How to match column of a one matrix with another column of second matrix?
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I have three matrices like
A =
0 0 3
0 2 3
1 2 0
1 0 0
B =
0 3 0
2 3 0
2 0 1
0 0 1
C =
3 0 0
3 0 2
0 1 2
0 1 0
I want arrange matrix B and C like matrix A means column wise they all should like similar.
Could you please help me?
4 个评论
John D'Errico
2021-11-28
So you are asking someone to show how to rearrange the columns of B and C, to match A? What if there is no exact rearrangement?
采纳的回答
John D'Errico
2021-11-28
编辑:John D'Errico
2021-11-28
A = [ 0 0 3
0 2 3
1 2 0
1 0 0];
B = [ 0 3 0
2 3 0
2 0 1
0 0 1];
C = [ 3 0 0
3 0 2
0 1 2
0 1 0];
If you will accept ONLY an exact rearrangement of the columns, then we might look for a permutation matrix. Does one exist?
PAB = (normalize(A,1,'norm',2)'*normalize(B,1,'norm',2)) > 1-10*eps
If the matrix PAB is a valid permutation matrix, then it will have exactly one unit element in every row and every column. Now we can transform B using the product:
B*PAB'
If you feel you really need to get the permutation vector itself, I could do this:
[~,pab] = max(PAB,[],2)
B(:,pab)
Similarly,
PAC = (normalize(A,1,'norm',2)'*normalize(C,1,'norm',2)) > 1-10*eps
C*PAC'
Now, suppose we have a matrix that fails to have an exact rearrangeent of the columns?
D = [ 3 0 0
0 2 3
1 2 0
1 0 0];
PAD = (normalize(A,1,'norm',2)'*normalize(D,1,'norm',2)) > 1-10*eps
No such permutation of the columns exists here. A simple test of that is:
sum(PAD,1)
sum(PAD,2)
Both such tests should result in vectors of purely ones if PAD were a permutation matrix.
3 个评论
John D'Errico
2021-11-30
But that does not mean there is any valid EXACT rearrangment for what you ask. I don't see your data, since you have not provided the actual data. So I cannot show you why what you are asking to do seems to have failed.
When I did ask you what was intended if an exact solution to the column rearrangement problem does not exist, all you said was "yes". Does "yes" tell me anything of significance?
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