Convert portion of matrix under the diagonal to column vector

Hi all,

I currently have a matrix with correlational information that is structured this:

A = [ 1 2 3 4 5

2 1 2 3 4

3 4 1 2 5

4 3 2 1 2

5 4 3 2 1]

I want to turn the portion of the matrix UNDER the diagonal into a vector, i.e. B = [2 3 4 4 3 5 4 3 2]

Hoever, so far the only way of extracting the data under the diagonal I've found so far is the tril(A, -1) function which returns

A = [ 0 0 0 0 0

2 0 0 0 0

3 4 0 0 0

4 3 2 0 0

5 4 3 2 0]

I can turn this into a vector from here but it will include all the extra zeros and I don't want those to be part of the final vector. Does anyone have any suggestions as to the best way to go about this? Thank you so much!

[Walter Roberson]

Perhaps you could make use of squareform() ? However it indexes down instead of across, so it is not directly suitable for your purpose.

[Isabel Nichoson]

Thanks everyone! I really appreciate the help!

Answer 1

[Stephen23]

Note that approaches which check if the data are zero (e.g. NONZEROS or ==0) are not robust, because you might have perfectly valid zero-values within that part of the matrix. Here is a more robust approach using indexing:

A = [1,2,3,4,5;2,1,2,3,4;3,4,1,2,5;0,3,2,1,2;5,4,3,2,1] % note the zero!

A = 5×5

1 2 3 4 5 2 1 2 3 4 3 4 1 2 5 0 3 2 1 2 5 4 3 2 1

B = A.';
V = B(tril(true(size(A)),-1).')

V = 10×1

2 3 4 0 3 2 5 4 3 2

[Fangjun Jiang]

Golfing ...

A=A';
B=A(triu(true(size(A)),1))

Answer 2

[Davide Masiello]

A = [ 1 2 3 4 5; 2 1 2 3 4; 3 4 1 2 5; 4 3 2 1 2; 5 4 3 2 1];
A = tril(A,-1)';
A = A(:);
A(A==0) = [];
A = A'

A = 1×10

2 3 4 4 3 2 5 4 3 2

[Jan]

This excludes zeros if they apear under the diagonal.

[Davide Masiello]

That's right, thanks for pointing it out.

Answer 3

[Fangjun Jiang]

Golfing... watch for any "holes" under the diagnal line

nonzeros(tril(A,-1)')

[Jan]

This excludes zeros if they apear under the diagonal.

[Fangjun Jiang]

I did mention watching for the holes

Answer 4

[Jan]

A = [ 1 2 3 4 5; ...
      2 1 2 3 4; ...
      3 4 1 2 5; ...
      0 3 2 1 2; ...   % 0 inserted
      5 4 3 2 1];

s = size(A);
m = cumsum(diag(ones(1, s(1)-1), -1)) == 1;
C = A(m)

% Or:
s = size(A);
C = A((1:s(1)).' > (1:s(2)))

% In modern Matlab without size():
C = A((1:height(A)).' > (1:width(A)))

Answer 5

[Image Analyst]

I think the easiest way is to just make a mask and use that to extract the values:

A = [ 1 2 3 4 5
    2 1 2 3 4
    3 4 1 2 5
    4 3 2 1 2
    5 4 3 2 1]
mask = tril(true(size(A)), -1)
columnVector = A(mask)

Note that this will work regardless if there are zeros in the lower diagonal or not.