问题
Im looking to align the following array of colours as precisely as possible.
After searching & trying many solutions suggested on Stackoverflow, the pusher.color library has the best solution, however, it too is far from perfect. I would like to hear solutions on how we can align them perfectly.
JSFIDDLE LINK : http://jsfiddle.net/dxux7y3e/
Code:
var coloursArray=['#FFE9E9','#B85958','#FFB1AE','#FFC2BF','#C55E58','#FFC7C4','#FF9A94','#FF9D96','#FA9790','#A78B88','#A78B88','#CE675B','#DB8073','#FF9D90','#FF7361','#FFD6D1','#F9A092','#FF7B67','#EBACA2','#FF806D','#DD6D5B','#D16654','#ED8673','#FFC4B8','#E2725B','#ED7A64','#8F3926','#BD492F','#9D3C27','#AD533E','#BF4024','#FFC9BC','#6B6766','#E1CDC8','#C2654C','#B3978F','#FFC7B8','#CE2B00','#C2654C','#A24D34','#FF926D','#E78667','#FFB198','#8C756D','#9E6D5B','#FFC7B0','#FFBEA4','#D2B9AF','#FFB193','#632710','#B26746','#976854','#F44900','#E79873','#EFA27F','#532510','#BC866B','#FDE5D9','#FF5B00','#D18C67','#FF5B00','#9E4312','#763713','#BB6B39','#B5622E','#CC7742','#6D4227','#B56B38','#FF7518','#F3B080','#995C30','#995C30','#FF6A00','#D89769','#71472A','#EDAC7B','#EEAB79','#EBCFB9','#FBE3D1','#E19255','#5E381B','#FFDCC1','#FFF0E4','#F68D39','#7B5B40','#FF8313','#FFCEA4','#AA8667','#975414','#CB9867','#8C5B2B','#FFCE9E','#7B4714','#FFF3E7','#FFA449','#CEAF90','#CDB69E','#EFD6BC','#DDA66B','#B27737','#B88A57','#CE9B61','#F4C38B','#543817','#BC9C78','#DBB07A','#FF8E04','#F6EADB','#DBC2A4','#C49B64','#CBA26B','#80551E','#FF9200','#FFECD3','#FFC87C','#FFB755','#DBB680','#D2D0CD','#EFDBBE','#E5C18B','#FFE5BC','#F2EADB','#885F12','#FFE7B6','#825A08','#906712','#F2D18E','#C8C6C2','#FFB000','#FFC243','#C6BEAD','#D0C3A4','#916800','#8C6700','#F4E9CA','#FFF0C5','#FFE080','#FFEBA8','#846600','#FFE692','#F5F0DB','#433F2F','#BBB394','#FFEFAA','#FFE76D','#FFFAE0','#3E3B28','#554900','#E1E0D8','#74725C','#605F54','#F8F7DD','#A5A467','#DDDDDA','#FFFFEE','#A3A39D','#E0E0D7','#BEBEB9','#E8E8E5','#454531','#ACACAA','#E9E9DF','#FFFFDC','#EBEBE7','#979831','#C5C6BE','#B9C866','#898D72','#F3FAD1','#616452','#CED5B0','#A1A787','#595C4E','#B0BB8C','#EEFFB6','#ACB78E','#8FA359','#858F6C','#86916E','#374912','#AEB0AA','#79904C','#627739','#747F60','#9FA98E','#E7F9CB','#E1F9BE','#495637','#8A9978','#4E5F39','#86996E','#C3CEB7','#78866B','#CEDDC1','#B5CEA2','#536149','#D6E6CC','#D6E6CC','#809873','#4F564C','#4F6C45','#555F52','#4F7942','#5F705B','#D0DFCD','#2B3929','#F0F7EF','#AAD5A4','#99BC95','#B6D4B4','#869E86','#618661','#006700','#E9EEE9','#739E73','#005B06','#EDF7EE','#D0E0D2','#809784','#ABCEB1','#C0E0C8','#3A5241','#435549','#E6ECE8','#E3EAE6','#3B604C','#00602F','#92B7A5','#2F5B49','#318061','#30745B','#316955','#00A275','#C2D1CE','#80A7A0','#00A082','#C2D1CF','#5C6E6C','#607473','#EDF7F7','#1E8285','#D5E7E8','#AADEE1','#188086','#107F87','#566364','#007B86','#66949A','#CAE2E5','#18656F','#004F61','#0C5B6C','#668E98','#BBD0DA','#91B4C5','#AFC3CD','#738A99','#3A5467','#476174','#244967','#556C80','#667A8C','#516D87','#1E4263','#7C8791','#849CB6','#738CAA','#1E3A5F','#1E3655','#9EB0CE','#B6BAC2','#67738D','#BEC1CD','#555559','#616180','#000049','#000031','#F8F8FC','#938BA4','#47375D','#F7F6F8','#3D0067','#514C53','#9566A2','#7F5482','#A279A4','#6D1261','#A06492','#925582','#945B80','#CE94BA','#ECCFE1','#A20058','#A6005B','#BC0061','#BB0061','#F3CEE1','#B3005B','#AB165F','#8A184D','#AA185B','#F3DAE4','#DB3779','#E71261','#E74F86','#FFD6E5','#BE9BA7','#D0396A','#DB1855','#F798B6','#9C294A','#D62B5B','#DE3969','#BC1641','#E7547A','#D52756','#9C7D85','#DB244F','#A1354F','#C22443','#FFBDCA','#8B6D73','#DC3D5B','#FF738C','#F13154','#BC4055','#FED4DB','#FFCFD6','#CB4E61','#ED455A','#F36C7B','#C94F5B','#F3959D','#A8444C','#FFCCD0','#735B5D','#D15D67','#B44B52','#FD868D','#FFD5D8','#C3767B','#FF8087','#C8242B','#FFEAEB','#F95A61','#E96D73','#E6656B','#FF6D73','#FF555B','#A35A5B','#FFD3D4','#B84B4D'];
var body=document.getElementsByTagName('body')[0];
function hexToRgb(hex) {
hex = hex.substring(1, hex.length);
