957
A2 = randn(5,10)                                    % X-Values
A2 = 5×10
    0.3850    0.4773    1.2600    0.0907   -2.9750    0.4598    0.7622   -0.7591   -0.6958    0.8931
    0.6983   -0.6295    0.4358   -1.0046    0.9418    0.5156   -1.1073   -0.2977    1.2468   -1.4143
    0.6650   -0.4260   -0.1401    0.0958   -1.3038   -1.2276   -0.3434   -1.5117    0.3252    1.2938
   -0.0940   -1.1737    0.9630   -1.8431    1.1628   -0.3611    0.3994   -0.5574   -0.6054    1.2328
    1.1454   -1.2291   -1.3081   -0.2680   -0.6106   -0.1132   -0.3333   -0.9338    0.7663    0.4511
Rows = size(A1,1);
for k = 1:size(A1,2)
    B(:,k) = [A2(:,k) ones(Rows,1)] \ A1(:,k);
end
Results = array2table(B.', 'VariableNames',{'Slope','Intercept'}, 'RowNames',compose('Col %4d',1:size(A1,2)))
Results = 10×2 table
                  Slope       Intercept
                __________    _________

    Col    1        1.4123      -1.4249
    Col    2        1.1448      0.70382
    Col    3      0.023978      0.80266
    Col    4       0.27999     -0.49571
    Col    5       0.19567       0.2243
    Col    6    -0.0015634     -0.41174
    Col    7      -0.16755    0.0012506
    Col    8        1.2526      0.32805
    Col    9      -0.85766      0.54969
    Col   10     -0.059937    -0.092407