filter a specific color from image

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Ah
Ah el 3 de Dic. de 2016
Comentada: Image Analyst el 21 de Dic. de 2016
Hello everybody
i have a problem and it would be great if someone helps i am trying to cut a color from an image where the result image is BW .and the color specified i want to chose is black and the rest is white .the main problem is since there was more light of the left side of the image the pixel value pattern changes for the same color over the same image,for example blue of the left has value of RGB= 106 156 183 where on the right side RGB=106 126 133 ,is there a way where i can make the brightness the same all over the image ?I mean a way of making the values of the color i see to have the same values all over the image or a same pattern?

Respuesta aceptada

Image Analyst
Image Analyst el 3 de Dic. de 2016
See color segmentation demos in my File Exchange. http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862
Or use the ColorThresholder App on the Apps tab of the tool ribbon.
  7 comentarios
Ah
Ah el 21 de Dic. de 2016
so i have been reading about x-rite ColorChecker chart,and tested the Method you recommended regarding the the white spot,is there a way to calibrate a camera using information already exsisted on the image? for example i can easly detect red color on my image feed and can i use a saved value to calibrate the camera ?multiplaying by a factor doesnot seems to help me a lot since i picked a point and multiplyed the whole image by factors one for each RGB .
Image Analyst
Image Analyst el 21 de Dic. de 2016
If you don't want to do calibrated color analysis then you can filter on whatever you have. I do that in several different ways (different color spaces) in my File Exchange: http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862.
Or you can try the Color Thresholder on the Apps tab of the MATLAB tool ribbon.

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Ah
Ah el 12 de Dic. de 2016
very intresting way i will try it in the next weeks,i will read about x-rite ColorChecker chart and report back.thank you again for help

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