I am trying to visualize some values on a form. They range from 0 to 200 and I would like the ones around 0 be green and turn bright red as they go to 200.
Basically the function should return color based on the value inputted. Any ideas ?
Basically, the general method for smooth transition between two values is the following function:
function transition(value, maximum, start_point, end_point):
return start_point + (end_point - start_point)*value/maximum
That given, you define a function that does the transition for triplets (RGB, HSV etc).
function transition3(value, maximum, (s1, s2, s3), (e1, e2, e3)):
r1= transition(value, maximum, s1, e1)
r2= transition(value, maximum, s2, e2)
r3= transition(value, maximum, s3, e3)
return (r1, r2, r3)
Assuming you have RGB colours for the s and e triplets, you can use the transition3 function as-is. However, going through the HSV colour space produces more "natural" transitions. So, given the conversion functions (stolen shamelessly from the Python colorsys module and converted to pseudocode :):
function rgb_to_hsv(r, g, b):
maxc= max(r, g, b)
minc= min(r, g, b)
v= maxc
if minc == maxc then return (0, 0, v)
diff= maxc - minc
s= diff / maxc
rc= (maxc - r) / diff
gc= (maxc - g) / diff
bc= (maxc - b) / diff
if r == maxc then
h= bc - gc
else if g == maxc then
h= 2.0 + rc - bc
else
h = 4.0 + gc - rc
h = (h / 6.0) % 1.0 //comment: this calculates only the fractional part of h/6
return (h, s, v)
function hsv_to_rgb(h, s, v):
if s == 0.0 then return (v, v, v)
i= int(floor(h*6.0)) //comment: floor() should drop the fractional part
f= (h*6.0) - i
p= v*(1.0 - s)
q= v*(1.0 - s*f)
t= v*(1.0 - s*(1.0 - f))
if i mod 6 == 0 then return v, t, p
if i == 1 then return q, v, p
if i == 2 then return p, v, t
if i == 3 then return p, q, v
if i == 4 then return t, p, v
if i == 5 then return v, p, q
//comment: 0 <= i <= 6, so we never come here
, you can have code as following:
start_triplet= rgb_to_hsv(0, 255, 0) //comment: green converted to HSV
end_triplet= rgb_to_hsv(255, 0, 0) //comment: accordingly for red
maximum= 200
… //comment: value is defined somewhere here
rgb_triplet_to_display= hsv_to_rgb(transition3(value, maximum, start_triplet, end_triplet))
You don't say in what environment you're doing this. If you can work with HSV colors, this would be pretty easy to do by setting S = 100 and V = 100, and determining H by:
H = 0.4 * value + 120
Converting from HSV to RGB is also reasonably easy.
[EDIT] Note: in contrast to some other proposed solutions, this will change color green -> yellow -> orange -> red.
red = (float)val / 200 * 255;
green = (float)(200 - val) / 200 * 255;
blue = 0;
return red << 16 + green << 8 + blue;
Pick a green that you like (RGB1 = #00FF00, e.g.) and a Red that you like (RGB2 = #FF0000, e.g.) and then calculate the color like this
R = R1 * (200-i)/200 + R2 * i/200
G = G1 * (200-i)/200 + G2 * i/200
B = B1 * (200-i)/200 + B2 * i/200
For best controllable and accurate effect, you should use the HSV color space. With HSV, you can easily scale Hue, Saturation and/or Brightness seperate from each other. Then, you do the transformation to RGB.
extending upon #tzot's code... you can also set up a mid-point in between the start and end points, which can be useful if you want a "transition color"!
//comment: s = start_triplet, m = mid_triplet, e = end_triplet
function transition3midpoint = (value, maximum, s, m, e):
mid = maximum / 2
if value < mid
return transition3(value, mid, s, m)
else
return transition3(value - mid, mid, m, e)
Looking through this wikipedia article I personally would pick a path through a color space, and map the values onto that path.
But that's a straight function. I think you might be better suited to a javascript color chooser you can find with a quick color that will give you the Hex, and you can store the Hex.
If you use linear ramps for Red and Green values as Peter Parker suggested, the color for value 100 will basically be puke green (127, 127, 0). You ideally want it to be a bright orange or yellow at that midpoint. For that, you can use:
Red = max(value / 100, 1) * 255
Green = (1 - max(value / 100, 1)) * 255
Blue = 0
Related
I am trying to get more comfortable with the math behind fractal coloring and understanding the coloring algorithms much better. I am the following paper:
http://jussiharkonen.com/files/on_fractal_coloring_techniques%28lo-res%29.pdf
The paper gives specific parameters to each of the functions, however when I use the same, my results are not quite right. I have no idea what could be going on though.
