Generate Random Color distinguishable to Humans - javascript

I am trying to randomly generate a color in hex in javascript.
However the colors generated are almost indistinguishable from eachother.
Is there a way to improve it?
Here is the code I am using:
function randomColor(){
var allowed = "ABCDEF0123456789", S = "#";
while(S.length < 7){
S += allowed.charAt(Math.floor((Math.random()*16)+1));
}
return S;
}
I heard something about HSL and HSV color model but can't get
it to work in my code.

The easiest way to pick maximally different colors would be to to use HSL values instead of RGB and then manipulate Hue, as it has a value from 0 to 360 value and wraps around (0 is red, and so is 360);
if you need 10 distinguishable colors you can divide 360 by 10 and then pick the individual color by multiplying the value by index (zero based).
Here's an example function that allows you to pick a color from :
function selectColor(colorNum, colors){
if (colors < 1) colors = 1; // defaults to one color - avoid divide by zero
return "hsl(" + (colorNum * (360 / colors) % 360) + ",100%,50%)";
}
This way you can randomize the color selection by randomizing index, but colors will always be in the same palette.
This will select a random color from a palette of 10:
var color = selectColor(Math.floor(Math.random() * 10), 10);
and so will this:
var color = selectColor(Math.floor(Math.random() * 999), 10);
or you can select a specific color from the palette, like 9th color (index 8) out of palette of 13:
var color = selectColor(8, 13);
Here's a fiddle to play with: http://jsfiddle.net/2UE2B/
Update on 2020-02-23:
So, today I needed a solution to this same problem. Googling for this answer here (I know, a very weird way to look for stuff on SO) I ran into the Golden Angle concept. It would make the above example even more trivial, and would not require a predetermined number of colors to be provided:
function selectColor(number) {
const hue = number * 137.508; // use golden angle approximation
return `hsl(${hue},50%,75%)`;
}
This answers the #netoperator-wibby's question

