Comparing Colors

[DefinedV]

Hello, I was wondering if there was a simple way of comparing two hex values with regard to 2 shades of variation. Also is there a way to check if a color is darker than another?

i.e

if $color1hex == $color2hex , however instead of a direct match, it takes into account 2 shades of variation.

-Thank you

Edited March 30, 2010 by DefinedV

[JohnOne]

I dont know how to answer your question sorry.

But (and forgive me if i'm wrong) == is used to compare strings as an exact match (case sensitive)

Just for reference.

[DefinedV]

Sorry, I apologize if I came across as unclear. But let me clarify; the PixelGetColor function returns a hex value, but I would like to account for color variation, 2 shades in this case. I would like to compare the value with a previous value I retrieved and if it's within 2 shades of each other, than return true, otherwise return false.

Thanks

[omikron48]

I'd guess you'd need to define a function for that one.

Can't help you here because I don't know how to quantify what "within 2 shades" means? The hex value return would be RGB values, I'd assume, but I have no idea what calculation you'd need to say that one color is within two shades of another.

[JohnOne]

If it means anything, The helpfile says Pixelgetcolor returns a decimal value.

From experience I know you can compare it to a hex and get a correct result, but I too dont know enough about colours to work out which wold be darker.

[jchd]

You probably need to do a little search about color matching to define your own appreciation of what you're ready to consider "close enough" for your needs.

Anyway and as just stated, colors you read are RGB values which, for elementary practice, can bee seen as a tridimensionnal space with coordinates bounded in [0..255], which is the so-called color-cube.

The simplest way to estimate how close are two points in N-space is to compute their (Euclidean) distance. In your case, N=3 and if your pixels' colors, once split in Red, Green and Blue components, are:

Local $pixel1[3] = [$r1, $g1, $b1]

Local $pixel2[3] = [$r2, $g2, $b2]

then the "simple" colorimetric distance separing them is:

Local $ColorDist = Sqrt(($r2 - $r1) ^ 2 + ($g2 - $g1) ^ 2 + ($b2 - $b1) ^ 2)

The smallest result you get, the closest the two colors are. The Sqrt() isn't even meaningful in your case so you can get along without it and without loosing any significance.

In RGB space, the higher values are the most intense: black is {0, 0, 0} white is {255, 255, 255}.

There are a large number of ways to appreciate and quantitize colors, but RGB is by far the simplest.

[DefinedV]

Thanks everyone, really appreciate it.

Edited March 31, 2010 by DefinedV