INTERSECTION BETWEEN 2 IMAGES

12 vues (au cours des 30 derniers jours)
francesco
francesco le 27 Fév 2014
Modifié(e) : Image Analyst le 15 Sep 2021
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.

Réponses (3)

Image Analyst
Image Analyst le 27 Fév 2014
Modifié(e) : Image Analyst le 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 commentaires
Diah Junaidi
Diah Junaidi le 1 Août 2019
what should I do for sir? i have 2 binary images that I wanna compare.
labeledImage = bwlabel(binaryImage);
Image Analyst
Image Analyst le 2 Août 2019
Modifié(e) : Image Analyst le 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.

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Azzi Abdelmalek
Azzi Abdelmalek le 27 Fév 2014
Modifié(e) : Azzi Abdelmalek le 27 Fév 2014
[M,ii,jj]=intersect(A,Am)

Mohsin Khalid
Mohsin Khalid le 20 Sep 2016
Modifié(e) : Mohsin Khalid le 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 commentaire
Walter Roberson
Walter Roberson le 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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