INTERSECTION BETWEEN 2 IMAGES
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Hi everyone (sorry for bad english).
I need code to find the intersection between two images. I have image A and image Am (is the modified version of image A) and I have to find points in common or the area (teacher suggest I use the intersection). Then repeat the process for 100 images in a folder.
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Answers (3)
Image Analyst
on 27 Feb 2014
Edited: Image Analyst
on 15 Sep 2021
You can subtract them and look for 0's. For grayscale images:
matchingPixels = (double(image1) - double(image2)) == 0;
imshow(matchingPixels);
To find area of matching pixels, you can use sum(), bwarea() or regionprops():
area1 = sum(matchingPixels);
area2 = bwarea(matchingPixels);
% To use regionprops to find areas of all matching regions.
labeledImage = bwlabel(binaryImage);
measurements = regionprops(labeledImage, 'Area');
area3 = sum([measurements.Area]);
For color images, you have to check each color channel for matches.
For binary images use the dice() function, introduced in R2017b:
Description
similarity = dice(BW1,BW2) computes the Sørensen-Dice similarity coefficient between binary images BW1 and BW2.
2 Comments
Diah Junaidi
on 1 Aug 2019
what should I do for sir? i have 2 binary images that I wanna compare.
labeledImage = bwlabel(binaryImage);
Image Analyst
on 2 Aug 2019
Edited: Image Analyst
on 15 Sep 2021
In what way do you want to compare these binary images? Maybe:
Description
similarity = dice(BW1,BW2) computes the Sørensen-Dice similarity coefficient between binary images BW1 and BW2.
Azzi Abdelmalek
on 27 Feb 2014
Edited: Azzi Abdelmalek
on 27 Feb 2014
[M,ii,jj]=intersect(A,Am)
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Mohsin Khalid
on 20 Sep 2016
Edited: Mohsin Khalid
on 20 Sep 2016
1) Convert them in to gray scale images 2) Then convert them in to binary images
Below is a command for intersection of two images
intersectedImage=bitand(binaryImage1,binaryImage2);
For 100 images take the intersectedImage and bitand it with binaryImage3 and so on..
1 Comment
Walter Roberson
on 20 Sep 2016
If you take that approach, the output will only be set at places where both images had intensity greater than or equal to 128 (out of 255), which is not what the user is asking. The user is asking for points that are in common, which includes places where both are both the same low intensity.
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