I know I'm very late to this party, but I wrote up a more elaborate function to generate a set of contrasting random colors for another project. They are both (at least somewhat) attractive and genuinely random (not based on predefined colors) but my code is a bit more complicated than some of the other responses (so it's not for just getting the basics)
This is for users who want to have more than one random color on their page, and want to make sure no two colors are too similar.
Fiddle
var generateRandomColors=function(number){
/*
This generates colors using the following algorithm:
Each time you create a color:
Create a random, but attractive, color{
Red, Green, and Blue are set to random luminosity.
One random value is reduced significantly to prevent grayscale.
Another is increased by a random amount up to 100%.
They are mapped to a random total luminosity in a medium-high range (bright but not white).
}
Check for similarity to other colors{
Check if the colors are very close together in value.
Check if the colors are of similar hue and saturation.
Check if the colors are of similar luminosity.
If the random color is too similar to another,
and there is still a good opportunity to change it:
Change the hue of the random color and try again.
}
Output array of all colors generated
*/
//if we've passed preloaded colors and they're in hex format
if(typeof(arguments[1])!='undefined'&&arguments[1].constructor==Array&&arguments[1][0]&&arguments[1][0].constructor!=Array){
for(var i=0;i<arguments[1].length;i++){ //for all the passed colors
var vals = /^#?([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})$/i.exec(arguments[1][i]); //get RGB values
arguments[1][i]=[parseInt(vals[1], 16),parseInt(vals[2], 16),parseInt(vals[3], 16)]; //and convert them to base 10
}
}
var loadedColors=typeof(arguments[1])=='undefined'?[]:arguments[1],//predefine colors in the set
number=number+loadedColors.length,//reset number to include the colors already passed
lastLoadedReduction=Math.floor(Math.random()*3),//set a random value to be the first to decrease
rgbToHSL=function(rgb){//converts [r,g,b] into [h,s,l]
var r=rgb[0],g=rgb[1],b=rgb[2],cMax=Math.max(r,g,b),cMin=Math.min(r,g,b),delta=cMax-cMin,l=(cMax+cMin)/2,h=0,s=0;if(delta==0)h=0;else if(cMax==r)h=60*((g-b)/delta%6);else if(cMax==g)h=60*((b-r)/delta+2);else h=60*((r-g)/delta+4);if(delta==0)s=0;else s=delta/(1-Math.abs(2*l-1));return[h,s,l]
},hslToRGB=function(hsl){//converts [h,s,l] into [r,g,b]
var h=hsl[0],s=hsl[1],l=hsl[2],c=(1-Math.abs(2*l-1))*s,x=c*(1-Math.abs(h/60%2-1)),m=l-c/2,r,g,b;if(h<60){r=c;g=x;b=0}else if(h<120){r=x;g=c;b=0}else if(h<180){r=0;g=c;b=x}else if(h<240){r=0;g=x;b=c}else if(h<300){r=x;g=0;b=c}else{r=c;g=0;b=x}return[r,g,b]
},shiftHue=function(rgb,degree){//shifts [r,g,b] by a number of degrees
var hsl=rgbToHSL(rgb); //convert to hue/saturation/luminosity to modify hue
hsl[0]+=degree; //increment the hue
if(hsl[0]>360){ //if it's too high
hsl[0]-=360 //decrease it mod 360
}else if(hsl[0]<0){ //if it's too low
hsl[0]+=360 //increase it mod 360
}
return hslToRGB(hsl); //convert back to rgb
},differenceRecursions={//stores recursion data, so if all else fails we can use one of the hues already generated
differences:[],//used to calculate the most distant hue
values:[]//used to store the actual colors
},fixDifference=function(color){//recursively asserts that the current color is distinctive
if(differenceRecursions.values.length>23){//first, check if this is the 25th recursion or higher. (can we try any more unique hues?)
//if so, get the biggest value in differences that we have and its corresponding value
var ret=differenceRecursions.values[differenceRecursions.differences.indexOf(Math.max.apply(null,differenceRecursions.differences))];
differenceRecursions={differences:[],values:[]}; //then reset the recursions array, because we're done now
return ret; //and then return up the recursion chain
} //okay, so we still have some hues to try.
var differences=[]; //an array of the "difference" numbers we're going to generate.
for(var i=0;i<loadedColors.length;i++){ //for all the colors we've generated so far
var difference=loadedColors[i].map(function(value,index){ //for each value (red,green,blue)
return Math.abs(value-color[index]) //replace it with the difference in that value between the two colors
}),sumFunction=function(sum,value){ //function for adding up arrays
return sum+value
},sumDifference=difference.reduce(sumFunction), //add up the difference array
loadedColorLuminosity=loadedColors[i].reduce(sumFunction), //get the total luminosity of the already generated color
currentColorLuminosity=color.reduce(sumFunction), //get the total luminosity of the current color
lumDifference=Math.abs(loadedColorLuminosity-currentColorLuminosity), //get the difference in luminosity between the two
//how close are these two colors to being the same luminosity and saturation?
