Convert xy Coordinates to Matrix

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John
John el 26 de Mayo de 2013
Comentada: Image Analyst el 24 de Mzo. de 2021
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!

Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 27 de Mayo de 2013
Editada: Andrei Bobrov el 27 de Mayo de 2013
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;
  4 comentarios
John
John el 27 de Mayo de 2013
Now I get it, thanks for the clarification.
cecilia dip
cecilia dip el 28 de Nov. de 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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Más respuestas (2)

Image Analyst
Image Analyst el 26 de Mayo de 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
  6 comentarios
Luigi Izzo
Luigi Izzo el 24 de Mzo. de 2021
what about if xy(k,1) or xy(k,1) are negative numbers?
Thanks
Image Analyst
Image Analyst el 24 de Mzo. de 2021
Try using a scatteredInterpolant. Demo attached.
If you still need help, attach your x, y, and z data in a new question (not here) so people can help you.

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Annisa Dewanti Putri
Annisa Dewanti Putri el 10 de Jul. de 2018
thanks for the help

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