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

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?

[Star Strider]

What do the output matrices look like?

[AS]

B and C will be

0 0 3

0 2 3

1 2 0

1 0 0

[John D'Errico]

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]

yes

Answer 1

[John D'Errico]

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

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.

[AS]

Thank you for your help. but in my data it is not working fully. data length is 3374 * 8. I can only rearrange cloumn 2 to column 6. But Columns 1, 7, 8 is not rearranged.

[John D'Errico]

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]

yes it is solved.thank you.