Matlab在2个矩阵中找到最相似的行 [英] matlab find most similar rows in 2 matrices

问题描述

我有2个矩阵

A = [66 1 29.2;
     80 0 29.4;
     80 0 29.4;
     79 1 25.6];

B = [66 1 28.2;
     79 0 28.4;
     66 1 27.6;
     80 0 22.4]

我想找到匹配行的索引.

I would like to find the indeces of the matching rows.

indx = [1 1;
        2 4;
        3 2;
        4 3]

idx表示A的row1与B的row1，A的row2与B的row4等匹配. 它应该成对匹配(A的1行与B的仅1行) 对于第2列中的值，应严格匹配.对于第1列和第3列的值，它应该是最匹配的.(即，如果存在一对具有相同值的对，则为好，否则我们应该选择最接近的值).

idx means that row1 of A matches with row1 of B, row2 of A with row4 of B etc. It should be a pairwise matching (1 row of A with only 1 row of B) For the values in column 2 it should be a strict match. For the values of columns 1 and 3 it should be the best match..(i.e. if it exist a pair with the same values good, otherwise we should pick the closest).

你能帮我吗? Tnx

Can you help me? Tnx

有关安德鲁评论的问题的更多见解

A的第3行无法匹配第4 B行，因为第4 B行已经与A的第2行匹配.A的第2行与B的第4行匹配，因为前两个元素80,0匹配，然后存在一个小错误在最后一个元素中(29.4-22.4 = 7). 可以说，正确匹配A和B的第二列比匹配第一列比匹配第三列更重要. 我

row 3 of A cant match row 4 B because row 4 B was already matched with row 2 of A. Row 2 of A matches row 4 of B because the first two elements 80,0 match and then there is a small error in the last element (29.4-22.4=7). We can say that matching properly the 2nd column of A and B is more important than matching the 1st column that is more important than matching the 3rd column. I

推荐答案

解决方案

感谢提供的评论，我设法提出了一个不优雅"但可行的解决方案.

Thanks to the comments provided I managed to come up with a "not elegant" but working solution.

B_rem = B;
weights_error = [2 4 1];
match = zeros(size(A,1),2);

for i = 1 : size(A,1)
    score = zeros(size(B_rem,1),1);
    for j =1 : size(B_rem,1)
        score(j) = sum(abs(A(i,:) - B_rem(j,:)).*weights_error);
    end
    [~,idxmin] = min(score);
    match(i,:) = [i,idxmin];
    B_rem(idxmin,:)=[1000 1000 1000];

end

indx = match;

table_match = zeros(size(A,1),7);
table_match(:,1:3) = A(match(:,1),:);
table_match(:,5:7) = B(match(:,2),:);

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