differenceRange=Math.max.apply(null,difference)-Math.min.apply(null,difference),
luminosityFactor=50, //how much difference in luminosity the human eye should be able to detect easily
rangeFactor=75; //how much difference in luminosity and saturation the human eye should be able to dect easily
if(luminosityFactor/(lumDifference+1)*rangeFactor/(differenceRange+1)>1){ //if there's a problem with range or luminosity
//set the biggest difference for these colors to be whatever is most significant
differences.push(Math.min(differenceRange+lumDifference,sumDifference));
}
differences.push(sumDifference); //otherwise output the raw difference in RGB values
}
var breakdownAt=64, //if you're generating this many colors or more, don't try so hard to make unique hues, because you might fail.
breakdownFactor=25, //how much should additional colors decrease the acceptable difference
shiftByDegrees=15, //how many degrees of hue should we iterate through if this fails
acceptableDifference=250, //how much difference is unacceptable between colors
breakVal=loadedColors.length/number*(number-breakdownAt), //break down progressively (if it's the second color, you can still make it a unique hue)
totalDifference=Math.min.apply(null,differences); //get the color closest to the current color
if(totalDifference>acceptableDifference-(breakVal<0?0:breakVal)*breakdownFactor){ //if the current color is acceptable
differenceRecursions={differences:[],values:[]} //reset the recursions object, because we're done
return color; //and return that color
} //otherwise the current color is too much like another
//start by adding this recursion's data into the recursions object
differenceRecursions.differences.push(totalDifference);
differenceRecursions.values.push(color);
color=shiftHue(color,shiftByDegrees); //then increment the color's hue
return fixDifference(color); //and try again
},color=function(){ //generate a random color
var scale=function(x){ //maps [0,1] to [300,510]
return x*210+300 //(no brighter than #ff0 or #0ff or #f0f, but still pretty bright)
},randVal=function(){ //random value between 300 and 510
return Math.floor(scale(Math.random()))
},luminosity=randVal(), //random luminosity
red=randVal(), //random color values
green=randVal(), //these could be any random integer but we'll use the same function as for luminosity
blue=randVal(),
rescale, //we'll define this later
thisColor=[red,green,blue], //an array of the random values
/*
#ff0 and #9e0 are not the same colors, but they are on the same range of the spectrum, namely without blue.
Try to choose colors such that consecutive colors are on different ranges of the spectrum.
This shouldn't always happen, but it should happen more often then not.
Using a factor of 2.3, we'll only get the same range of spectrum 15% of the time.
*/
valueToReduce=Math.floor(lastLoadedReduction+1+Math.random()*2.3)%3, //which value to reduce
/*
Because 300 and 510 are fairly close in reference to zero,
increase one of the remaining values by some arbitrary percent betweeen 0% and 100%,
so that our remaining two values can be somewhat different.
*/
valueToIncrease=Math.floor(valueToIncrease+1+Math.random()*2)%3, //which value to increase (not the one we reduced)
increaseBy=Math.random()+1; //how much to increase it by
lastLoadedReduction=valueToReduce; //next time we make a color, try not to reduce the same one
thisColor[valueToReduce]=Math.floor(thisColor[valueToReduce]/16); //reduce one of the values
thisColor[valueToIncrease]=Math.ceil(thisColor[valueToIncrease]*increaseBy) //increase one of the values
rescale=function(x){ //now, rescale the random numbers so that our output color has the luminosity we want
return x*luminosity/thisColor.reduce(function(a,b){return a+b}) //sum red, green, and blue to get the total luminosity
};
thisColor=fixDifference(thisColor.map(function(a){return rescale(a)})); //fix the hue so that our color is recognizable
if(Math.max.apply(null,thisColor)>255){ //if any values are too large
rescale=function(x){ //rescale the numbers to legitimate hex values
return x*255/Math.max.apply(null,thisColor)
}
thisColor=thisColor.map(function(a){return rescale(a)});
}
return thisColor;
};
for(var i=loadedColors.length;i<number;i++){ //Start with our predefined colors or 0, and generate the correct number of colors.
loadedColors.push(color().map(function(value){ //for each new color
return Math.round(value) //round RGB values to integers
}));
}
//then, after you've made all your colors, convert them to hex codes and return them.
return loadedColors.map(function(color){
var hx=function(c){ //for each value
var h=c.toString(16);//then convert it to a hex code
return h.length<2?'0'+h:h//and assert that it's two digits
}
return "#"+hx(color[0])+hx(color[1])+hx(color[2]); //then return the hex code
});
}
Please note, although I don't do so in my example, that this can also be used to add new distinct, random colors to a set:
generateRandomColors(1,generateRandomColors(10))

