How to match column of a one matrix with another column of second matrix?

Question

AS 2021-11-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1597844-how-to-match-column-of-a-one-matrix-with-another-column-of-second-matrix

评论： AS 2021-12-1

采纳的回答： John D'Errico

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
显示 2更早的评论隐藏 2更早的评论

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?

AS 2021-11-28

yes

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

John D'Errico 2021-11-28

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1597844-how-to-match-column-of-a-one-matrix-with-another-column-of-second-matrix#answer_842289

编辑：John D'Errico 2021-11-28

A = [ 0     0     3
   2     3
   2     0
   0     0];
B = [ 0     3     0
   3     0
   0     1
   0     1];
C = [ 3     0     0
   0     2
   1     2
   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
PAB = 3×3 logical array
   0   0   1
   1   0   0
   0   1   0

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'
ans = 4×3
     0     0     3
     0     2     3
     1     2     0
     1     0     0

If you feel you really need to get the permutation vector itself, I could do this:

[~,pab] = max(PAB,[],2)
pab = 3×1
     3
     1
     2
B(:,pab)
ans = 4×3
     0     0     3
     0     2     3
     1     2     0
     1     0     0

Similarly,

PAC =  (normalize(A,1,'norm',2)'*normalize(C,1,'norm',2)) > 1-10*eps
PAC = 3×3 logical array
   0   1   0
   0   0   1
   1   0   0
C*PAC'
ans = 4×3
     0     0     3
     0     2     3
     1     2     0
     1     0     0

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
PAD = 3×3 logical array
   0   0   0
   0   1   0
   0   0   0

No such permutation of the columns exists here. A simple test of that is:

sum(PAD,1)
ans = 1×3
     0     1     0
sum(PAD,2)
ans = 3×1
     0
     1
     0

Both such tests should result in vectors of purely ones if PAD were a permutation matrix.

3 个评论
显示 1更早的评论隐藏 1更早的评论

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?

AS 2021-12-1