MATLAB Answers

## How to compare two matrices of diffrent dimensions by looping.

Asked by Sonny Sood

### Sonny Sood (view profile)

on 27 Jul 2013

Basically, i want to scan a small matrix over bigger matrix and then compare it's elements one by one for greater or less than bigger matrix.The comparison can be done in either way through column or through row.

Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 27 Jul 2013

If you have

```a=[1 2;3 4]  % the small matrix
%and
b=[1 5 6;4 0 8;11 12 0]  % the big matrix
```

What should be the result?

Sonny Sood

### Sonny Sood (view profile)

on 28 Jul 2013

the small matrix should be compared with that of big matrix for greater than or equal to. The result is to find a match between the elements of two matrix's and store their positions(index) after being compared. The result i want is; 1=1,match found,store it 5>2,store it's position and the element 4>3,store it 0<4,neglect it

Jan Simon

### Jan Simon (view profile)

on 28 Jul 2013

@Sonny: Please answer Azzi's question. Do not describe how you want to create the output but post manually created results. Such unequivocal examples are very useful. I do not understand the explanation "1=1,match found,store it 5>2,store it's position and the element 4>3,store it 0<4,neglect it". Neither "store it" not "neglect it" reveal any details about the wanted result.

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## 4 Answers

### per isakson (view profile)

Answer by per isakson

### per isakson (view profile)

on 27 Jul 2013
Edited by per isakson

### per isakson (view profile)

on 28 Jul 2013

Not by looping, however try

```    %%
a=[1,2;3,4];
b=[1,5,6;4,0,8;11,12,0];
%%
sza = size( a );
is_greater_equal = ( a >=  b(1:sza(1),1:sza(2)) )```

it returns

```    is_greater_equal =
1     0
0     1```

and to find the indicies

```    >> [ixr,ixc]=find(is_greater_equal)
ixr =
1
2
ixc =
1
2```

Sonny Sood

### Sonny Sood (view profile)

on 28 Jul 2013

Thanks, this is correct answer but i want to continue the comparison for the whole matrices by looping sine i am comparing two images, that's why i need this, like for the first column it does, now for the second and third column, then next row, then first and second column, so on.

per isakson

### per isakson (view profile)

on 28 Jul 2013

Something like this might do it?

```    sza = ...
for ii = 1 : ...
for jj = 1 : ...
is_greater_equal = ( a(1:sza(1),1:sza(2)) >=  b(ii:sza(1),jj:sza(2)) )
...
end
end ```

### Image Analyst (view profile)

Answer by Image Analyst

### Image Analyst (view profile)

on 28 Jul 2013

Please explain what you're trying to do, rather than how you want to solve it. Because I'm thinking that normalized cross correlation (with function normxcorr2) will help you but I'm not really sure what you want to do. And don't just say you want to scan a large matrix with a small one, say WHY you want to do that so we know whether to recommend normxcorr2() or blockproc(), or if you really need nested looping.

Sonny Sood

### Sonny Sood (view profile)

on 29 Jul 2013

The first 3x3 are the first nine elements of the bigger matrix,second 3x3 means next 9 elements of he matrix with next column when column is incremented .