I am using the iteration count coloring algorithm to start and using the following julia set:
c = 0.5 + 0.25i and p = 2
with the coloring algorithm:
The coloring function simply returns the number of
elements in the truncated orbit divided by 20
And the palette function:
I(u) = k(u − u0),
where k = 2.5 and u0 = 0, was used.
And with a palette being white at 0 and 1, and interpolating to black in-between.
and following this algorithm:
Set z0 to correspond to the position of the pixel in the complex plane.
Calculate the truncated orbit by iterating the formula zn = f(zn−1) starting
from z0 until either
• |zn| > M, or
• n = Nmax,
where Nmax is the maximum number of iterations.
Using the coloring and color index functions, map the resulting truncated
orbit to a color index value.
Determine an RGB color of the pixel by using the palette function
Using this my code looks like the following:
float izoom = pow(1.001, zoom );
vec2 z = focusPoint + (uv * 4.0 - 2.0) * 1.0 / izoom;
vec2 c = vec2(0.5f, 0.25f) ;
const float B = 2.0;
float l;
for( int i=0; i<100; i++ )
{
z = vec2( z.x*z.x - z.y*z.y, 2.0*z.x*z.y ) + c;
if( length(z)>10.0) break;
l++;
}
float ind = basicindex(l);
vec4 col = color(ind);
and have the following index and coloring functions:
float basicindex(float val){
return val / 20.0;
}
vec4 color(float index){
float r = 2.5 * index;
float g = r;
float b = g;
vec3 v = 0.5 - 0.5 * sin(3.14/2.0 + 3.14 * vec3(r, g, b));
return vec4(1.0 - v, 1.0) ;
}
The paper provides the following image:
https://imgur.com/YIZMhaa
While my code produces:
https://imgur.com/OrxdMsN
I get the correct results by using k = 1.0 instead of 2.5, however I would prefer to understand why my results are incorrect. When extending this to the smooth coloring algorithms, my results are still incorrect so I would like to figure this out first.
Let me know if this isn't the correct place for this kind of question and I can move it to the math stack exchange. I wasn't sure which place was more appropriate.
Your image is perfectly implemented for Figure 3.3 in the paper. The other image you posted uses a different routine.
Your figure seems to have that bit of perspective code there at top, but remove that and they should be the same.
If your objection is the color extremes you set that with the "0.5 - 0.5 * ..." part of your code. This makes the darkest black originally 0.5 when in the example image you're trying to duplicate the darkest black should be 1 and the lightest white should be 0.
You're making the whiteness equal to the distance from 0.5
If you ignore the fractal all together you are getting a bunch of values that can be normalized between 0 and 1 and you're coloring those in some particular ways. Clearly the image you are duplicating is linear between 0 and 1 so putting black as 0.5 cannot be correct.
o = {
length : 500,
width : 500,
c : [.5, .25], // c = x + iy will be [x, y]
maxIterate : 100,
canvas : null
}
function point(pos, color){
var c = 255 - Math.round((1 + Math.log(color)/Math.log(o.maxIterate)) * 255);
c = c.toString(16);
if (c.length == 1) c = '0'+c;
o.canvas.fillStyle="#"+c+c+c;
o.canvas.fillRect(pos[0], pos[1], 1, 1);
}
function conversion(x, y, R){
var m = R / o.width;
var x1 = m * (2 * x - o.width);
var y2 = m * (o.width - 2 * y);
return [x1, y2];
}
function f(z, c){
return [z[0]*z[0] - z[1] * z[1] + c[0], 2 * z[0] * z[1] + c[1]];
}
function abs(z){
return Math.sqrt(z[0]*z[0] + z[1]*z[1]);
}
function init(){
var R = (1 + Math.sqrt(1+4*abs(o.c))) / 2,
z, x, y, i;
o.canvas = document.getElementById('a').getContext("2d");
for (x = 0; x < o.width; x++){
for (y = 0; y < o.length; y++){
i = 0;
z = conversion(x, y, R);
while (i < o.maxIterate && abs(z) < R){
z = f(z, o.c);
if (abs(z) > R) break;
i++;
}
if (i) point([x, y], i / o.maxIterate);
}
}
}
init();
<canvas id="a" width="500" height="500"></canvas>
via: http://jsfiddle.net/3fnB6/29/
I know the following:
P (x,y coordinates).
A1 and A2 (angles in degrees)
R (radius of circle).
Now I need to calculate a center for the green circle, as to make it 'tangentize?' the two blue lines (it will later be minkowski'ed with other shapes to form a point of a rounded triangle).
Any help will be appreciated!
