Views
386
Replies
4
Status
Closed
Hi:
I’m working with these images and I have to remove only the pixels in which the blue rgb values is less than .9 the red RGB and green value. After I have detected all of the pixels that meet this condition I want to replace them with black. First of all I want to know if it is possible for photoshop to make this type of decision making process and then how do I select those pixels and replace or remove them. I have tried using the selecting a particular blue color to remove it but It does not get the trick done. I have been able to implement this in python but I want to know if it is possible photoshop. I have photoshop CS. If you can referred me to another source of information that you think deals with this issue. I will be very gratefully.
Example RGB(0,0,255) : 255 > .9*0 and 255 > .9*0 This pixel would be chosen.
RGB(35,45,25): 25 > .9*35 and 25 > .9*45 This pixel would not be chosen.
I’m working with these images and I have to remove only the pixels in which the blue rgb values is less than .9 the red RGB and green value. After I have detected all of the pixels that meet this condition I want to replace them with black. First of all I want to know if it is possible for photoshop to make this type of decision making process and then how do I select those pixels and replace or remove them. I have tried using the selecting a particular blue color to remove it but It does not get the trick done. I have been able to implement this in python but I want to know if it is possible photoshop. I have photoshop CS. If you can referred me to another source of information that you think deals with this issue. I will be very gratefully.
Example RGB(0,0,255) : 255 > .9*0 and 255 > .9*0 This pixel would be chosen.
RGB(35,45,25): 25 > .9*35 and 25 > .9*45 This pixel would not be chosen.
Master Retouching Hair
Learn how to rescue details, remove flyaways, add volume, and enhance the definition of hair in any photo. We break down every tool and technique in Photoshop to get picture-perfect hair, every time.