surf indices reversed?

33 visualizaciones (últimos 30 días)
Craig
Craig el 29 de Dic. de 2021
Editada: Stephen23 el 30 de Dic. de 2021
Quick question: It seems surf and surfl have indices reversed. Is this correct? For example, if I type
x=0:pi/10:pi
y = 0:pi/10:2*pi
for i = 1:11
for j = 1:21
z(i,j) = sin(x(i))*sin(y(j))
end
end
surfl(x,y,z)
I get the standard: Error using surfl (line 94)
The lengths of X and Y must match the size of Z.
error. If I use z' in place of z it runs. But shouldn't the first indiex be the x and the second the y? This is completely unintuitive to me.
  4 comentarios
Image Analyst
Image Analyst el 30 de Dic. de 2021
Editada: Image Analyst el 30 de Dic. de 2021
Craig, if the axis direction defaults don't do it the way you'd like, you can flip the direction of the y axis with
axis ij
axis xy
Use whichever does it the way you want.
Craig
Craig el 30 de Dic. de 2021
Thanks!

Iniciar sesión para comentar.

Respuesta aceptada

Matt J
Matt J el 29 de Dic. de 2021
Editada: Matt J el 29 de Dic. de 2021
Yes, the x-axis traverses the rows of z and the y-axis traverses the columns.
For some reason, many people prefer their x-axis to be horizontal.
  3 comentarios
Matt J
Matt J el 29 de Dic. de 2021
Editada: Matt J el 30 de Dic. de 2021
As a matrix, x should traverse the rows (the first index). But it's not.
When I say "traverse the rows", I mean you move along a row as x varies.
but anyone familiar with plotting matrix data figures out that you rotate 90 degrees from matrix to plot.
I'm not sure what language requires that you rotate 90 degrees. Regardless, here you have a very similar situation, except that you are required to transpose instead of rotate. The need to do either one is unappealing, IMHO.
The bottom line is, your conjecture is right. The surf() routines view z(i,j) such that y varies with i and x varies with j. The same is true for interp2(), meshgrid(), and nearly all Image Processing Toolbox functions.
Craig
Craig el 29 de Dic. de 2021
Thanks, Matt. And thanks for your first answer, too. My first comment was a little short.
Yes, when I read "traverse the rows" I parsed that as moving from row to another ("traverse a row" I would have read as going down one row, from one column to the next. Don't know if that is right, but what you said makes sense now).
The rotation is not a language, just mathematical conventions. When you access the i-jth entry of a matrix z ( z(i,j) in Matlab), you go down to the ith row and across to the jth column. In a function z(x,y), for z(x_i, y_j) you would go across the x-axis to the ith entry on our grid, then up to the jth entry on the grid. Anyway, the different conventions for plotting and matrix entries cause some confusion, but I understand what Matlab is doing now. Not intuitive to me, but I can use it.

Iniciar sesión para comentar.

Más respuestas (2)

Sean de Wolski
Sean de Wolski el 30 de Dic. de 2021
I think you're seeing the difference between meshgrid and ndgrid. Meshgrid is used for plotting, ndgrid for matrix/tensor work
[rr,cc] = ndgrid(1:3,1:4)
rr = 3×4
1 1 1 1 2 2 2 2 3 3 3 3
cc = 3×4
1 2 3 4 1 2 3 4 1 2 3 4
[xx,yy] = meshgrid(1:3,1:4)
xx = 4×3
1 2 3 1 2 3 1 2 3 1 2 3
yy = 4×3
1 1 1 2 2 2 3 3 3 4 4 4
When I was heavily involved in 3d image processing in grad school I tried to be very very consistent and always use row/col as: the convention, notation, and variable names.
  2 comentarios
Image Analyst
Image Analyst el 30 de Dic. de 2021
Oh, so that's the difference. I never knew. I just always used meshgrid() for everything since I knew that one and didn't want to learn another function. But I can see how ndgrid() could be useful in some situations.
Craig
Craig el 30 de Dic. de 2021
Thanks, Sean!

Iniciar sesión para comentar.


Image Analyst
Image Analyst el 29 de Dic. de 2021
Craig, you do know that matrices in MATLAB are indexed (row, column), right? Apparently not since if I
choose better names for your loop iterators we get this:
x = 0:pi/10:pi
y = 0:pi/10:2*pi
for xIndex = 1:11
for yIndex = 1:21
z(xIndex,yIndex) = sin(x(xIndex))*sin(y(yIndex))
end
end
Why are you indexing z like that? Well since row is y, M(row, column) is M(y, x). Matrices are not indexed like M(x, y). You should have z(yIndex, xIndex). The obvious solution is to just label the axes:
xlabel('X', 'FontSize', 25);
ylabel('y', 'FontSize', 25);
zlabel('Z', 'FontSize', 25);
axis equal
But make sure you do it right. You can also use meshgrid(), which you should. I don't have surfl() so I used surf():
x = 0:pi/10:pi;
y = 0:pi/10:2*pi;
for xIndex = 1: length(x)
for yIndex = 1: length(y)
z(yIndex,xIndex) = sin(x(xIndex))*sin(y(yIndex));
end
end
[X, Y] = meshgrid(x, y);
surf(X, Y, z);
% Label the first argument to surf(), which is the columns or x values.
xlabel('X', 'FontSize', 25);
% Label the second argument to surf(), which is the rows or y values.
ylabel('Y', 'FontSize', 25);
zlabel('Z', 'FontSize', 25);
axis equal
g = gcf;
g.WindowState = 'maximized';
  1 comentario
Craig
Craig el 29 de Dic. de 2021
Editada: Craig el 29 de Dic. de 2021
Thanks, image analyst. I do know that matrices are labeled row, column. The "apparently not" comment sounds to me condescending, but it's easy to misinterpret things online so I won't worry about it.
Read the last response to Matt above. The issue is with the different conventions for matrices compared to plotting. If I give you a point (4,2,3), you x=4, y=2, and z =3 (across 4 units, then up, then out of the page in my mind, or across, back, up). But in matrix z, z(4,2)=3 you go down to the 4th row, the across to the seocnd column, and input 3.

Iniciar sesión para comentar.

Categorías

Más información sobre Surface and Mesh Plots en Help Center y File Exchange.

Etiquetas

Productos


Versión

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by