My code for the depicted example:
//KNOWNS
P=[-3.0,1.0,0.0];
A1=60; A2=-5;
R=5;
//UNKNOWN
SECRET_CALCULATION = [8.2,4.3,0];//???
//ILLUSTRATION
C0=[0,0,0,1]; C1=[0,1,1,0.3]; C2=[0,1,0,0.4]; C3=[1,0,0,0.4];//Colors
translate(P){
color(C1) rotate(A1) translate([0,-0.5,0]) square([250,1],0);
color(C1) rotate(A2) translate([0,-0.5,0]) square([250,1],0);
color(C2) translate(SECRET_CALCULATION) circle(R);
}
//EXPLANATIONS
color(C0) translate(P) {
translate(SECRET_CALCULATION){
translate([0,-0.2,0]) square([R,0.4],0);
translate([R+1,0,0]) rotate(-90) text(str("R:",R),halign="center",valign="top",0.75);
translate([0,0.2,0]) text("[x?,y?,0]",halign="center",valign="bottom",0.75);
}
rotate(((A1-A2)/2)-90) translate([0,-1,0]) text(str("P: ",P),halign="center",valign="top",0.75);
rotate(A1) text(str(" A1: ",A1,"°"),halign="left",valign="center",0.75);
rotate(A2) text(str(" A2: ",A2,"°"),halign="left",valign="center",0.75);
}
Center of circle lies at the bisector of rays A1 and A2. So we can find direction for this bisector as
B = (A1 + A2) / 2 = 27.5 here
(but don't forget about angle normalization - angle -5 might be represented as 355 and so on)
And angle between bisector and A1 or A2 is
D = (A1 - A2) / 2 = 32.5 here
We can see that points C(enter)-T(angentpoint)-P form right triangle with relation
L (P-C, DistAlongBisector) = R / Sin(D)
So we can find center coordinates as
C.X = P.X + Cos(B) * L
C.Y = P.Y + Sin(B) * L
Quick hand-check gives
L=5/Sin(32.5)=9.3
C.X = -3 + Cos(27.5)*9.3 = 5.24
C.Y = 1 + Sin(27.5)*9.3 = 5.3
I need to convert 24-bit colors to 4-bit RGBI (1 bit for Red, Green, Blue + Intensity).
Converting to 3-bit RGB is rather simple: set color bit if greater than 127, clear otherwise. However, there's only one intensity bit for all three channels, so what's the correct way to set it (if any)?
First I thought about dividing 8-bit channel to three parts like below:
if 0 <= color <= 85, then clear rgbi-color bit
if 86 <= color <= 170, then set rgbi-color bit
if 171 <= color <= 255, then set rgbi-color bit and intensity
But then I thought that probably the correct way would be to set intensity bit only if two of three channels are greater than 127. But in that case pure R, G, or B will not have intensity ever set (for example, in case of rbg(0,0,200)
Any advice is highly appreciated
A simple way to find the closest 4-bit RGBI approximation of a color is to consider the two possibilities for the intensity bit separately. That is to say, first find the closest RGB0 and RGB1 approximations for the color (which is easy to do, just by dividing each color axis at the appropriate point), and the determine which of these approximations is better.
Here's a simple C-ish pseudocode description of this algorithm:
// find the closest RGBx approximation of a 24-bit RGB color, for x = 0 or 1
function rgbx_approx(red, green, blue, x) {
threshold = (x + 1) * 255 / 3;
r = (red > threshold ? 1 : 0);
g = (green > threshold ? 1 : 0);
b = (blue > threshold ? 1 : 0);
return (r, g, b);
}
// convert a 4-bit RGBI color back to 24-bit RGB
function rgbi_to_rgb24(r, g, b, i) {
red = (2*r + i) * 255 / 3;
green = (2*g + i) * 255 / 3;
blue = (2*b + i) * 255 / 3;
return (red, green, blue);
}
// return the (squared) Euclidean distance between two RGB colors
function color_distance(red_a, green_a, blue_a, red_b, green_b, blue_b) {
d_red = red_a - red_b;
d_green = green_a - green_b;
d_blue = blue_a - blue_b;
return (d_red * d_red) + (d_green * d_green) + (d_blue * d_blue);
}
// find the closest 4-bit RGBI approximation (by Euclidean distance) to a 24-bit RGB color
function rgbi_approx(red, green, blue) {
// find best RGB0 and RGB1 approximations:
(r0, g0, b0) = rgbx_approx(red, green, blue, 0);
(r1, g1, b1) = rgbx_approx(red, green, blue, 1);
// convert them back to 24-bit RGB:
(red0, green0, blue0) = rgbi_to_rgb24(r0, g0, b0, 0);
(red1, green1, blue1) = rgbi_to_rgb24(r1, g1, b1, 1);
// return the color closer to the original:
d0 = color_distance(red, green, blue, red0, green0, blue0);
d1 = color_distance(red, green, blue, red1, green1, blue1);
if (d0 <= d1) return (r0, g0, b0, 0);
else return (r1, g1, b1, 1);
}
Alternatively, you could simply use any generic fixed-palette color quantization algorithm. This may yield better results if your actual color palette is not a pure evenly spaced RGBI palette like the code above assumes, but rather something like e.g. the CGA tweaked RGBI palette.
