Mean of hexadecimal numbers
On 16/04/2016 12:33 PM, Atte Tenkanen wrote:
Hm..., Should these two versions produce the same solution?
I wouldn't expect them to. Duncan Murdoch Unfortunately and
shame to confess, I don't know much about the colors in R:
myColors <- c("#FF7C00","#00BF40","#FFFF00")
Colors=rgb2hsv(col2rgb(myColors))
apply(Colors,1,mean)
h s v
0.2122974 1.0000000 0.9163399
* * * * *
# Average the 1st two by taking the middle colour of a 3 colour palette
x <- colorRampPalette(c("#FF7C00","#00BF40"), space = "Lab")(3)[2]
# Average in the third by taking the 2nd of a 4 colour palette, so x
# gets twice the weight
colorRampPalette(c(x, "#FFFF00"), space = "Lab")(4)[2]
rgb2hsv(col2rgb(colorRampPalette(c(x, "#FFFF00"), space = "Lab")(4)[2]))
[,1]
h 0.1597633
s 0.8407960
v 0.7882353
Atte T.
16.4.2016, 19.03, Duncan Murdoch kirjoitti:
On 16/04/2016 8:47 AM, Atte Tenkanen wrote:
Hi,
How would you calculate the "mean colour" of several colours, for
example c("#FF7C00","#00BF40","#FFFF00")?
Bert answered your subject line question. Your text is asking
something else: if those are colours, you don't want to treat each of
them as a single integer.
A simple-minded approach would split them into 3 hex numbers, and
average those (using Bert's solution).
A more sophisticated approach would take into account that they are
really colours. You could probably put together something using the
colorRamp or colorRampPalette functions to average in perception
space. For example,
# Average the 1st two by taking the middle colour of a 3 colour palette
x <- colorRampPalette(c("#FF7C00","#00BF40"), space = "Lab")(3)[2]
# Average in the third by taking the 2nd of a 4 colour palette, so x
# gets twice the weight
colorRampPalette(c(x, "#FFFF00"), space = "Lab")(4)[2]
Duncan Murdoch