help with matrix concatenation

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James Kamwela 2021-10-8

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编辑： DGM 2021-10-10

Hi Everyone, i am a beginner and i would like to ask if its possible to concatenate more than two matrices,please try and correct my code;

i am trying to swap a matrix then concanate all into one into one big matrice that will contain all swapped matrices so i can access them again.Is that possible and if it is how can i access the matrices later? thank you.

Allmatrix=zeros(length(main),length(main)*length(main));
for i=0:length(main)
    for swap=i:length(main)-1
        Axb=main;
        swaprow=Axb(:,1);
        Axb(:,1)=Axb(:,swap+1);
        Axb(:,swap+1)=swaprow;
    end 
    Allmatrix=Axb(i);
end

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DGM 2021-10-9

编辑：DGM 2021-10-9

在 MATLAB Online 中打开

Well, let's start with that. This gives the permutation described by the example, as best fits the code provided:

A = [11 12 13; 21 22 23; 31 32 33];
s = size(A);
B = zeros([s(1)^2 s(2)]);
B(1:s(1),:) = A;
for r = 2:s(2)
    thisB = A;
    thisB(:,[1 r]) = thisB(:,[r 1]);
    B((r-1)*s(1)+1:r*s(1),:) = thisB;
end
B
B = 9×3
    11    12    13
    21    22    23
    31    32    33
    12    11    13
    22    21    23
    32    31    33
    13    12    11
    23    22    21
    33    32    31

But while that looks fine for 3 columns, it doesn't generalize. Say A is wider (using a vector for compactness):

A = [1 2 3 4];
% this is unchanged
s = size(A);
B = zeros([s(1)^2 s(2)]);
B(1:s(1),:) = A;
for r = 2:s(2)
    thisB = A;
    thisB(:,[1 r]) = thisB(:,[r 1]);
    B((r-1)*s(1)+1:r*s(1),:) = thisB;
end
B
B = 4×4
     1     2     3     4
     2     1     3     4
     3     2     1     4
     4     2     3     1

Which is probably not what you want. There could be other ways to incrementally flip row vectors, but I miss how they would adhere to the examples. What should the incremental permutations look like for a 1x4 or 1x5 vector input?

I would imagine a flip process would look like this:

% initial
1 2 3 4
% first pass
2 1 3 4
3 1 2 4
4 1 2 3
% second pass
4 2 1 3
4 3 1 2
% third pass
4 3 2 1

And the result would be a sample from each pass, say the last sample from each pass

The example seems to use the first sample from each pass instead.

2 3 % this one
1 3 % this one
1 2
2 1 % this one

Is that what's intended?

James Kamwela 2021-10-9

Thanks for investing your time to help, the first example is what i was looking for. However, you say it does not work if A is wider and that is also what i was wondering,if it can work if A was winder lets say 7*7,and if it can then how.

DGM 2021-10-9

编辑：DGM 2021-10-9

在 MATLAB Online 中打开

... That's what I was asking you. You have to define how the process should be generalized to wider arrays. In the last example I gave, there are two implied variations depending on how the loops are structured. For A = [1 2 3 4], you could either get

2 3 4 % case 1 (using last sample)
1 2 3
3 1 2
3 2 1

or you could get

2 3 4 % case 2 (using first sample)
1 3 4
2 1 3
3 2 1

or you could get something different yet depending on what you intended. This is an arbitrary subsampling of a process by which a vector is flipped by an arbitrary number of pairwise flips. It's not up to me to decide what the goals are.

EDIT:

If the first case meets the requirements, it simplifies very neatly:

A = [1 2 3 4 5 6 7]
s = size(A);
B = zeros([s(1)*s(2) s(2)]);
for rb = 1:s(2)
    B((rb-1)*s(1)+1:rb*s(1),:) = A(:,[s(2):-1:(s(2)-rb+2) 1:(s(2)-rb+1)]);
end
B

but this doesn't match the order in your smaller example.

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Answer 1

C B 2021-10-9

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@James Kamwela is this what you expect as answer please check

main =[1 2 ; 3 4];

main =

1 2

3 4

Allmatrix=[];
 for i=1:size(main,2)
    for swap=1:size(main,1)-1
        Axb=main;
        swaprow=Axb(i,swap);
        Axb(i,swap)=Axb(i,swap+1);
        Axb(i,swap+1)=swaprow;
    end 
     Allmatrix=[Allmatrix Axb];
 end

Allmatrix =

2 1 1 2

3 4 4 3

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James Kamwela 2021-10-9

Thanks for your help, however this doesn`t fully answer my question.

For example; we have 3*3 matrix A=[a1 a2 a3;b1 b2 b3;c1 c2 c3];

after swapping and concatenation A should look like

A=[a1 a2 a3;b1 b2 b3;c1 c2 c3;a2 a1 a3;b2 b1 b3;c2 c1 c3;a3 a2 a1;b3 b2 b1;c3 c2 c1];

i hope you can read the matrix and i hope i make sense

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Answer 2

DGM 2021-10-9

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I'm just going to post these two examples as an answer.

