Asked by John
on 26 May 2013

I have an xy coordinates positions (100x2) and another z vector (100x1) with values corresponding to each xy coordinate. How can I make a matrix of the coordinates with the position of each coordinate having the corresponding z value? Thanks!

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Answer by Andrei Bobrov
on 27 May 2013

Edited by Andrei Bobrov
on 27 May 2013

Accepted answer

after John's comment in Image Analyst's answer:

out = accumarray([x(:),y(:)],z(:),[10 10]);

or

out = zeros(10); out(sub2ind(size(out),x,y)) = z;

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Andrei Bobrov
on 27 May 2013

Hi John! NOTE: Second part my answer (code after 'or') is the same as the Image Analyst' answer...

John
on 27 May 2013

Now I get it, thanks for the clarification.

cecilia dip
on 28 Nov 2016

Hi, I have to do the same thing, and i've tried this, but my coordinates (x,y) are negative and non-integer numbers, as they are latitude,longitude.. how can i do this? I want a plot where for each(lat,long) i can have my Z value (in a color scale, as i will compare it with interpolation methos later). Thank you!

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Answer by Image Analyst
on 26 May 2013

Try this:

% Setup / initialization. % Start out matrix as zeros. m = zeros(20,10); % Generate 100 random coordinates. xy = int32(randi(10, 100, 2)); % Get matrix values for those x,y locations z = randi(255, 100, 1); % 100 values.

% Now, do what the poster, John, wants to do. % Assign the z values to the (x,y) coordinates at the corresponding row. % E.g. m at (x(1), y(1)) will have a value of z(1). % m at (x(2), y(2)) will have a value of z(2). And so on. indexes = sub2ind([20, 10], xy(:,1), xy(:,2)) m(indexes) = z

John
on 27 May 2013

Thanks for your help, unfortunately it didn't work. Here's my data. x and y are the coordinates and z is the corresponding value for each coordinate.

x [4 7 5 9 3 5 5 2 1 1 ]

y [9 2 9 7 9 9 1 4 7 6 ]

z [1 0 1 1 1 0 1 0 1 0 ]

I wish to create a 10x10 matrix (since my xy coordinates are between 1 and 10). In this matrix, I want each coordinate, say for instance, the first xy coordinate (4,9) whose z value is 1 is shown as 1 in column 4 row 9 of the matrix.

Image Analyst
on 27 May 2013

Of course it *DOES work* if you adapt it to a 10 by 10.

m=zeros(10); indexes = sub2ind([10, 10], x, y) m(indexes) = z

When I did my example, I picked random numbers for x, y, and z, and random sizes. You were supposed to know that and be able to make the simple adaptations yourself. But anyway, andrei did it for you so you're all set now.

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