You could use a fixed set of colors, such as the ones listed in the jquery.color.js plugin.
List of colors from jquery.color.js plugin:
Colors = {};
Colors.names = {
aqua: "#00ffff",
azure: "#f0ffff",
beige: "#f5f5dc",
black: "#000000",
blue: "#0000ff",
brown: "#a52a2a",
cyan: "#00ffff",
darkblue: "#00008b",
darkcyan: "#008b8b",
darkgrey: "#a9a9a9",
darkgreen: "#006400",
darkkhaki: "#bdb76b",
darkmagenta: "#8b008b",
darkolivegreen: "#556b2f",
darkorange: "#ff8c00",
darkorchid: "#9932cc",
darkred: "#8b0000",
darksalmon: "#e9967a",
darkviolet: "#9400d3",
fuchsia: "#ff00ff",
gold: "#ffd700",
green: "#008000",
indigo: "#4b0082",
khaki: "#f0e68c",
lightblue: "#add8e6",
lightcyan: "#e0ffff",
lightgreen: "#90ee90",
lightgrey: "#d3d3d3",
lightpink: "#ffb6c1",
lightyellow: "#ffffe0",
lime: "#00ff00",
magenta: "#ff00ff",
maroon: "#800000",
navy: "#000080",
olive: "#808000",
orange: "#ffa500",
pink: "#ffc0cb",
purple: "#800080",
violet: "#800080",
red: "#ff0000",
silver: "#c0c0c0",
white: "#ffffff",
yellow: "#ffff00"
};
The rest is simply picking a random property from a Javascript object.
Colors.random = function() {
var result;
var count = 0;
for (var prop in this.names)
if (Math.random() < 1/++count)
result = prop;
return result;
};
Using Colors.random() might get you a human-readable color. I even powered an example below.
(function(){
Colors = {};
Colors.names = {
aqua: "#00ffff",
azure: "#f0ffff",
beige: "#f5f5dc",
black: "#000000",
blue: "#0000ff",
brown: "#a52a2a",
cyan: "#00ffff",
darkblue: "#00008b",
darkcyan: "#008b8b",
darkgrey: "#a9a9a9",
darkgreen: "#006400",
darkkhaki: "#bdb76b",
darkmagenta: "#8b008b",
darkolivegreen: "#556b2f",
darkorange: "#ff8c00",
darkorchid: "#9932cc",
darkred: "#8b0000",
darksalmon: "#e9967a",
darkviolet: "#9400d3",
fuchsia: "#ff00ff",
gold: "#ffd700",
green: "#008000",
indigo: "#4b0082",
khaki: "#f0e68c",
lightblue: "#add8e6",
lightcyan: "#e0ffff",
lightgreen: "#90ee90",
lightgrey: "#d3d3d3",
lightpink: "#ffb6c1",
lightyellow: "#ffffe0",
lime: "#00ff00",
magenta: "#ff00ff",
maroon: "#800000",
navy: "#000080",
olive: "#808000",
orange: "#ffa500",
pink: "#ffc0cb",
purple: "#800080",
violet: "#800080",
red: "#ff0000",
silver: "#c0c0c0",
white: "#ffffff",
yellow: "#ffff00"
};
Colors.random = function() {
var result;
var count = 0;
for (var prop in this.names)
if (Math.random() < 1/++count)
result = prop;
return { name: result, rgb: this.names[result]};
};
var $placeholder = $(".placeholder");
$placeholder.click(function(){
var color = Colors.random();
$placeholder.css({'background-color': color.rgb});
$("#color").html("It's " + color.name);
});
})();
.placeholder {
width: 150px;
height: 150px;
border: 1px solid black;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="placeholder"></div>
<span id="color">Click the square above.</span>

