Determining vector Xq and Yq for interp2 with scale

Hello,

I am studying how to make interpolation in image processing.

My error: Undefined function or variable 'scale'. Error in PL7ex3 (line 4) X1 = 1:1/scale:size(I0grey,2);

%PL7. Compare different methods for interpolation.
I0 = imread('glomeruli.tif'); %MxNx3
I0grey = im2double(rgb2gray(I0)); %grey, MxN
%interp2
X1 = 1:1/scale:size(I0grey,2);
Y1 = (1:1/scale:size(I0grey,1))';
X2 = 1:1/scale:size(I0grey,2);
Y2 = (1:1/scale:size(I0grey,1))';
I01 = interp2(I0grey, X1, Y1);
I02 = interp2(I0grey, X1, Y1, 'nearest');
I03 = interp2(I0grey, X1, Y1, 'cubic');
I04 = interp2(I0grey, X1, Y1, 'spline');
%Vq = interp2(___,method) specifies an optional, trailing input argument 
%that you can pass with any of the previous syntaxes. The method argument 
%can be any of the following strings that specify alternative interpolation 
%methods: 'linear', 'nearest', 'cubic', or 'spline'. The default method is 
%'linear'.

figure(1) % create new figure
subplot(2,3,1) % first subplot
imshow(I0)
title('Original')
subplot(2,3,2)
imshow(I0gray)
title('Original in greyscale')
subplot(2,3,3)
imshow(I01)
title('Linear interpolation')
subplot(2,3,4)
imshow(I02)
title('Interpolation 2: Nearest sample grid point.')
subplot(2,3,5)
imshow(I03)
title('Interpolation 3: Cubic convolution')
subplot(2,3,7)
imshow(I04)
title('Interpolation 4: Cubic spline')

figure(2) % create new figure
subplot(2,3,1) % first subplot
plot(I0(:,50))
title('Original')
subplot(2,3,2)
plot(I0gray(:,50))
title('Original in greyscale')
subplot(2,3,3)
plot(I01(:,50))
title('Linear interpolation')
subplot(2,3,4)
plot(I02(:,50))
title('Interpolation 2: Nearest sample grid point.')
subplot(2,3,5)
plot(I03(:,50))
title('Interpolation 3: Cubic convolution')
subplot(2,3,7)
plot(I04(:,50))
title('Interpolation 4: Cubic spline')

I rewrote this line from the example that prof. was showing us. Is there any misspelling? How can I correct it?

Thank you in advance.

Answer 1

[Star Strider]

Ask your professor what ‘scale’ is supposed to be, and how you would assign it or calculate it.

[Katarzyna Wieciorek]

If I could ask professor I wouldn't ask here :( is there any way to determine these vectors basing on size of original image? This is what this line was supposed to do.

[Star Strider]

In the absence of other information and since Image Analyst hasn’t opined thus far, assume ‘scale’ is some number greater than the dimensions of ‘I0grey’.

Choose:

scale = 5*max(size(I0grey));

or something more to your liking, and see what that does. The world will not come to an end if that is wrong.

[Katarzyna Wieciorek]

I solved my problem. If there is anyone with similar it may be useful:

%y = linspace(x1,x2,n), separation 0.5 = (x2-x1)/(n-1)
[sy,sx] = size(I0grey);
nx = 2*sx-1;
ny = 2*sy-1;
X1 = linspace(1,sx,nx);
Y1 = (linspace(1,sy,ny))';

[Star Strider]

Congratulations!