Convert portion of matrix under the diagonal to column vector

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Isabel Nichoson
Isabel Nichoson 2022-3-16
评论： Isabel Nichoson 2022-3-17
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!
  2 个评论
Walter Roberson
Walter Roberson 2022-3-16
Perhaps you could make use of squareform() ? However it indexes down instead of across, so it is not directly suitable for your purpose.

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采纳的回答

Stephen23
Stephen23 2022-3-16
编辑：Stephen23 2022-3-16
Ran in:
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

更多回答（4 个）

Davide Masiello
Davide Masiello 2022-3-16
Ran in:
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
Fangjun Jiang
Fangjun Jiang 2022-3-16
编辑：Fangjun Jiang 2022-3-16
Golfing... watch for any "holes" under the diagnal line
nonzeros(tril(A,-1)')
Jan
Jan 2022-3-16
编辑：Jan 2022-3-16
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)))
Image Analyst
Image Analyst 2022-3-16
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.

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