Try this:
function getRandomColor() {
var letters = '0123456789ABCDEF'.split('');
var color = '#';
for (var i = 0; i < 6; i++ ) {
color += letters[Math.round(Math.random() * 15)];
}
return color;
}
See it in action:
http://jsfiddle.net/3wjgG/1/

What you are saying is that you do not want to generate random colors, you are saying that you want to generate different colors.
You can find a good tutorial on how to do it here: http://krazydad.com/tutorials/makecolors.php .
I made this fiddle with the relevant code from the tutorial demonstrating how you would generate non-repeating colors:
http://jsfiddle.net/rGL52/
The only differnce from the tutorial code is that the makegradient() function returns an array of colors that you can later apply at will in your page.

For randomly generating colors, I tend to go for something simple like this:
​function randomColor () {
var max = 0xffffff;
return '#' + Math.round( Math.random() * max ).toString( 16 );
}
​
I'm not sure what you mean by unrecognizable. This method doesn't offer much customization, but at very least makes it easy to keep numbers from being too light or too dark.
If you want to give bigger gaps between the generated colors, you could try reducing the number of allowed characters. I've used a method like that in the past where I only used 0369cf as the pool of characters to pull from. Combining this with a check for duplicates tends to give more distinguishable colors, as well as only utilizing the #fff 3-character syntax.
Here's your original function modified to use this method:
function randomColor(){
var allowed = "0369cf".split( '' ), s = "#";
while ( s.length < 4 ) {
s += allowed.splice( Math.floor( ( Math.random() * allowed.length ) ), 1 );
}
return s;
}

I needed to solve this problem today for a new course I'm writing, so here is my solution:
function getUniqueColor(n) {
const rgb = [0, 0, 0];
for (let i = 0; i < 24; i++) {
rgb[i%3] <<= 1;
rgb[i%3] |= n & 0x01;
n >>= 1;
}
return '#' + rgb.reduce((a, c) => (c > 0x0f ? c.toString(16) : '0' + c.toString(16)) + a, '')
}
It "spreads" the bits from the input number through the RGB values, backwards. It's not perfect, but the code is compact and adjacent colors are distinguishable. Here is the fiddle.

I agree with all the answers, we don't really know what you expect here...
This is a possibility that can give you the choice between the rgb(r, g, b) output for css elements, and the hex output...
This is a quick example, you have just to adapt this draft but it works as it on Firefox :
<script type="text/javascript">
//<![CDATA[
function RndColor(){
var maximum = 255;
var minimum = 100;
var range = maximum - minimum;
var red = Math.floor(Math.random()*range)+minimum;
var green = Math.floor(Math.random()*range)+minimum;
var blue = Math.floor(Math.random()*range)+minimum;
var redToHex = red.toString(16);
var greenToHex = green.toString(16);
var blueToHex = blue.toString(16);
this.rgbValue = "rgb(" + red + "," + green + "," + blue + ")";
this.hexValue = "#" + redToHex + "" + greenToHex + "" + blueToHex;
}
RndColor.prototype.getRGB = function(){
return this.rgbValue;
}
RndColor.prototype.getHex = function(){
return this.hexValue;
}
//]]>
</script>
Then you can retrieve the value as here bellow :
<script type="text/javascript">
//<![CDATA[
rndCol = new RndColor();
document.write("<div style = width:150px;height:100px;background-color:" + rndCol.getHex() + ">" + rndCol.getHex() + "</div><br /><br />");
document.write("<div style = width:150px;height:100px;background-color:" + rndCol.getRGB() + ">" + rndCol.getRGB() + "</div>");
//]]>
</script>
I hope this can help you.
Best regards.

First of all, why are you building hex values from strings? Just use numbers for the values, then output with something like yourNumber.toString(16).
Then, to make the colours more distinct, don't use the full range of 0 to 255 for each colour component, but maybe go in jumps of 10, or 20, or whatever you need to generate wide enough differences.

I wrote up a small script called SwitchColors.js which can be found here: https://github.com/akulmehta/SwitchColors.js
The script produces more saturated colors and the brightness can be controlled. Although it may not produce visually distinguishable colors, it produces high saturation and bright colors which can also be attractive.

Related

Can not complete codewar task: "Which color is the brightest?"

