Index in position 2 exceeds array bounds. Index must not exceed 1024.

I am student writing my thesis and new to matlab ,i have posted the code below in which the particle image velocimetry was attempted to be done.

clear all;
clc;
close all;

imagea = imread('image1.tif');
imageb = imread('image2.tif');

[xmax, ymax]=size(imagea);

wsize = [64, 64];
w_width = wsize(1);
w_height = wsize(2);

xmin = w_width/2;
ymin = w_height/2;
xgrid = 200:w_width/2:864;
ygrid = 200:w_height/2:1696;

w_xcount = length(xgrid);
w_ycount = length(ygrid);

x_disp_max = w_width/2;
y_disp_max = w_height/2;

test_ima(w_width, w_height) = 0;
test_imb(w_width+2*x_disp_max, w_height+2*y_disp_max) =0;

dpx(w_xcount, w_ycount) = 0;
dpy(w_xcount, w_ycount) = 0;
xpeak1 = 0;
ypeak1 = 0;

for i=1:(w_xcount)
    for j=1:(w_ycount)
        max_correlation = 0;
    	test_xmin = xgrid(i)-w_width/2;
    	test_xmax = xgrid(i)+w_width/2;
    	test_ymin = ygrid(j)-w_height/2;
    	test_ymax = ygrid(j)+w_height/2;
    	x_disp = 0;
    	y_disp = 0;
    	test_ima = imagea(test_xmin:test_xmax, test_ymin:test_ymax);
    	test_imb = imageb((test_xmin-x_disp_max):(test_xmax+x_disp_max), (test_ymin-y_disp_max):(test_ymax+y_disp_max));
    	correlation = normxcorr2(test_ima, test_imb);
    	[xpeak, ypeak] = find(correlation==max(correlation(:)));
    	
    	xpeak1 = test_xmin + xpeak - wsize(1)/2  - x_disp_max;
    	ypeak1 = test_ymin + ypeak - wsize(2)/2  - y_disp_max;
    	dpx(i,j) = xpeak1 - xgrid(i);
    	dpy(i,j) = ypeak1 - ygrid(j);
    end
end

quiver(dpy, -dpx)

[Rik]

You didn't upload the images, you didn't format your code, you didn't write any comments explaining what you expect your code to do.

Have a read here and here. It will greatly improve your chances of getting an answer.

[DGM]

Without knowing what size your images are, it's hard to guess. If I had to guess, I'd say that this is potentially a broad problem:

[xmax, ymax]=size(imagea);

The first dimension is rows (y); the second is columns (x). I'm sure this mistake is repeated in other lines, but since you're also using literals in your addressing, I can't guess at where those come from or how they relate to the size of your images. You'll have to figure that out.

[ymax, xmax]=size(imagea);

The other hazard is that when you call size() like this, the last output is not the size of dim2. It's the product of the sizes of dim 2 and any trailing dimensions. If the image has more than one channel, the second output will be wrong. If you can't guarantee the dimensionality of inputs to size(), always discard the last output.

[ymax,xmax,~] = size(imagea);

[Sumanth Kokkula]

the size of both the images is 564 * 1024

[Sumanth Kokkula]

i am expecting a output picture showing the displacement of the particles from image 1 to image 2

[DGM]

Again, you're using literals in your addressing setup.

xgrid = 200:w_width/2:864;
ygrid = 200:w_height/2:1696;

These vectors determine the number of loop iterations, but they're independent of the image geometry.

This seems like a good job for blockproc(), but I'm not sure why you're using normxcorr() blockwise.

[Sumanth Kokkula]

i am using the normcoxx2() to do the cross co-relation.

i have taken a part of image1 and using this function i wanted to find the location of it in the image2

[Walter Roberson]

The images are indexed pseudocolor not grayscale. But that does not affect the size, and it looks like the colormap is in increments of 1/256 so it should turn out to make no color difference

[Image Analyst]

"i have taken a part of image1 and using this function i wanted to find the location of it in the image2" <= yes but in your comment below to me you said that is the reason why you did not use my code. Why the discrepancy? And like I said there, you did not have to use a part of the main image as a template -- you could use whatever image you wanted. You could synthesize it or read it in from a file or whatever you want.

Answer 1

[Image Analyst]

I do exactly that in my attached demo.

[Sumanth Kokkula]

thanks for the reply but,i dont want to extract a particular object from the picture i want to use the normxcorr to locate the displaced part of my first image in the second

[Image Analyst]

Yes, I know. I thought you'd realize that the template image could be anything. You could read it from a file of course. I didn't want to attach two image files with the demo and I wanted to just use built-in image for the demo, so I took the template image from the main image. But of course you can use whatever template image you want -- it doesn't have to come from the main image. I'm attaching a newer demo where I explicitly spell that out.

Answer 2

[Sumanth Kokkula]

i have the answer for the above mentioned error.

i have taken the wrong values in the room of intrest changing the values given the solution.

xgrid = 200:w_width/2:864;

ygrid = 200:w_height/2:1696;

the bold ones are the mistaken ones. by changing them i have found the solution.

thanks to all for your suggestions.