Image Analyst

### Image Analyst (view profile)

on 29 Jul 2013

Try normalized cross correlation - it does exactly that. See this demo where I applied it to a color image to find a small color subimage inside the larger color image:

```% Demo to use normxcorr2 to find a template (a white onion)
% in a larger image (of a pile of vegetables)
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
imtool close all;  % Close all imtool figures.
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 11;
```
```% Check that user has the Image Processing Toolbox installed.
hasIPT = license('test', 'image_toolbox');
if ~hasIPT
% User does not have the toolbox installed.
message = sprintf('Sorry, but you do not seem to have the Image Processing Toolbox.\nDo you want to try to continue anyway?');
reply = questdlg(message, 'Toolbox missing', 'Yes', 'No', 'Yes');
if strcmpi(reply, 'No')
% User said No, so exit.
return;
end
end
```
```% Read in a standard MATLAB color demo image.
folder = fullfile(matlabroot, '\toolbox\images\imdemos');
baseFileName = 'peppers.png';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
if ~exist(fullFileName, 'file')
% Didn't find it there.  Check the search path for it.
fullFileName = baseFileName; % No path this time.
if ~exist(fullFileName, 'file')
% Still didn't find it.  Alert user.
errorMessage = sprintf('Error: %s does not exist.', fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
rgbImage = imread(fullFileName);
% Get the dimensions of the image.  numberOfColorBands should be = 3.
[rows columns numberOfColorBands] = size(rgbImage);
% Display the original color image.
subplot(2, 2, 1);
imshow(rgbImage, []);
axis on;
title('Original Color Image', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0, 0, 1, 1]);
```
```% Let's get our template by extracting a small portion of the original image.
templateWidth = 71
templateHeight = 49
smallSubImage = imcrop(rgbImage, [192, 82, templateWidth, templateHeight]);
subplot(2, 2, 2);
imshow(smallSubImage, []);
axis on;
title('Template Image to Search For', 'FontSize', fontSize);
```
```% Ask user which channel to search for a match.
% channelToCorrelate = menu('Correlate which color channel?', 'Red', 'Green', 'Blue');
% It actually finds the same location no matter what channel you pick,
% for this image anyway, so let's just go with red (channel #1).
channelToCorrelate = 1;
correlationOutput = normxcorr2(smallSubImage(:,:,1), rgbImage(:,:, channelToCorrelate));
subplot(2, 2, 3);
imshow(correlationOutput, []);
axis on;
title('Normalized Cross Correlation Output', 'FontSize', fontSize);
```
```% Find out where the normalized cross correlation image is brightest.
[maxCorrValue, maxIndex] = max(abs(correlationOutput(:)));
[yPeak, xPeak] = ind2sub(size(correlationOutput),maxIndex(1))
% Because cross correlation increases the size of the image,
% we need to shift back to find out where it would be in the original image.
corr_offset = [(xPeak-size(smallSubImage,2)) (yPeak-size(smallSubImage,1))]
```
```% Plot it over the original image.
subplot(2, 2, 4); % Re-display image in lower right.
imshow(rgbImage);
axis on; % Show tick marks giving pixels
hold on; % Don't allow rectangle to blow away image.
% Calculate the rectangle for the template box.  Rect = [xLeft, yTop, widthInColumns, heightInRows]
boxRect = [corr_offset(1) corr_offset(2) templateWidth, templateHeight]
% Plot the box over the image.
rectangle('position', boxRect, 'edgecolor', 'g', 'linewidth',2);
% Give a caption above the image.
title('Template Image Found in Original Image', 'FontSize', fontSize);
uiwait(helpdlg('Done with demo!'));
```
Sonny Sood

on 29 Jul 2013

Thank you Sir,

### Azzi Abdelmalek (view profile)

Answer by Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 28 Jul 2013
```B=randi(100,9)  % the big matrix
[n,m]=size(B);
S=randi(100,3) % the small matrix
out=[];
for id1=1:3:n-3
for id2=1:3:m-3
a=B(id1:id1+2,id2:id2+2);
out=[out;a(a>=S)];
end
end
```

Sonny Sood

### Sonny Sood (view profile)

on 29 Jul 2013

I am getting error in this code at

line out=[out;a(a>=S)] of 'matrix dimensions must agree'

Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 29 Jul 2013

In your case what are the sizes of your two matrices ?

Sonny Sood

### Sonny Sood (view profile)

on 30 Jul 2013

512x512 and 256x20, basically it can be any two different sized matrices since i am using two different sized images.

### Azzi Abdelmalek (view profile)

Answer by Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 30 Jul 2013
Edited by Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 30 Jul 2013
```B=randi(100,512,512) ; % the big matrix
[n,m]=size(B);
S=randi(100,256,20) % the small matrix
[n1,m1]=size(S);
q1=fix(n/n1);
q2=fix(m/m1);
idxr= [ n1*ones(1,q1) n-n1*q1];
idxc=[m1*ones(1,q2) m-m1*q2];
idxc(~idxc)=[];
idxr(~idxr)=[];
out=[];
ii0=1;
for id1=1:numel(idxr)
ii1=ii0+idxr(id1)-1;
jj0=1;
for id2=1:numel(idxc)
jj1=jj0+idxc(id2)-1;
a=B(ii0:ii1,jj0:jj1);
[nn,mm]=size(a);
b=S(1:nn,1:mm);
out=[out;a(a>=b)];
jj0=jj1+1;
end
ii0=ii1+1;
end
```

Sonny Sood

### Sonny Sood (view profile)

on 5 Aug 2013

Can you please add comments for each line, as I am beginner in this, and also will be easy to grab the logic.

Thanks

Image Analyst

### Image Analyst (view profile)

on 5 Aug 2013

I had comments in my answer. Don't be intimidated by all the fancy stuff I put in there to make a fancy demo. The basic heart of the code is only two lines: one line to call normxcorr2() and one to call max(). The rest is just comments and well-commented things to make a nice gui.

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