DESCRIPTION for a task:
One of the common ways of representing color is the RGB color model, in which the Red, Green, and Blue primary colors of light are added together in various ways to reproduce a broad array of colors.
One of the ways to determine brightness of a color is to find the value V of the alternative HSV (Hue, Saturation, Value) color model. Value is defined as the largest component of a color:
V = max(R,G,B)
You are given a list of colors in 6-digit hexidecimal notation #RRGGBB. Return the brightest of these colors!
For example,
brightest(["#001000", "#000000"]) == "#001000"
brightest(["#ABCDEF", "#123456"]) == "#ABCDEF"
If there are multiple brightest colors, return the first one:
brightest(["#00FF00", "#FFFF00", "#01130F"]) == "#00FF00"
Note that both input and output should use upper case for characters A, B, C, D, E, F.
My solution:
function brightest(colors){
let colorIndex = 0,
maxValue = 0
for (let i = 0; i < colors.lenght; i++) {
let color = colors[i],
r = ParseInt(color.slise(1,3), 16),
g = ParseInt(color.slise(3,5), 16),
b = ParseInt(color.slise(5,7), 16),
value = Math.max(r,g,b)
if (value > maxValue) {
maxValue = value
colorIndex = i
}
}
return colors [colorIndex]
}
Result:
Test failed with colors = #CAA365,#1861D3,#E8E2C6,#3D3548,#F19BBF,#BF12C3: expected '#CAA365' to deeply equal '#F19BBF'
What is wrong? And how can I fix this?
declare a variable
let value = Math.max(r,g,b)
length not lenght
let/var to variables
parseInt not ParseInt
slice not slise
And all will working

Getting A Color Range From Percentage

What I'm Attempting
I am pulling from an API that gets a number between 1 and 32. I'm attempting to turn that return into a certain background color, ranging from green to red, for good to bad respectively. Basically, if the returning number is 32, the background turns bright red, while if it's 1, it returns bright green.
What I've Tried
Here's the Code currently.
function colorChange(x) {
let y = x.substring(0, x.length - 2); <<<----THIS IS TO TAKE THE NUMBER FROM 32ND TO JUST 32.
let z = y / 2
return "#" + z+z+z
}
<div
className="team-stat"
style={{backgroundColor: colorChange(teamStats.stats[1].splits[0].stat.pts)}}
>
<h4>Points:</h4>
<h1>{teamStats.stats[0].splits[0].stat.pts}</h1>
<h3>Rank: {teamStats.stats[1].splits[0].stat.pts}</h3>
</div>
Currently in the code I'm calling an API that results in a result from 1st to 32nd. I've already taken out the last two characters, so no th or nd are included. I managed to make it so it returns basically a hex code of #171717, but the current range goes from black to slightly-less-black. With the hex codes going into letters, I'm a little stumped on how to proceed.
Any help is appreciated,and I'll happily clarify on any misunderstandings.
You could use rgb() for the background-color then scale the red and green values depending on the x value. Assuming 32 is the most green, something like:
function colorChange(x) {
...
const maxPossibleScore = 32;
const individualPointValue = 255 / maxPossibleScore;
return {
red: 255 - individualPointValue * x,
green: individualPointValue * x
}
}
const colors = colorChange(10);
style={{backgroundColor: `rgb(${colors.red}, ${colors.green}, 0)`}}
This doesn't answer your question directly using hex but you in theory could slot in a decimal to hex translation and get the hex result you're looking for.
Use HSLA model for the background color. Then change the first parameter (hue) based on your value from 1 to 32 multiplied by 3.75 (120/32)
Here's an example: https://codepen.io/akost/pen/ZEJWZVx
<script>
let list='';
for(let i = 1; i <= 32; i++) {
let h = i*3.75; // we need hue values from 0 to 120
let row = `<div style='background:hsla(${h}, 100%, 50%, 1)'></div>`;
list += row;
}
document.getElementById('foo').innerHTML = list;
</script>
<style>
#foo div {
width:100px;
height:10px;
border-bottom:solid 1px #fff;
}
</style>
<div id="foo"></div>

