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pixelpusher94608
CorvusCroax

Posts: 1227
Filters: 18
So, what is the most efficient way to filter out just one color?

Suppose I have an image input. I just want to filter out the pixels of a particular shade of yellow, and make an alpha map out of it.

What is the most *processor efficient* way to do this in FF?

-RGB math?
-Blend using difference?
-gradient map?
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Velho
Velho
Posts: 53
Filters: 6
I'd put my money on blend using difference --> threshold. Two fast components and you can output anything from threshold.
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Sphinx.
Filter Optimizer

Posts: 1750
Filters: 39
The RGB Math If is probably the fastest for exact comparison. If you want a bit more flexibility, I'd use a distance measure and a curve/threshold on that..
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Velho
Velho
Posts: 53
Filters: 6
Quote
The RGB Math If is probably the fastest for exact comparison


Maybe, but in this case using RGB math would require six components to achieve the same result what I did with only two. If you have a special trick in your sleeve, please share smile;)

What comes to the original question, I doubt the method for filtering a specific color is the first thing to optimize.
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pixelpusher94608
CorvusCroax

Posts: 1227
Filters: 18
So, what i'm trying to do is to figure out angles relative to a normal map. (Which is created from a height map.) I want to place some bomber stuff on the part of the shape which if 'UP' and some different stuff on the part which is 'DOWN' etc. I'd like to be able to broaden or narrow my range based on a control. Since I have to make a bunch of these, I want to do it efficiently.

But it got me wondering, since this seems to come up a lot in filter construction. (I can imagine a comic book filter which flattens out specific colors for example.)

Sphinx: what specific RGB map construction were you thinking about?


Quote
I'd put my money on blend using difference --> threshold. Two fast components and you can output anything from threshold.


Yeah, that's sort of what I started with. But I can only specify a green component for the comparison input, which seems inefficient.
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