Let Matrices = [A B C D] be a set of square matrices. I want to construct
H = [A B C D;
B A B C;
C B A B;
D C B A]
If all I was interested in was the 4 by 4 case, then this would be enough. However, I want to construct the analogous matrix when Matrices = [A B C D E F] etc. What code could I write to do this?
This can be done as follows:
Concatenate the matrices into a 3D array;
Build the basic with numbers instead of matrices using toeplitz;
Extend this structure so that using linear indexing into the 3D array produces the desired result.
N = 2; % matrix size
M = 3; % number of matrices
matrices = arrayfun(#(x) {randi(9,N)}, 1:M); % input matrices
matrices_3D = cat(3, matrices{:}); % concatenate the matrices along the third dim
ind1 = repmat(reshape(1:N^2, N, N), M, M); % linear indices within each matrix
ind2 = (repelem(toeplitz(1:M), N, N)-1)*N^2; % linear indices to select the matrices
result = matrices_3D(ind1 + ind2); % build result
Example run:
>> celldisp(matrices)
matrices{1} =
1 4
2 6
matrices{2} =
6 4
5 9
matrices{3} =
5 4
6 5
>> result
result =
1 4 6 4 5 4
2 6 5 9 6 5
6 4 1 4 6 4
5 9 2 6 5 9
5 4 6 4 1 4
6 5 5 9 2 6
I have a matrix as follows:
A= 1 2
3 4
5 6
7 8
I want to arrange the elements of this matrix in such a way that it will give me the following output:
B= 1
2
3
4
5
6
7
8
Any kind of suggestion will be helpful. Thanks!
Take the transpose of A and unroll it into a vector:
B = A.';
B = B(:);
Alternatively, you can use reshape:
B = reshape(A.', [], 1);
The reason why you transpose the matrix A first is because MATLAB does the unrolling in column-major format, which means that the columns are traversed first. You are trying to do this row-wise, and so you'd need to transpose the input to achieve the same effect.
Here's what the output looks like (using both):
>> A= [1 2
3 4
5 6
7 8];
>> B = A.';
>> B = B(:);
>> B
B =
1
2
3
4
5
6
7
8
Also:
>> A= [1 2
3 4
5 6
7 8];
>> B = reshape(A.', [], 1)
B =
1
2
3
4
5
6
7
8
suppose that we are creating following matrix from given signal
function [ x ]=create_matrix1(b,l)
n = length(b);
m = n-l+1;
x = zeros(m,l);
for i=1:m
x(i,:)=b(i:i+l-1);
end;
end
with some window length,for example
X=[2;1;3;4;5;7]
X =
2
1
3
4
5
7
>> B=create_matrix1(X,3)
B =
2 1 3
1 3 4
3 4 5
4 5 7
if we have given matrix and windows length ,how can i reconstruct original signal?let say i know that windows length is 3,thanks in advance,i think i should sum elements on anti diagonal and divide by number of elements in this anti diagonal ,but how can i do it by code?thanks in advance
Your original vector is located along the top and right edge of your matrix B and can be reconstructed like so:
>> X_reconstructed = [B(1,1:end-1).'; B(:,end)]
X_reconstructed =
2
1
3
4
5
7
In case the matrix B is some noisy matrix and you actually want to do the averages along the diagonals:
>> BB = fliplr(B);
>> X_mean = arrayfun(#(i) mean(diag(BB,i)), size(B,2)-1:-1:-size(B,1)+1).'
X_mean =
2
1
3
4
5
7
Lets say I have matrix such that A(:,:,1)=[1,2,3;2,3,4], A(:,:,2)=[3,4,5;4,5,6].
How is the easiest way of accessing and plotting the vectors (1,2,3),(2,3,4),(3,4,5),(4,5,6). I tried creating B=[A(:,:,1);A(:,:,2)], but i need a procedure to arbitrary number of A's.
Hope this isn't trivial and I've formulated myself satisfactory.
You should think 'vertically'. This will allow you to use colon indexing:
>> A(:,:,1) = [1,2,3;2,3,4].'; %'// NOTE: transpose of your original
>> A(:,:,2) = [3,4,5;4,5,6].'; %'// NOTE: transpose of your original
>> A(:,:)
ans =
1 2 3 4
2 3 4 5
3 4 5 6
The colon indexing with two colons works for any dimension A:
>> A(:,:,:,:,1,1) = [1 2 3; 2 3 4].'; %'
>> A(:,:,:,:,2,1) = [3 4 5; 4 5 6].'; %'
>> A(:,:,:,:,1,2) = [5 6 7; 6 7 8].'; %'
>> A(:,:,:,:,2,2) = [7 8 9; 8 9 0].'; %'
>> A(:,:)
ans =
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 4 5 6 7 8 9 0
Colon indexing in MATLAB is quite interesting and really powerful once you master it. For example, if you use fewer colons than there are dimensions in the array (like above), MATLAB will automatically concatenate the remainder of the data along the dimension equal to the colon count.
So, if A has 48 dimensions, but you index with just 2 colons: you'll get a 2D array, that is the concatenation of the remaining 46 dimensions along the 2nd dimension.
In general: if A has N dimensions, but you index with just M ≤ N colons: you'll get an M-D array, that is the concatenation of the remaining N-M dimensions along the Mth dimension.
So as long as you are free to define your A to contain vectors on the columns rather than the rows (you should advise everyone to do this, as virtually everything in MATLAB is a bit faster that way), I think this is the fastest and most elegant way to do what you want.
If not, well, then just reshape like Dan :)
Assuming the order does not matter, here is how you can do it for vectors of length 3:
B = reshape(shiftdim(A,2), [], 3)
plot(B')
For vectors of arbitrary dimensions, replace 3 by size(A,2)
For an m-by-m (square) array, how do you concatenate all the rows into a column vector with size m^2 ?
There are a couple of different ways you can collapse your matrix into a vector, depending upon how you want the contents of your matrix to fill that vector. Here are two examples, one using the function reshape (after first transposing the matrix) and one using the colon syntax (:):
>> M = [1 2 3; 4 5 6; 7 8 9]; % Sample matrix
>> vector = reshape(M.', [], 1) % Collect the row contents into a column vector
vector =
1
2
3
4
5
6
7
8
9
>> vector = M(:) % Collect the column contents into a column vector
vector =
1
4
7
2
5
8
3
6
9
A very important note in changing a matrix to a vector is that , MATLAB produce the output vector form the columns of the matrix, if you use A(:)
for example :
A = [1 2 3 ; 4 5 6]
B = A (:)
B = [1 4 2 5 3 6]
You can see the direction of changing in the following image.