I'm creating a Color Picker tool and for the HSL slider, I need to be able to convert RGB to HSL. When I searched SO for a way to do the conversion, I found this question HSL to RGB color conversion.
While it provides a function to do conversion from RGB to HSL, I see no explanation to what's really going on in the calculation. To understand it better, I've read the HSL and HSV on Wikipedia.
Later, I've rewritten the function from the "HSL to RGB color conversion" using the calculations from the "HSL and HSV" page.
I'm stuck at the calculation of hue if the R is the max value. See the calculation from the "HSL and HSV" page:
This is from another wiki page that's in Dutch:
and this is from the answers to "HSL to RGB color conversion":
case r: h = (g - b) / d + (g < b ? 6 : 0); break; // d = max-min = c
I've tested all three with a few RGB values and they seem to produce similar (if not exact) results. What I'm wondering is are they performing the same thing? Will get I different results for some specific RGB values? Which one should I be using?
hue = (g - b) / c; // dutch wiki
hue = ((g - b) / c) % 6; // eng wiki
hue = (g - b) / c + (g < b ? 6 : 0); // SO answer
function rgb2hsl(r, g, b) {
// see https://en.wikipedia.org/wiki/HSL_and_HSV#Formal_derivation
// convert r,g,b [0,255] range to [0,1]
r = r / 255,
g = g / 255,
b = b / 255;
// get the min and max of r,g,b
var max = Math.max(r, g, b);
var min = Math.min(r, g, b);
// lightness is the average of the largest and smallest color components
var lum = (max + min) / 2;
var hue;
var sat;
if (max == min) { // no saturation
hue = 0;
sat = 0;
} else {
var c = max - min; // chroma
// saturation is simply the chroma scaled to fill
// the interval [0, 1] for every combination of hue and lightness
sat = c / (1 - Math.abs(2 * lum - 1));
switch(max) {
case r:
// hue = (g - b) / c;
// hue = ((g - b) / c) % 6;
// hue = (g - b) / c + (g < b ? 6 : 0);
break;
case g:
hue = (b - r) / c + 2;
break;
case b:
hue = (r - g) / c + 4;
break;
}
}
hue = Math.round(hue * 60); // °
sat = Math.round(sat * 100); // %
lum = Math.round(lum * 100); // %
return [hue, sat, lum];
}
I've been reading several wiki pages and checking different calculations, and creating visualizations of RGB cube projection onto a hexagon. And I'd like to post my understanding of this conversion. Since I find this conversion (representations of color models using geometric shapes) interesting, I'll try to be as thorough as I can be. First, let's start with RGB.
RGB
Well, this doesn't really need much explanation. In its simplest form, you have 3 values, R, G, and B in the range of [0,255]. For example, 51,153,204. We can represent it using a bar graph:
RGB Cube
We can also represent a color in a 3D space. We have three values R, G, B that corresponds to X, Y, and Z. All three values are in the [0,255] range, which results in a cube. But before creating the RGB cube, let's work on 2D space first. Two combinations of R,G,B gives us: RG, RB, GB. If we were to graph these on a plane, we'd get the following:
These are the first three sides of the RGB cube. If we place them on a 3D space, it results in a half cube:
If you check the above graph, by mixing two colors, we get a new color at (255,255), and these are Yellow, Magenta, and Cyan. Again, two combinations of these gives us: YM, YC, and MC. These are the missing sides of the cube. Once we add them, we get a complete cube:
And the position of 51,153,204 in this cube:
Projection of RGB Cube onto a hexagon
Now that we have the RGB Cube, let's project it onto a hexagon. First, we tilt the cube by 45° on the x, and then 35.264° on the y. After the second tilt, black corner is at the bottom and the white corner is at the top, and they both pass through the z axis.
As you can see, we get the hexagon look we want with the correct hue order when we look at the cube from the top. But we need to project this onto a real hexagon. What we do is draw a hexagon that is in the same size with the cube top view. All the corners of the hexagon corresponds to the corners of the cube and the colors, and the top corner of the cube that is white, is projected onto the center of the hexagon. Black is omitted. And if we map every color onto the hexagon, we get the look at right.
