How to count the number of object present in binary image?

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Hanuman Wable
Hanuman Wable el 20 de Dic. de 2013
Movida: DGM el 13 de Feb. de 2023
Hello sir, i am going to count the number of object present in binary image. so any of you please suggest me regarding counting object.because i have problem during count such object. thank you
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sanhita das
sanhita das el 7 de Abr. de 2018
Movida: DGM el 13 de Feb. de 2023
sir,i want to show paragraphs from document image.pls tell the algoritm name. I attach a input file.and how to extract paragraph from this image.please help me

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Respuestas (3)

Sabarinathan Vadivelu
Sabarinathan Vadivelu el 20 de Dic. de 2013
Editada: Sabarinathan Vadivelu el 20 de Dic. de 2013
You can use bwconncomp.
cc = bwconncomp(yourBinaryImage,4);
number = cc.NumObjects;

Image Analyst
Image Analyst el 20 de Dic. de 2013
Here's yet another way
[labeledImage, numberOfObject] = bwlabel(binaryImage);
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romaisa sabih
romaisa sabih el 18 de Mayo de 2016
sir,i have to count the active fingers in my image.how can i do this ? please help.

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David Sanchez
David Sanchez el 20 de Dic. de 2013
If you are planning to do something with those objects, you could be interested in using regionprops. Besides the number of objects within the image, you can retrieve any other property of the objects, such as the area, centroid, orientation,....
BW = imread('text.png');
s = regionprops(BW, 'Area');
N_objects = numel(C);
For all the options, take a look at
doc regionprops
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muhamad zulhairi zailan
muhamad zulhairi zailan el 7 de Nov. de 2018
hello sir, how to count the object using area parameter, which mean i have the image and i want to count the object in that image using range of area parameter and i want to count the object refer to it area parameter? thanks sir your reply very appreciated.
Image Analyst
Image Analyst el 7 de Nov. de 2018
I don't know what you want. By the way, the code above is not correct. It should say
N_objects = length(s)
That number will be the same regardless which attributes you measure (area, perimeter, or whatever).

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