How to Copy Labeled Object Area Value and Paste on Other Object
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Hello,
I am not sure if it is possible or not but would like to ask if anybody has a knowledge on copying the properties of any labeled objects to another oobject. I mean Regionprops values with properties.
An example, I am using below properties on regionprops:
regionprops=regionprops('Centroid','Area','Eccentricity','Perimeter')
and my aim is to copy the area value to pasted point.
2 Commenti
Walter Roberson
il 7 Set 2018
So at the target location, the code should check to see if there is an existing object there and if so it should delete or add randomly to the object until the area of the resulting object was the same as the first object?
Risposte (3)
Walter Roberson
il 7 Set 2018
regionprops returns a struct. You can assign the struct or parts of it to another variable.
0 Commenti
Image Analyst
il 7 Set 2018
Why would you want to do that? It sounds like a very bad and deceptive idea and would lead to wrong, or at best, misleading results.
You forgot to attach your input binary image, and what you want to see as the result of this (in my opinion, unwise) operation.
What is a "pasted point"? What are you pasting and how and where?
Do you simply want to display some text (in the overlay) of the area of one object but above some different object? Again, this is wrong.
0 Commenti
Murat Kocaman
il 10 Set 2018
Modificato: Murat Kocaman
il 10 Set 2018
4 Commenti
Image Analyst
il 11 Set 2018
You haven't given the "use case" for this application, and that would help guide our replies.
First of all, see if you can take photos without overexposing the subject.
Then, I'm guessing that you want to compare this thing (ring or jewelry or whatever) to a perfect version, so do you have a perfect version to compare it to?
Next, if the subject is the same for every photo, then why can't you just use imfreehand() to hand draw a template - a set of regions that define where every region should be? If all your subjects will be the same shape/structure, then a template would be the simplest approach.
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