Combining color hex blending algorithm with standard CMYK colored buttons

I am trying to blend any combination of colors using colored buttons which output a specific hex number in combination with a slider bar from bootstrap that allows the user to indicate the percentage of color they want to use.
I couldn't get the slider to properly run and I'm not sure why.
I do need some help combining the pieces of code into a working algorithm that I can use for my art class. The full code can be found here:
https://jsfiddle.net/mw7optL5/289/
//Hex blending algorithm
var mix = function(color_1, color_2, weight) {
function d2h(d) { return d.toString(16); } // convert a decimal value to hex
function h2d(h) { return parseInt(h, 16); } // convert a hex value to decimal
weight = (typeof(weight) !== 'undefined') ? weight : 50; // set the weight to 50%, if that argument is omitted
var color = "#";
for(var i = 0; i <= 5; i += 2) { // loop through each of the 3 hex pairs—red, green, and blue
var v1 = h2d(color_1.substr(i, 2)), // extract the current pairs
v2 = h2d(color_2.substr(i, 2)),
// combine the current pairs from each source color, according to the specified weight
val = d2h(Math.floor(v2 + (v1 - v2) * (weight / 100.0)));
while(val.length < 2) { val = '0' + val; } // prepend a '0' if val results in a single digit
color += val; // concatenate val to our new color string
}
return color; // PROFIT!
};
//buttons
<button class="Yellow" onclick="mix('#ffff00')">Yellow</button>
<button class="Magenta" onclick="mix('#D9017A')">Magenta</button>
<button class="Cyan" onclick="mix('#009FDF')">Cyan</button>
<button class="Black" onclick="mix('#2D2926')">Black</button>
<button class="Orange021" onclick="mix('#FE5000')">Orange</button>
//slider bar
<input id="ex1" data-slider-id='ex1Slider' type="text" data-slider-min="0" data-slider-max="20" data-slider-step="1" data-slider-value="14"/>
var slider = new Slider('#ex1', {
formatter: function(value) {
return 'Current value: ' + value;
}
});
This just solves the mechanics I think you want (run and save the slider value properly), and it could be used as starter point. This doesn't solve the formula of any combination of colors as you want.
Create an object based in the buttons with all percentages (0 default) like { Yellow: 0, Magenta: 0, Cyan: 0, Black: 0 }
Add an event listener to each button .click (jquery) or .addEventListener (pure js)
In the callback of each button, save the color and percentage into the object, like colorPercentages[colors[i].getAttribute("id")] = $("#ex1").val(); where $("#ex1").val() is the slider's value.
Draw the color based in some formula, like $("body").css('background-color', 'rgb(' + r + ', ' + g + ', ' + b + ')');
In this case I've used this formula:
Red = 255 x ( 1 - Cyan / 100 ) x ( 1 - Black / 100 )
Green = 255 x ( 1 - Magenta / 100 ) x ( 1 - Black / 100 )
Blue = 255 x ( 1 - Yellow / 100 ) x ( 1 - Black / 100 )
Here a full functional example based on #barbsan modifications: https://jsfiddle.net/5dsgbne9/
That example is using the color button click event, but you could use the inverse logic as well, changing the color when user moves the slider with something like:
$("#ex1").slider().on('slideStop', function(ev){
//code
});
As I understand, a blending algorithm it's pretty difficult to achieve, this question and the answers give some good approaches: Mixing two colors "naturally" in javascript, but you still need to mix n colors, not just 2, which is even more complex. Fortunately it seems to be not impossible.

How to generate skewed random numbers in Javascript?

