Extract only diagonal elements from matrix

I have a matrix in one variable and a list of coordinates in another variable. Is there a way to extract only the matching pairs of coordinate from the matrix? I.e. X(1),Y(1); X(2), Y(2)...

I can extract all of the permutations (X1, Y1; X1 Y2 ...X2,Y1 ... etc) and then take the diagonal, but I was wondering if there was a simple solution I'm missing to only extract the matched pairs.

Thanks,

Will

%Data
mat = rand(100);

%Coordinates
x_coord = round(rand(10,1)*100);
y_coord = round(rand(10,1)*100);

%Extract coordinates
extracted_coord = diag(mat(x_coord,y_coord));

Answer 1

[Image Analyst]

You can do this, where m is your matrix, and v is the vector of extracted values from the x,y locations

v = zeros(1, length(X));
for k = 1 : length(X)
    row = Y(k);
    col = X(k);
    v(k) = m(row, col);
end

Then just repeat with a scrambled x and y - I'm sure there is some function for that. Put the various v into a matrix and use eye() to extract the diagonal.

[Will Bassett]

I was hoping to find something without a for loop for speed, but this works well and gets the job done. Thanks!

It feels like the kind of thing that has a one line function for it, but perhaps not.

[Image Analyst]

How long was it taking?

I did a million points and it took about 78 milliseconds. If I use 1000 points, it's 0.0003 seconds. Just how many points are you going to use, and what kind of speed do you require?

mat = rand(100);
[rows, columns] = size(mat)
numPoints = 1000000
X = ceil(rand(numPoints,1)*columns);
Y = ceil(rand(numPoints,1)*rows);
tic
v = zeros(1, length(X));
for k = 1 : length(X)
    row = Y(k);
    col = X(k);
    v(k) = mat(row, col);
end
toc

Elapsed time is 0.078826 seconds.

[Will Bassett]

It was taking several seconds for me, but I was using a 2k x 2k matrix and the time consuming thing for that (in my testing) was generating the matrix. Pulling the actual points took about the same as yours. I think the speed is just fine this way.

This is much faster than making the full matrix and diagonalizing as I was doing before.

Thanks

Answer 2

[Florian]

For a one line solution you can use linear indices via sub2ind:

v = mat(sub2ind(size(mat), 1:length(mat), 1:length(mat));

However i don't know about performance in comparison to the for-loop.

Answer 3

[max]

For a square matrix A, if coordinates are taken from vectors say, xc -> row index and yc -> column index, then all the required elements can be accessed by the single command A((yc-1)*size(A,1)+xc)

Example:

n = 4;

A = round(rand(n)*10)

xc = mod(ceil(rand(1,n)*10),n)+1

yc = mod(ceil(rand(1,n)*10),n)+1

A((yc-1)*size(A,1)+xc)

[max]

It works for non square matrix also...