And the position of 51,153,204 on the hexagon would be:
Calculating the Hue
Before we make the calculation, let's define what hue is.
Hue is roughly the angle of the vector to a point in the projection, with red at 0°.
... hue is how far around that hexagon’s edge the point lies.
This is the calculation from the HSL and HSV wiki page. We'll be using it in this explanation.
Examine the hexagon and the position of 51,153,204 on it.
First, we scale the R, G, B values to fill the [0,1] interval.
R = R / 255 R = 51 / 255 = 0.2
G = G / 255 G = 153 / 255 = 0.6
B = B / 255 B = 204 / 255 = 0.8
Next, find the max and min values of R, G, B
M = max(R, G, B) M = max(0.2, 0.6, 0.8) = 0.8
m = min(R, G, B) m = min(0.2, 0.6, 0.8) = 0.2
Then, calculate C (chroma). Chroma is defined as:
... chroma is roughly the distance of the point from the origin.
Chroma is the relative size of the hexagon passing through a point ...
C = OP / OP'
C = M - m
C = 0.8- 0.2 = 0.6
Now, we have the R, G, B, and C values. If we check the conditions, if M = B returns true for 51,153,204. So, we'll be using H'= (R - G) / C + 4.
Let's check the hexagon again. (R - G) / C gives us the length of BP segment.
segment = (R - G) / C = (0.2 - 0.6) / 0.6 = -0.6666666666666666
We'll place this segment on the inner hexagon. Starting point of the hexagon is R (red) at 0°. If the segment length is positive, it should be on RY, if negative, it should be on RM. In this case, it is negative -0.6666666666666666, and is on the RM edge.
Next, we need to shift the position of the segment, or rather P₁ towars the B (because M = B). Blue is at 240°. Hexagon has 6 sides. Each side corresponds to 60°. 240 / 60 = 4. We need to shift (increment) the P₁ by 4 (which is 240°). After the shift, P₁ will be at P and we'll get the length of RYGCP.
segment = (R - G) / C = (0.2 - 0.6) / 0.6 = -0.6666666666666666
RYGCP = segment + 4 = 3.3333333333333335
Circumference of the hexagon is 6 which corresponds to 360°. 53,151,204's distance to 0° is 3.3333333333333335. If we multiply 3.3333333333333335 by 60, we'll get its position in degrees.
H' = 3.3333333333333335
H = H' * 60 = 200°
In the case of if M = R, since we place one end of the segment at R (0°), we don't need to shift the segment to R if the segment length is positive. The position of P₁ will be positive. But if the segment length is negative, we need to shift it by 6, because negative value means that the angular position is greater than 180° and we need to do a full rotation.
So, neither the Dutch wiki solution hue = (g - b) / c; nor the Eng wiki solution hue = ((g - b) / c) % 6; will work for negative segment length. Only the SO answer hue = (g - b) / c + (g < b ? 6 : 0); works for both negative and positive values.
JSFiddle: Test all three methods for rgb(255,71,99)
JSFiddle: Find a color's position in RGB Cube and hue hexagon visually
Working hue calculation:
console.log(rgb2hue(51,153,204));
console.log(rgb2hue(255,71,99));
console.log(rgb2hue(255,0,0));
console.log(rgb2hue(255,128,0));
console.log(rgb2hue(124,252,0));
function rgb2hue(r, g, b) {
r /= 255;
g /= 255;
b /= 255;
var max = Math.max(r, g, b);
var min = Math.min(r, g, b);
var c = max - min;
var hue;
if (c == 0) {
hue = 0;
} else {
switch(max) {
case r:
var segment = (g - b) / c;
var shift = 0 / 60; // R° / (360° / hex sides)
if (segment < 0) { // hue > 180, full rotation
shift = 360 / 60; // R° / (360° / hex sides)
}
hue = segment + shift;
break;
case g:
var segment = (b - r) / c;
var shift = 120 / 60; // G° / (360° / hex sides)
hue = segment + shift;
break;
case b:
var segment = (r - g) / c;
var shift = 240 / 60; // B° / (360° / hex sides)
hue = segment + shift;
break;
}
}
return hue * 60; // hue is in [0,6], scale it up
}
This page provides a function for conversion between color spaces, including RGB to HSL.
