R help - Aug 2006 - rgb and col2rgb color conversion/modification/shading

I want to get a lighter shade of a color...I have a lot of colored objects and
want each one printed as a foreground against a slightly lighter background.

I thought I could try something like changing the alpha channel by first
converting it to rgb. 

But prior to trying that, I'm stuck with how to get the color after
converting
using col2rgb() to be interpreted again as a color, rather than a simple
vector?

Anyone have any help/ or alternative suggestion...

Thanks, -c
----------------------
TRYING WITH A SINGLE COLOR:

mycol<-"red"
> col2rgb(mycol)
      [,1]
red    255
green    0
blue     0
> rgb(col2rgb(mycol),maxColorValue=255)
Error in rgb(col2rgb("red")) : argument "green" is missing,
with no default

ccarey at fhcrc.org wrote:
> I want to get a lighter shade of a color...I have a lot of colored objects
and
> want each one printed as a foreground against a slightly lighter
background.
> 
> I thought I could try something like changing the alpha channel by first
> converting it to rgb. 
> 
> But prior to trying that, I'm stuck with how to get the color after
converting
> using col2rgb() to be interpreted again as a color, rather than a simple
> vector?
> 
> Anyone have any help/ or alternative suggestion...
> 
> Thanks, -c
> ----------------------
> TRYING WITH A SINGLE COLOR:
> 
> mycol<-"red"
> 
>> col2rgb(mycol)
>       [,1]
> red    255
> green    0
> blue     0
> 
>> rgb(col2rgb(mycol),maxColorValue=255)

rgb() required separate arguments for red green and blue.

Hence saying

   mycol2 <- col2rgb(mycol)
   rgb(mycol2[1,], mycol2[2,], mycol2[3,], maxColorValue=255)

does not look like beautiful code, but is probably easier than the other 
"cool" tricks.

Uwe Ligges


> Error in rgb(col2rgb("red")) : argument "green" is
missing, with no default
> 
> ______________________________________________
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http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

<ccarey at fhcrc.org> wrote in message 
news:1154392412.44cea15c0c1fa at webmail.fhcrc.org...
>I want to get a lighter shade of a color...I have a lot of colored objects 
>and
> want each one printed as a foreground against a slightly lighter 
> background.
>
> I thought I could try something like changing the alpha channel by first
> converting it to rgb.
I'm not sure what you want to do with the alpha channel - it's sometimes
used for transparency, especially on Macs, but is not used much on PCs 
(AFAIK).



Let's say you want different shades of gold:
> colors()[142]
[1] "gold"



Instead of RGB color space perhaps you should consider HSV 
(Hue-Saturation-Value) color space.



Let's convert "gold" to rgb to hsv:


> col2rgb( colors()[142] )
      [,1]

red    255

green  215

blue     0


> rgb2hsv( col2rgb( colors()[142] ) )
       [,1]

h 0.1405229

s 1.0000000

v 1.0000000



The "hue" (h) is the color ranging from 0 to 1 around a color circle
(with
red= 0 or 1).  Find h = 0.140 ("gold") in this color circle:



hue <- seq(0.0, 1.0, by=1/40)



pie(rep(1,40),

    labels=formatC(hue, digits=3, format="f"), cex=0.75,

    col=hsv(hue, 1.0, 1.0),

    radius=1.0,

    main="HSV (S=1, V=1)" )




Hues range from 0.0 to 1.0.




A color is saturated (s=1) when it is "far" from a shade of gray
(ranging
from black to white).  Grays are unsaturated (no color) colors with s = 0. 
Saturation ranges from 0.0 to 1.0.



The value (v) is the brightness of the color.  Low values appear quite dark 
but still have color.  v=1 is as bright as possible.   Values range from 0.0 
to 1.0.



You can get different "shades" of the same color by varying changing
the
saturation and value for a given hue.  The hsv function returns the RGB 
color in hex form.



Consider:
> hsv(0.1405, 1, 1)
[1] "#FFD700"



Hex FF = decimal 255 = red

Hex D7 = decimal 215 = green

Hex 00 = decimal 0 = blue



Let's vary Saturation from 0.0 to 1.0 and Value from 0.0 to 1.0 in this 
plot:





MakeHSVRectangle <- function(saturation, value)

{

  GoldHue <- 0.140

  color <- hsv(GoldHue, saturation, value)

  rect(100*saturation, 100*value, 100*saturation+4, 100*value+4, col=color)

}





plot(0:110,0:110, type="n",

     xlab="Saturation[%]", ylab="Value[%]",

     main="Shades of Gold, H=0.140")

outer(seq(0.0, 1.0, 0.05), seq(0.0, 1.0, 0.05), MakeHSVRectangle)





With Value = 0, all colors are "black".  With Saturation=0, the only 
"colors" along the y axis are the shades of gray.  The original
"gold"
rectangle is at the upper right.



So, given a starting color, you have a number of "shades" (various 
saturations and values) with the same color hue.



I hope this helps.



efg

Earl F. Glynn

Scientific Programmer

Stowers Institute for Medical Research