"Data dimensions must agree" Error

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Amr
Amr on 14 May 2011
Commented: Alexandra Roxana on 18 Jul 2021
[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 ?
  3 Comments
Amr
Amr on 14 May 2011
when i used another image which is colored one
Jan
Jan on 14 May 2011
Please post the Matlab code. Bugs in the code cannot be found in a text description of the method.

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Accepted Answer

Andrew Newell
Andrew Newell on 15 May 2011
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.
  2 Comments
Walter Roberson
Walter Roberson on 15 May 2011
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
Andrew Newell on 15 May 2011
Good point. I had a feeling I'd be hearing from you!

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More Answers (3)

Andrew Newell
Andrew Newell on 14 May 2011
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.
  2 Comments
Amr
Amr on 14 May 2011
i convert it to gray scale and it became matrix like the other X and Y
Walter Roberson
Walter Roberson on 16 May 2011
Amr, I am not sure whether you are still having difficulty with this matter?

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guj
guj on 15 May 2011
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
  1 Comment
Andrew Newell
Andrew Newell on 15 May 2011
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.

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Alexandra Roxana
Alexandra Roxana on 18 Jul 2021
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)
  8 Comments
Torsten
Torsten on 18 Jul 2021
Then replace
U = u(1,:,:)
by
U(:,:) = u(1,:,:)
Alexandra Roxana
Alexandra Roxana on 18 Jul 2021
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.

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