function RGBToHSL(r,g,b) {
// Make r, g, and b fractions of 1
r /= 255;
g /= 255;
b /= 255;
// Find greatest and smallest channel values
let cmin = Math.min(r,g,b),
cmax = Math.max(r,g,b),
delta = cmax - cmin,
h = 0,
s = 0,
l = 0;
// Calculate hue
// No difference
if (delta == 0)
h = 0;
// Red is max
else if (cmax == r)
h = ((g - b) / delta) % 6;
// Green is max
else if (cmax == g)
h = (b - r) / delta + 2;
// Blue is max
else
h = (r - g) / delta + 4;
h = Math.round(h * 60);
// Make negative hues positive behind 360°
if (h < 0)
h += 360;
// Calculate lightness
l = (cmax + cmin) / 2;
// Calculate saturation
s = delta == 0 ? 0 : delta / (1 - Math.abs(2 * l - 1));
// Multiply l and s by 100
s = +(s * 100).toFixed(1);
l = +(l * 100).toFixed(1);
return "hsl(" + h + "," + s + "%," + l + "%)";
}
Hue in HSL is like an angle in a circle. Relevant values for such angle reside in the 0..360 interval. However, negative values might come out of the calculation. And that's why those three formulas are different. They do the same in the end, they just handle differently the values outside the 0..360 interval. Or, to be precise, the 0..6 interval which is then eventually multiplied by 60 to 0..360
hue = (g - b) / c; // dutch wiki
does nothing with negative values and presumes the subsequent code can handle negative H values.
hue = ((g - b) / c) % 6; // eng wiki uses the % operator to fit the values inside the 0..6 interval
hue = (g - b) / c + (g < b ? 6 : 0); // SO answer takes care of negative values by adding +6 to make them positive
You see that these are just cosmetic differences. Either the second or the third formula will work fine for you.
Continuing from my comment, the English version looks correct, but I'm not sure what's happening in the Dutch version as I don't understand the WIKI page.
Here is an ES6 version that I made from the English WIKI page, along with some sample data that appear to match the WIKI examples (give or take Javascript's numeric accuracy). Hopefully it may be of use while creating your own function.
// see: https://en.wikipedia.org/wiki/RGB_color_model
// see: https://en.wikipedia.org/wiki/HSL_and_HSV
// expects R, G, B, Cmax and chroma to be in number interval [0, 1]
// returns undefined if chroma is 0, or a number interval [0, 360] degrees
function hue(R, G, B, Cmax, chroma) {
let H;
if (chroma === 0) {
return H;
}
if (Cmax === R) {
H = ((G - B) / chroma) % 6;
} else if (Cmax === G) {
H = ((B - R) / chroma) + 2;
} else if (Cmax === B) {
H = ((R - G) / chroma) + 4;
}
H *= 60;
return H < 0 ? H + 360 : H;
}
// returns the average of the supplied number arguments
function average(...theArgs) {
return theArgs.length ? theArgs.reduce((p, c) => p + c, 0) / theArgs.length : 0;
}
// expects R, G, B, Cmin, Cmax and chroma to be in number interval [0, 1]
// type is by default 'bi-hexcone' equation
// set 'luma601' or 'luma709' for alternatives
// see: https://en.wikipedia.org/wiki/Luma_(video)
// returns a number interval [0, 1]
function lightness(R, G, B, Cmin, Cmax, type = 'bi-hexcone') {
if (type === 'luma601') {
return (0.299 * R) + (0.587 * G) + (0.114 * B);
}
if (type === 'luma709') {
return (0.2126 * R) + (0.7152 * G) + (0.0772 * B);
}
return average(Cmin, Cmax);
}
// expects L and chroma to be in number interval [0, 1]
// returns a number interval [0, 1]
function saturation(L, chroma) {
return chroma === 0 ? 0 : chroma / (1 - Math.abs(2 * L - 1));
}
// returns the value to a fixed number of digits
function toFixed(value, digits) {
return Number.isFinite(value) && Number.isFinite(digits) ? value.toFixed(digits) : value;
}
// expects R, G, and B to be in number interval [0, 1]
// returns a Map of H, S and L in the appropriate interval and digits
function RGB2HSL(R, G, B, fixed = true) {
const Cmin = Math.min(R, G, B);
const Cmax = Math.max(R, G, B);
const chroma = Cmax - Cmin;
// default 'bi-hexcone' equation
const L = lightness(R, G, B, Cmin, Cmax);
// H in degrees interval [0, 360]
// L and S in interval [0, 1]
return new Map([
['H', toFixed(hue(R, G, B, Cmax, chroma), fixed && 1)],
['S', toFixed(saturation(L, chroma), fixed && 3)],
['L', toFixed(L, fixed && 3)]