var r = parseInt((hex).substring(0, 2), 16);
var g = parseInt((hex).substring(2, 4), 16);
var b = parseInt((hex).substring(4, 6), 16);
return r + "," + g + "," + b;
}
function rgbToHex(r, g, b) {
return "#" + ((1 << 24) + (r << 16) + (g << 8) + b).toString(16).slice(1);
}
var rgbArr=new Array();
var div=document.createElement('div');
div.id='Original';
body.appendChild(div);
for(var color in coloursArray){
color=coloursArray[color];
displayColor(color,div);
rgbArr.push(hexToRgb(color));
}
var hslArr=new Array();
for(var i=0;i<rgbArr.length;i++){
//Transforming rgb to hsl
//`hslArr[i][1]` (`i`) is a reference to the rgb color, in order to retrieve it later
hslArr[i]=[rgbToHsl(rgbArr[i]),i];
}
var sortedHslArr=new Array();
//Sorting `hslArr` into `sortedHslArr`
outerloop:
for(var i=0;i<hslArr.length;i++){
for(var j=0;j<sortedHslArr.length;j++){
if(sortedHslArr[j][0][0]>hslArr[i][0][0]){
sortedHslArr.splice(j,0,hslArr[i]);
continue outerloop;
}
}
sortedHslArr.push(hslArr[i]);
}
var sortedRgbArr=new Array();
//Retrieving rgb colors
for(var i=0;i<sortedHslArr.length;i++){
sortedRgbArr[i]=rgbArr[sortedHslArr[i][1]];
}
function displayColor(color,parent){
var div;
div=document.createElement('div');
div.style.backgroundColor=color;
div.style.width='22px';
div.style.height='22px';
div.style.cssFloat='left';
div.style.position='relative';
parent.appendChild(div);
}
var finalArray=new Array();
var div=document.createElement('div');
div.id='Sorted';
body.appendChild(div);
for(var color in sortedRgbArr){
color=sortedRgbArr[color];
color=color.split(',');
color=rgbToHex(parseInt(color[0]),parseInt(color[1]),parseInt(color[2]));
displayColor(color,div);
finalArray.push(color);
}
function rgbToHsl(c){
var r = c[0]/255, g = c[1]/255, b = c[2]/255;
var max = Math.max(r, g, b), min = Math.min(r, g, b);
var h, s, l = (max + min) / 2;
if(max == min){
h = s = 0; // achromatic
}else{
var d = max - min;
s = l > 0.5 ? d / (2 - max - min) : d / (max + min);
switch(max){
case r: h = (g - b) / d + (g < b ? 6 : 0); break;
case g: h = (b - r) / d + 2; break;
case b: h = (r - g) / d + 4; break;
}
h /= 6;
}
return new Array(h * 360, s * 100, l * 100);
}
var sorted = coloursArray.sort(function(colorA, colorB) {
return pusher.color(colorA).hue() - pusher.color(colorB).hue();
});
// console.log(sorted);
var div=document.createElement('div');
div.id='Pusher';
body.appendChild(div);
for(var color in sorted){
color=sorted[color];
displayColor(color,div);
}
var div=document.createElement('div');
body.appendChild(div);
var str='';
for(var color in sorted){
color=sorted[color];
str+='\''+color+'\',';
}
div.innerHTML=str;
function sorthueColors (colors) {
for (var c = 0; c < colors.length; c++) {
/* Get the hex value without hash symbol. */
var hex = colors[c].substring(1);
//var hex = colors[c].hex.substring(1);
/* Get the RGB values to calculate the Hue. */
var r = parseInt(hex.substring(0,2),16)/255;
var g = parseInt(hex.substring(2,4),16)/255;
var b = parseInt(hex.substring(4,6),16)/255;
/* Getting the Max and Min values for Chroma. */
var max = Math.max.apply(Math, [r,g,b]);
var min = Math.min.apply(Math, [r,g,b]);
/* Variables for HSV value of hex color. */
var chr = max-min;
var hue = 0;
var val = max;
var sat = 0;
if (val > 0) {
/* Calculate Saturation only if Value isn't 0. */
sat = chr/val;
if (sat > 0) {
if (r == max) {
hue = 60*(((g-min)-(b-min))/chr);
if (hue < 0) {hue += 360;}
} else if (g == max) {
hue = 120+60*(((b-min)-(r-min))/chr);
} else if (b == max) {
hue = 240+60*(((r-min)-(g-min))/chr);
}
}
}
/* Modifies existing objects by adding HSV values. */
colors[c].hue = hue;
colors[c].sat = sat;
colors[c].val = val;
}
/* Sort by Hue. */
return colors.sort(function(a,b){return a.hue - b.hue;});
}
回答1:
Problem is, sorting requires a well-defined order - in other words, you need to map all colors to one dimension. While there are some approaches to display the color space in two dimensions, I am not aware of any that would display it in one dimension and still make sense to the human eye.