If you want to sample the process at the end of each pass, the structure is more simple:

A = [1 2 3 4 5 6 7];
s = size(A);
B = zeros([s(1)*s(2) s(2)]);
for rb = 1:s(2)
    B((rb-1)*s(1)+1:rb*s(1),:) = A(:,[s(2):-1:(s(2)-rb+2) 1:(s(2)-rb+1)]);
end
B
B = 7×7
     1     2     3     4     5     6     7
     7     1     2     3     4     5     6
     7     6     1     2     3     4     5
     7     6     5     1     2     3     4
     7     6     5     4     1     2     3
     7     6     5     4     3     1     2
     7     6     5     4     3     2     1

but this doesn't match the order in your smaller example.

A = [11 12 13; 21 22 23; 31 32 33];
s = size(A);
B = zeros([s(1)^2 s(2)]);
for rb = 1:s(2)
    B((rb-1)*s(1)+1:rb*s(1),:) = A(:,[s(2):-1:(s(2)-rb+2) 1:(s(2)-rb+1)]);
end
B
B = 9×3
    11    12    13
    21    22    23
    31    32    33
    13    11    12
    23    21    22
    33    31    32
    13    12    11
    23    22    21
    33    32    31

If you want to sample the process at the beginning of each pass, the pattern isn't as neat and the code accordingly isn't as simple.

A = [1 2 3 4 5 6 7];
s = size(A);
B = zeros([s(1)*s(2) s(2)]);
B(1:s(1),:) = A;
B(end-s(1)+1:end,:) = fliplr(A);
for rb = 2:s(2)-1
    B((rb-1)*s(1)+1:rb*s(1),:) = A(:,[s(2):-1:(s(2)-rb+3) 2 1 3:(s(2)-rb+2)]);
end
B
B = 7×7
     1     2     3     4     5     6     7
     2     1     3     4     5     6     7
     7     2     1     3     4     5     6
     7     6     2     1     3     4     5
     7     6     5     2     1     3     4
     7     6     5     4     2     1     3
     7     6     5     4     3     2     1

but this does match your example...

A = [11 12 13; 21 22 23; 31 32 33];
s = size(A);
B = zeros([s(1)*s(2) s(2)]);
B(1:s(1),:) = A;
B(end-s(1)+1:end,:) = fliplr(A);
for rb = 2:s(2)-1
    B((rb-1)*s(1)+1:rb*s(1),:) = A(:,[s(2):-1:(s(2)-rb+3) 2 1 3:(s(2)-rb+2)]);
end
B
B = 9×3
    11    12    13
    21    22    23
    31    32    33
    12    11    13
    22    21    23
    32    31    33
    13    12    11
    23    22    21
    33    32    31

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Answer 3

James Kamwela 2021-10-9

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https://ww2.mathworks.cn/matlabcentral/answers/1469966-help-with-matrix-concatenation#answer_805191

so i guess it doesnt continue after the third swap,So now that i cant concatenate more than 3 matrices i would like to ask another question, lets suppose A = [11 12 13; 21 22 23; 31 32 33]; we swap each column with column one however after the swap i want the product of the diagonal stored in B, so that should make B a (1,3) vector,

this is actually what i was struggling with, i am able to swap the columns but B only stores the value of the last swap diagonal product, so what i think happens is that it finishes all the swaps then calculates the diagonal product of the last matrix

i hope i made sense and thank you for your help

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James Kamwela 2021-10-10

No thats not what i asked, i will try to explain again A=[11 12 13;21 22 23;31 32 33]; B=zeros(1,length(n))

step1--->so the first action will be B=A(i,i)-->B1=prod(B)--->the product of diagonal coefficints in this case (11,22,33)

step2--->then it will swap columns-->column 1 will be column 2 and column 2 will be column 1

A will look the same but columns 1 and 2 will have swapped places

calculate B2 again

step 3--->A will default to its original

then it will swap columns-->column 1 will be column 3 and column 3 will be column 1

then it will calculate B3 again

step4---> this process will continue to the length(A)-1(if matrix was larger)

calculate B(n)

step5---> so taking A as reference,after the process finishesat step 3.that will mean that B has three values since A is a 3*3 matrix.my problem is that i am not able to store the value of of B somewhere after every calculation

this is what i want to happen

step1---find B1

step2---find B2

step2---find B3

Barray=(B1 B2 B3);

i think i am clear now,sorry ive taken so long.

DGM 2021-10-10

编辑：DGM 2021-10-10

在 MATLAB Online 中打开

Assuming that A is square, then

% these are copied from the main example so i don't ahve to repaste everything
s = [3 3];
B = [11 12 13;21 22 23;31 32 33;12 11 13;22 21 23;32 31 33;13 12 11;23 22 21;33 32 31]
B = 9×3
    11    12    13
    21    22    23
    31    32    33
    12    11    13
    22    21    23
    32    31    33
    13    12    11
    23    22    21
    33    32    31
Bd = cellfun(@diag,mat2cell(B,repmat(s(1),[s(2) 1]),s(2)),'uniform',false);
Bd = prod(cell2mat(Bd.'),1)
Bd = 1×3
        7986        8316        8866

Those are the products of the diagonals of the 3x3 sub-blocks within B.

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help with matrix concatenation

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