overlapped objects counting in matlab

hi. i m working on a project which is related to digital image processing but i can't count the overlapped objects it take multiple overlapped objects as one object. please tell me which method i can use for this problem.... and overlapped objects may be two or more than two..... so please tell me that method which is suitable for this......
best regards FAISAL SALEH KHATTAK

3 Commenti

please don't refer me watershed method and otsu algorithm. i have already worked on them. These are not suitable for more than two overlapping objects. please help me out of this.
Just a second... Do you have multiple images, or just one?
i have one image in which there are lot small objects and noise and objects are overlapped fully and somewhere object is also overlapped with noise and the overlapped objects are more than two...... i have work a lot on it i can't make it.... if you have any idea about that please help me out....
thanks in advance...

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Risposte (2)

Image Analyst
Image Analyst il 8 Apr 2012

0 voti

If you can't use watershed or similar object splitting methods, then you can try some shaped based methods if you know something about the shape but they will be complicated. For example, you know that all your shapes are hexagons or something. But let's say that you have discs. Now let's say you want to know how many discs (let's say DVD or CD discs) are overlapping in a stack of 50 of them. All you have is an overhead photo of the 50 disc stack. How are you going to know how many discs are in the stack? You can't because they're hidden from view.

11 Commenti

all objects have different size and shape that's why you can't specify one shape.
If I know that there are 50 discs in the image. How can I count these?
Md, segment them, then call bwareafilt() on the binary image:
mask = bwareafilt(mask, 50); % Extract the 50 largest blobs only.
Md Hafizur Rahman
Md Hafizur Rahman il 12 Feb 2020
Modificato: Image Analyst il 12 Feb 2020
Thank you for your kind information. I tried that but for my image, that isn't worked.
Here my image link
0000 Screenshot.png
I want to count all objects [ the red circle] from the image. How can I do that?
Md, that's the gray scale image. bwareafilt() only works on the segmented image. You forgot to attach that. Let's see your binary image.
0000 Screenshot.png
Md Hafizur Rahman
Md Hafizur Rahman il 12 Feb 2020
Modificato: Md Hafizur Rahman il 12 Feb 2020
I have this image. Can I count every objects from the image? Is there any methods to determine the number of the objects?
I, as a human, cannot count every object in that image, so I doubt that a computer could do so. Not without a strict definition of what an "object" is for this purpose.
I know it's quite hard to count all objects- rod-shaped bacteria. At least, I need to count as much as possible the bacteria from the image. Can I do that by using any methods?
Often you would start by using a binary threshold on the grayscale image. Image Analyst asked you to show the result of using the binary threshold.
The Main Code
A = imread('1_03.png');
figure(1)
imshow(A)
I = rgb2gray(A);
I = adapthisteq(I);
I = wiener2(I, [3 3]);
bw = im2bw(I, graythresh(I));
bw2 = imfill(bw,'holes');
bw3 = imopen(bw2, strel('disk',1));
bw4 = bwareaopen(bw3, 500);
L = bwlabel(bw4);
s = regionprops(L, 'Centroid');
figure(9)
imshow(bw4)
for k = 1:numel(s)
c = s(k).Centroid;
text(c(1), c(2), sprintf('%d', k), 'Color', 'r', ...
'HorizontalAlignment', 'center', ...
'VerticalAlignment', 'middle');
end
bw4_perim = bwperim(bw4);
overlay1 = imoverlay(I, bw4_perim, [1 .3 .3]);
figure(11)
imshow(overlay1)
Output of the code: (Please see the attachment)
The first image is input image the second image is output image
I just don't see how all the bacteria can be automatically identified in this. Even we as humans won't necessarily get the "correct" answer. I'd suggest you just call drawpoint() in a loop to have the user keep dropping down points until he's dropped a point on every single bacteria.

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Richiesto:

il 5 Mar 2012

Commentato:

il 13 Feb 2020

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