]);
}
// expects value to be number in interval [0, 255]
// returns normalised value as a number interval [0, 1]
function colourRange(value) {
return value / 255;
};
// expects R, G, and B to be in number interval [0, 255]
function RGBdec2HSL(R, G, B) {
return RGB2HSL(colourRange(R), colourRange(G), colourRange(B));
}
// converts a hexidecimal string into a decimal number
function hex2dec(value) {
return parseInt(value, 16);
}
// slices a string into an array of paired characters
function pairSlicer(value) {
return value.match(/../g);
}
// prepend '0's to the start of a string and make specific length
function prePad(value, count) {
return ('0'.repeat(count) + value).slice(-count);
}
// format hex pair string from value
function hexPair(value) {
return hex2dec(prePad(value, 2));
}
// expects R, G, and B to be hex string in interval ['00', 'FF']
// without a leading '#' character
function RGBhex2HSL(R, G, B) {
return RGBdec2HSL(hexPair(R), hexPair(G), hexPair(B));
}
// expects RGB to be a hex string in interval ['000000', 'FFFFFF']
// with or without a leading '#' character
function RGBstr2HSL(RGB) {
const hex = prePad(RGB.charAt(0) === '#' ? RGB.slice(1) : RGB, 6);
return RGBhex2HSL(...pairSlicer(hex).slice(0, 3));
}
// expects value to be a Map object
function logIt(value) {
console.log(value);
document.getElementById('out').textContent += JSON.stringify([...value]) + '\n';
};
logIt(RGBstr2HSL('000000'));
logIt(RGBstr2HSL('#808080'));
logIt(RGB2HSL(0, 0, 0));
logIt(RGB2HSL(1, 1, 1));
logIt(RGBdec2HSL(0, 0, 0));
logIt(RGBdec2HSL(255, 255, 254));
logIt(RGBhex2HSL('BF', 'BF', '00'));
logIt(RGBstr2HSL('008000'));
logIt(RGBstr2HSL('80FFFF'));
logIt(RGBstr2HSL('8080FF'));
logIt(RGBstr2HSL('BF40BF'));
logIt(RGBstr2HSL('A0A424'));
logIt(RGBstr2HSL('411BEA'));
logIt(RGBstr2HSL('1EAC41'));
logIt(RGBstr2HSL('F0C80E'));
logIt(RGBstr2HSL('B430E5'));
logIt(RGBstr2HSL('ED7651'));
logIt(RGBstr2HSL('FEF888'));
logIt(RGBstr2HSL('19CB97'));
logIt(RGBstr2HSL('362698'));
logIt(RGBstr2HSL('7E7EB8'));
<pre id="out"></pre>
I've been trying to use ImageMagick to replicate Photoshops Colour Blend Mode. I found the following formulas in an online guide but I don't know what they mean. Do I just need to swap certain channels?
A while ago I reversed engineered Photoshop blending modes.
Have a look here:
http://www.kineticsystem.org/?q=node/13
And here below the code I use to convert between HSY (Hue, Saturation, Luminosity) to and from RGB (Red, Green, Blue). Photoshop use something called Hexacones to calculate the saturation.
Giovanni
/**
* This is the formula used by Photoshop to convert a color from
* RGB (Red, Green, Blue) to HSY (Hue, Saturation, Luminosity).
* The hue is calculated using the exacone approximation of the saturation
* cone.
* #param rgb The input color RGB normalized components.
* #param hsy The output color HSY normalized components.
*/
public static void rgbToHsy(double rgb[], double hsy[]) {
double r = Math.min(Math.max(rgb[0], 0d), 1d);
double g = Math.min(Math.max(rgb[1], 0d), 1d);
double b = Math.min(Math.max(rgb[2], 0d), 1d);
double h;
double s;
double y;
// For saturation equals to 0 any value of hue are valid.
// In this case we choose 0 as a default value.
if (r == g && g == b) { // Limit case.
s = 0d;
h = 0d;
} else if ((r >= g) && (g >= b)) { // Sector 0: 0° - 60°
s = r - b;
h = 60d * (g - b) / s;
} else if ((g > r) && (r >= b)) { // Sector 1: 60° - 120°
s = g - b;
h = 60d * (g - r) / s + 60d;
} else if ((g >= b) && (b > r)) { // Sector 2: 120° - 180°
s = g - r;
h = 60d * (b - r) / s + 120d;
} else if ((b > g) && (g > r)) { // Sector 3: 180° - 240°
s = b - r;
h = 60d * (b - g) / s + 180d;
} else if ((b > r) && (r >= g)) { // Sector 4: 240° - 300°
s = b - g;
h = 60d * (r - g) / s + 240d;
} else { // Sector 5: 300° - 360°
s = r - g;
h = 60d * (r - b) / s + 300d;
}
y = R * r + G * g + B * b;
// Approximations erros can cause values to exceed bounds.
hsy[0] = h % 360;
hsy[1] = Math.min(Math.max(s, 0d), 1d);
hsy[2] = Math.min(Math.max(y, 0d), 1d);
}
/**
* This is the formula used by Photoshop to convert a color from
* HSY (Hue, Saturation, Luminosity) to RGB (Red, Green, Blue).