However, if you don't insist on some universal ordering and you merely want to place a given list of colors in a way that looks nice, then some clustering approach might produce better results. I tried the naive approach, the idea here is to put similar colors into the same cluster and to merge these clusters until you only have one. Here is the code I've got:
function colorDistance(color1, color2) {
// This is actually the square of the distance but
// this doesn't matter for sorting.
var result = 0;
for (var i = 0; i < color1.length; i++)
result += (color1[i] - color2[i]) * (color1[i] - color2[i]);
return result;
}
function sortColors(colors) {
// Calculate distance between each color
var distances = [];
for (var i = 0; i < colors.length; i++) {
distances[i] = [];
for (var j = 0; j < i; j++)
distances.push([colors[i], colors[j], colorDistance(colors[i], colors[j])]);
}
distances.sort(function(a, b) {
return a[2] - b[2];
});
// Put each color into separate cluster initially
var colorToCluster = {};
for (var i = 0; i < colors.length; i++)
colorToCluster[colors[i]] = [colors[i]];
// Merge clusters, starting with lowest distances
var lastCluster;
for (var i = 0; i < distances.length; i++) {
var color1 = distances[i][0];
var color2 = distances[i][1];
var cluster1 = colorToCluster[color1];
var cluster2 = colorToCluster[color2];
if (!cluster1 || !cluster2 || cluster1 == cluster2)
continue;
// Make sure color1 is at the end of its cluster and
// color2 at the beginning.
if (color1 != cluster1[cluster1.length - 1])
cluster1.reverse();
if (color2 != cluster2[0])
cluster2.reverse();
// Merge cluster2 into cluster1
cluster1.push.apply(cluster1, cluster2);
delete colorToCluster[color1];
delete colorToCluster[color2];
colorToCluster[cluster1[0]] = cluster1;
colorToCluster[cluster1[cluster1.length - 1]] = cluster1;
lastCluster = cluster1;
}
// By now all colors should be in one cluster
return lastCluster;
}
Complete fiddle
The colorDistance() function works with RGB colors that are expressed as arrays with three elements. There are certainly much better approaches to do this and they might produce results that look better. Note also that this isn't the most efficient algorithm because it calculates the distance between each and every color (O(n2)) so if you have lots of colors you might want to optimize it.
回答2:
Assuming there is no best solution. Here is how i would find what i like the most.
First i would use color distance function:
var balance = [10, 0, 0.01];
function colorDistance(color1, color2) {
var result = 0;
color1 = rgbToHsl(color1[0], color1[1], color1[2]);
color2 = rgbToHsl(color2[0], color2[1], color2[2]);
for (var i = 0; i < color1.length; i++)
result += (color1[i] - color2[i]) * (color1[i] - color2[i]) * balance[i];
return result;
}
I would use balance function :
- hue distance
var balance = [1, 0, 0];
- saturation distance
var balance = [0, 1, 0]; lightness distancevar balance = [0, 0, 1];
After that you can just experiment and pick what you like. Note that values are not normalized to appear between 0 to 1 but other then that it should work well. You can also you another distance function this is just most trivial and well known one that gets decent results.
This approach does not give you the best solution, but describes a way to fiddle with it and get result you where looking for. Have fun.
回答3:
Using the Hue of a color is usually a pleasant way to map the color on a one-dimensional scale. You could convert the rgb color to hsv using the answers here: RGB to HSV color in javascript?
回答4:
This is an improved code based on pusher.color lib.
https://jsfiddle.net/kt1zv6g4/3/
来源：https://stackoverflow.com/questions/25822124/sort-colour-color-values