"Data dimensions must agree" Error

[code]im = im2double(imread('rice.png'));

[X Y]= meshgrid(1:size(im,1),1:size(im,2));

surf(zeros(size(im)),X,Y,im,'EdgeColor','none');[/code]

when i run this script it worked me fine but when i tried to change the image to RGB image it gives me this 2 errors

??? Error using ==> surf at 78 Data dimensions must agree.

Error in ==> CoOrdinating at 6 surf(zeros(size(im)),X,Y,im,'EdgeColor','none');

i tried to convert the image to grayscale but it didn't work with me and gave me the same errors

any help ?

[Jan]

Please post, what exactly "when i tried to change the image to RGB" means. If the problem depends on this details, it is worth to show it.

[Amr]

when i used another image which is colored one

[Jan]

Please post the Matlab code. Bugs in the code cannot be found in a text description of the method.

Answer 1

[Andrew Newell]

Here is a demo that works:

rgb_img = imread('ngc6543a.jpg');
imshow(rgb_img)

I = .2989*rgb_img(:,:,1)...
+.5870*rgb_img(:,:,2)...
+.1140*rgb_img(:,:,3);
figure;
imshow(I)

I = double(I);
[X,Y]= meshgrid(1:size(I,1),1:size(I,2));
figure; 
surf(zeros(size(I')),X,Y,I','EdgeColor','none');
colormap('gray')

Probably you didn't take the transpose of I in the surf command. I don't know why you need to do this, but you do.

[Walter Roberson]

Taking the transpose or not would be the difference between using meshgrid() or using ndgrid().

>> size(meshgrid(1:2,1:3))

ans =

3 2

>> size(ndgrid(1:2,1:3))

ans =

2 3

[Andrew Newell]

Good point. I had a feeling I'd be hearing from you!

Answer 2

[Andrew Newell]

If you enter the command

I = imread('rice.png');

you get a matrix. If you enter

I = imread('ngc6543a.jpg');

you get a 3D array with one page for each of the R, G and B components of the color. This does not have the same dimensions as the matrices X and Y.

[Amr]

i convert it to gray scale and it became matrix like the other X and Y

[Walter Roberson]

Amr, I am not sure whether you are still having difficulty with this matter?

Answer 3

[guj]

am also stuck in same thing I have samples point of 5:5:85 missing samples ..so i have 17 numbers which i have chosen has missing %.

so my vector is of length 17 by 1

Number of iteration = 500;

number of iteration carried out for reconstruction is 500 but 100 seems to be enough...so i have just zeros after 100 row my matrix is 500 by 17 in my error)matrix

I am using [X,Y]=meshgrid(iteration,missing sample) surf(X,Y,errormatrix(1:100,:))

But i am getting error...

whole objective is to show 3d graph having iteration on one axis, missing % of samples on other axis and error on the third one

This is what i am doing

[X,Y]=meshgrid(x,y) x=1 by 17 (decimation factors) y=1 by 100 (number of iteration)

errormatrix=[100 by 17] whch shows error during 100 iteration for each decimation %.

now surf(X',Y,errormatrix(1:100,:))

Error dimension at 78

[Andrew Newell]

This is not an answer to this question and shouldn't be here. If my answer above doesn't solve your problem, you could make a new question. Please delete this one.

Answer 4

[Alexandra Roxana]

Hello! I'm having the same problem. I want to plot the results from the following code:

clc
clear all
alpha = 2;
L=50;
dx = 1;
dt = (dx^2)/(4*alpha);
gamma = (alpha*dt)/(dx^2);
itert = 1000;
u=zeros(itert,L,L);
uinit = 0;
utop=1;
uleft=0;
ubottom=0;
uright=0;
%Boundary conditions
u(:,L,:)=utop;
u(:,:,1)=uleft;
u(:,1,1)=ubottom;
u(:,:,L)=uright;
for k=1:itert-1
    for i=2:L-1
        for j=2:L-1
            u(k+1,i,j) = gamma*(u(k,i+1,j) + u(k,i-1,j) + ...
                u(k,i,j+1) + u(k,i,j-1) - 4*u(k,i,j)) + u(k,i,j);
        end
    end
end
%display(u)
[x,y]=meshgrid(0:.01:pi);
mesh(x,y,u)

[Torsten]

The region where you solved Poisson's equation is [0:50] x [0:50] with step 1, not something with pi.

Further

u(:,1,:) = ubottom,

not

u(:,1,1) = ubottom.

[Alexandra Roxana]

Modified it with

 [x,y]=meshgrid(0:0.1:50);
 mesh(x,y,u)

Still doesn't work.

[Torsten]

[X,Y] = meshgrid(1:50,1:50);

U = u(1,:,:);

U = U(:);

surf(X,Y,U)

But I'm unsure whether the order of U corresponds to the order of the (X,Y) pairs.

[Alexandra Roxana]

Now it says that Z must be a matrix, not a scalar or vector.

[Torsten]

Then delete the line

U = U(:)

[Alexandra Roxana]

I already did. No result. X,Y = 50x50 double, U = 1x50x50 double, u = 100x50x50 double, I think it has to do with the dimensions.

[Torsten]

Then replace

U = u(1,:,:)

by

U(:,:) = u(1,:,:)

[Alexandra Roxana]

Now it works! Thank you! I guess I have to work on the code because the graph isn't quite what I expected it to be.