* The hue is calculated using the exacone approximation of the saturation
* cone.
* #param hsy The input color HSY normalized components.
* #param rgb The output color RGB normalized components.
*/
public static void hsyToRgb(double hsy[], double rgb[]) {
double h = hsy[0] % 360;
double s = Math.min(Math.max(hsy[1], 0d), 1d);
double y = Math.min(Math.max(hsy[2], 0d), 1d);
double r;
double g;
double b;
double k; // Intermediate variable.
if (h >= 0d && h < 60d) { // Sector 0: 0° - 60°
k = s * h / 60d;
b = y - R * s - G * k;
r = b + s;
g = b + k;
} else if (h >= 60d && h < 120d) { // Sector 1: 60° - 120°
k = s * (h - 60d) / 60d;
g = y + B * s + R * k;
b = g - s;
r = g - k;
} else if (h >= 120d && h < 180d) { // Sector 2: 120° - 180°
k = s * (h - 120d) / 60d;
r = y - G * s - B * k;
g = r + s;
b = r + k;
} else if (h >= 180d && h < 240d) { // Sector 3: 180° - 240°
k = s * (h - 180d) / 60d;
b = y + R * s + G * k;
r = b - s;
g = b - k;
} else if (h >= 240d && h < 300d) { // Sector 4: 240° - 300°
k = s * (h - 240d) / 60d;
g = y - B * s - R * k;
b = g + s;
r = g + k;
} else { // Sector 5: 300° - 360°
k = s * (h - 300d) / 60d;
r = y + G * s + B * k;
g = r - s;
b = r - k;
}
// Approximations erros can cause values to exceed bounds.
rgb[0] = Math.min(Math.max(r, 0d), 1d);
rgb[1] = Math.min(Math.max(g, 0d), 1d);
rgb[2] = Math.min(Math.max(b, 0d), 1d);
}
Wikipedia has a good article on blend modes
http://en.wikipedia.org/wiki/Blend_modes
They give formulas for Multiply, Screen and Overlay modes.
Multiply
Formula: Result Color = (Top Color) * (Bottom Color) /255
Screen
Formula: Result Color = 255 - [((255 - Top Color)*(255 - Bottom Color))/255]
Overlay
Formula: Result Color = if (Bottom Color < 128)
then (2 * Top Color * Bottom Color / 255)
else (255 - 2 * (255 - Top Color) * (255 - Bottom Color) / 255)
A is the Foreground pixel, B is the Background pixel, C is the new pixel. H is the Hue value for each pixel, S is the Saturation value, L is the Luminance value, and Y is the Brightness value. (Not sure what the difference is between luminance and brightness though.
Anyway, in the first example the Hue(H) and Saturation(S) values of the new pixel(C) are copied from the Foreground pixel(A) while the Brightness(Y) value of the new pixel is taken from the Luminance(L) value of the Background(B) pixel.
These color blending formulas are quite tricky if you need to incorporate also the alpha channel. I was not able to reproduce the blending of Photoshop, but Gimp works like this:
Color mix_hsv(
ColorMixMode::Enum mode, // blending mode
Color cd, // destination color (bottom pixel)
Color cs) // source color (top pixel)
{
// Modify the source color
float dh, ds, dv; // destination hsv
float sh, ss, sv; // source hsv
cd.GetHsv(dh, ds, dv);
cs.GetHsv(sh, ss, sv);
switch (mode) {
case HUE: cs.InitFromHsv(sh, ds, dv); break;
case SATURATION: cs.InitFromHsv(dh, ss, dv); break;
case COLOR: cs.InitFromHsv(sh, ss, dv); break;
case LUMINOSITY: cs.InitFromHsv(dh, ds, sv); break;
}
cs.A = std::min(cd.A, cs.A);
// Blend the modified source color onto the destination color
unsigned char cd_A_orig = cd.A;
cd = mix(NORMAL, cd, cs); // normal blending
cd.A = cd_A_orig;
return cd;
}
If you use premultiplied alpha, don't forget to correctly handle it in the above code. I was not able to find the code for blending in Gimp's source code, but the resulting images are very similar.
Photoshop's color blending is clearly different, so if anyone finds a way to implement it, please let us all know :o)
Miso