Using Javascript, how can I generate random numbers that are skewed towards one end or the other of the distribution? Or ideally an point within the range?
For context: I'm creating a UI that has uses a grid of random grey squares. I'm generating the grey's RGB values using Math.random() but would like to be able to skew the greys to be on average darker or lighter while still having the full range from black to white represented.
(I think this is a similar question to Skewing java random number generation toward a certain number but I'm working with Javascript...)
Any help greatly appreciated.
Raise Math.random() to a power to get a gamma curve - this changes the distribution between 0 and 1, but 0 and 1 stay constant endpoints.
var r= Math.pow(Math.random(), 2);
var colour= 'rgb('+r*255+', '+r*255+', '+r*255+')';
For gamma>1, you will get darker output; for 0<gamma<1 you get lighter. (Here, '2' gives you the x-squared curve; the equidistant lightness would be '0.5' for the square-root curve.)
This seems a little crude and less graceful than #bobince's answer, but what the hell.
//setup
var colours = [], num_colours = 10, skew_to = 255, skew_chance = 20;
//get as many RGB vals as required
for (var i=0; i<num_colours; i++) {
//generate random grey
var this_grey = Math.floor(Math.random() * 256);
//skew it towards the #skew_to endpoint, or leave as-is?
if (Math.floor(Math.random() * 100) >= skew_chance && this_grey != skew_to) {
//skew by random amount (0 - difference between curr val and endpoint)
var skew_amount = Math.floor(Math.random() * Math.abs(this_grey - skew_to));
this_grey += ' (skewed to '+(skew_to < this_grey ? this_grey - skew_amount : this_grey + skew_amount)+')';
}
colours.push(this_grey);
}
console.log(colours);
Essentially it generates random greys then decides, based on probably specified (as a percentage) in skew_chance, whether to skew it or not. (In case you wanted to make this occasional, not constant). If it decides to skew, a random number is then added or subtracted from the grey value (depending on whether the skew endpoint is under or above the current value).
This random number is a number between 0 and the absolute difference between the current value and the endpoint, e.g. if current value is 40, and the endpoint is 100, the number added would be between 0 and 60.
Like I say, #bobince's answer is somewhat, er, more graceful!
[This might be a little different approach.]
This approach deals with getting the number in the following fashion:
random = numberToSkewTo + random(-1,1)*stdDeviation
Where:
numberToSkewTo is the number you want to skew towards.
stdDeviation is the deviation from numberToSkewTo
numberToSkewTo + abs(stdDeviation) <= MAX_NUMBER and
numberToSkewTo - abs(stdDeviation) >= MIN_NUMBER
What the following code does is, it pick a random number around the given number with constantly increasing standard deviations. It returns the average of results.
function skew(skewTo,stdDev){
var rand = (Math.random()*2 - 1) + (Math.random()*2 - 1) + (Math.random()*2 - 1);
return skewTo + rand*stdDev;
}
function getRandom(skewTo){
var difference = Math.min(skewTo-MIN_NUMBER, MAX_NUMBER-skewTo);
var steps = 5;
var total = 0.0;
for(var i=1; i<=steps; i++)
total += skew(skewTo, 1.0*i*difference/steps);
return total/steps
}

How to figure out all colors in a gradient? [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Javascript color gradient
I have color one (let's say yellow) and color two (blue) - they make up a gradient.
Based on a value of 0 to 100, (0 being yellow and 100 being blue), I'd like to represent a mixture of color one and two.
I am trying to do this in a mobile browser (safari specifically).
Is there a way to do this in javascript?
If what you're trying to do is to create a color that is some percentage (0-100) between two other colors, you can do that with this javascript:
function makeGradientColor(color1, color2, percent) {
var newColor = {};
function makeChannel(a, b) {
return(a + Math.round((b-a)*(percent/100)));
}
function makeColorPiece(num) {
num = Math.min(num, 255); // not more than 255
num = Math.max(num, 0); // not less than 0
var str = num.toString(16);
if (str.length < 2) {
str = "0" + str;
}
return(str);
}
newColor.r = makeChannel(color1.r, color2.r);
newColor.g = makeChannel(color1.g, color2.g);
newColor.b = makeChannel(color1.b, color2.b);
newColor.cssColor = "#" +
makeColorPiece(newColor.r) +
makeColorPiece(newColor.g) +
makeColorPiece(newColor.b);
return(newColor);
}
This function assumes the gradient is made with linear interpolation between each r, g and b channel value of the two endpoint colors such that the 50% gradient value is the midpoint of each r,g,b value (halfway between the two colors presented). Once could make different types of gradients too (with different interpolation functions).
To assign this result to a background, you use the CSS color value I've added to the return result like this:
// sample usage
var yellow = {r:255, g:255, b:0};
var blue = {r:0, g:0, b:255};
var newColor = makeGradientColor(yellow, blue, 79);
element.style.backgroundColor = newColor.cssColor;

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