C# 在图形中生成明显不同的 RGB 颜色
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原文地址: http://stackoverflow.com/questions/309149/
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Generate distinctly different RGB colors in graphs
提问by Riri
When generating graphs and showing different sets of data it usually a good idea to difference the sets by color. So one line is red and the next is green and so on. The problem is then that when the number of datasets is unknown one needs to randomly generate these colors and often they end up very close to each other (green, light green for example).
在生成图表并显示不同的数据集时，通过颜色区分数据集通常是一个好主意。所以一条线是红色的，下一条线是绿色的，依此类推。问题是，当数据集的数量未知时，需要随机生成这些颜色，而且它们通常彼此非常接近（例如绿色、浅绿色）。
Any ideas on how this could be solved and how it would be possibler to generate distinctly different colors?
关于如何解决这个问题以及如何产生明显不同的颜色的任何想法？
I'd be great if any examples (feel free to just discuss the problem and solution without examples if you find that easier) were in C# and RGB based colors.
如果任何示例（如果您觉得更简单，请随意讨论没有示例的问题和解决方案）是基于 C# 和 RGB 的颜色，我会很棒。
采纳答案by Sam Meldrum
You have three colour channels 0 to 255 R, G and B.
您有 0 到 255 R、G 和 B 三个颜色通道。
First go through
首先通过
0, 0, 255
0, 255, 0
255, 0, 0
Then go through
然后通过
0, 255, 255
255, 0, 255
255, 255, 0
Then divide by 2 => 128 and start again:
然后除以 2 => 128 并重新开始：
0, 0, 128
0, 128, 0
128, 0, 0
0, 128, 128
128, 0, 128
128, 128, 0
Divide by 2 => 64
除以 2 => 64
Next time add 64 to 128 => 192
下次将 64 添加到 128 => 192
follow the pattern.
根据规律。
Straightforward to program and gives you fairly distinct colours.
直接编程并为您提供相当不同的颜色。
EDIT: Request for code sample
编辑：请求代码示例
Also - adding in the additional pattern as below if gray is an acceptable colour:
另外 - 如果灰色是可接受的颜色，则添加如下附加图案：
255, 255, 255
128, 128, 128
There are a number of ways you can handle generating these in code.
有多种方法可以处理在代码中生成这些。
The Easy Way
简单的方法
If you can guarantee that you will never need more than a fixed number of colours, just generate an array of colours following this pattern and use those:
如果您可以保证您永远不会需要超过固定数量的颜色，只需按照此模式生成一组颜色并使用它们：
static string[] ColourValues = new string[] {
"FF0000", "00FF00", "0000FF", "FFFF00", "FF00FF", "00FFFF", "000000",
"800000", "008000", "000080", "808000", "800080", "008080", "808080",
"C00000", "00C000", "0000C0", "C0C000", "C000C0", "00C0C0", "C0C0C0",
"400000", "004000", "000040", "404000", "400040", "004040", "404040",
"200000", "002000", "000020", "202000", "200020", "002020", "202020",
"600000", "006000", "000060", "606000", "600060", "006060", "606060",
"A00000", "00A000", "0000A0", "A0A000", "A000A0", "00A0A0", "A0A0A0",
"E00000", "00E000", "0000E0", "E0E000", "E000E0", "00E0E0", "E0E0E0",
};
The Hard Way
艰难的道路
If you don't know how many colours you are going to need, the code below will generate up to 896 colours using this pattern. (896 = 256 * 7 / 2) 256 is the colour space per channel, we have 7 patterns and we stop before we get to colours separated by only 1 colour value.
如果您不知道需要多少种颜色，下面的代码将使用此图案生成多达 896 种颜色。(896 = 256 * 7 / 2) 256 是每个通道的颜色空间，我们有 7 个图案，我们在得到仅由 1 个颜色值分隔的颜色之前停止。
I've probably made harder work of this code than I needed to. First, there is an intensity generator which starts at 255, then generates the values as per the pattern described above. The pattern generator just loops through the seven colour patterns.
我可能比我需要的更努力地处理这段代码。首先，有一个从 255 开始的强度生成器，然后按照上述模式生成值。图案生成器只是循环遍历七种颜色图案。
using System;
class Program {
static void Main(string[] args) {
ColourGenerator generator = new ColourGenerator();
for (int i = 0; i < 896; i++) {
Console.WriteLine(string.Format("{0}: {1}", i, generator.NextColour()));
}
}
}
public class ColourGenerator {
private int index = 0;
private IntensityGenerator intensityGenerator = new IntensityGenerator();
public string NextColour() {
string colour = string.Format(PatternGenerator.NextPattern(index),
intensityGenerator.NextIntensity(index));
index++;
return colour;
}
}
public class PatternGenerator {
public static string NextPattern(int index) {
switch (index % 7) {
case 0: return "{0}0000";
case 1: return "00{0}00";
case 2: return "0000{0}";
case 3: return "{0}{0}00";
case 4: return "{0}00{0}";
case 5: return "00{0}{0}";
case 6: return "{0}{0}{0}";
default: throw new Exception("Math error");
}
}
}
public class IntensityGenerator {
private IntensityValueWalker walker;
private int current;
public string NextIntensity(int index) {
if (index == 0) {
current = 255;
}
else if (index % 7 == 0) {
if (walker == null) {
walker = new IntensityValueWalker();
}
else {
walker.MoveNext();
}
current = walker.Current.Value;
}
string currentText = current.ToString("X");
if (currentText.Length == 1) currentText = "0" + currentText;
return currentText;
}
}
public class IntensityValue {
private IntensityValue mChildA;
private IntensityValue mChildB;
public IntensityValue(IntensityValue parent, int value, int level) {
if (level > 7) throw new Exception("There are no more colours left");
Value = value;
Parent = parent;
Level = level;
}
public int Level { get; set; }
public int Value { get; set; }
public IntensityValue Parent { get; set; }
public IntensityValue ChildA {
get {
return mChildA ?? (mChildA = new IntensityValue(this, this.Value - (1<<(7-Level)), Level+1));
}
}
public IntensityValue ChildB {
get {
return mChildB ?? (mChildB = new IntensityValue(this, Value + (1<<(7-Level)), Level+1));
}
}
}
public class IntensityValueWalker {
public IntensityValueWalker() {
Current = new IntensityValue(null, 1<<7, 1);
}
public IntensityValue Current { get; set; }
public void MoveNext() {
if (Current.Parent == null) {
Current = Current.ChildA;
}
else if (Current.Parent.ChildA == Current) {
Current = Current.Parent.ChildB;
}
else {
int levelsUp = 1;
Current = Current.Parent;
while (Current.Parent != null && Current == Current.Parent.ChildB) {
Current = Current.Parent;
levelsUp++;
}
if (Current.Parent != null) {
Current = Current.Parent.ChildB;
}
else {
levelsUp++;
}
for (int i = 0; i < levelsUp; i++) {
Current = Current.ChildA;
}
}
}
}
回答by mmiika
I would start with a set brightness 100% and go around primary colors first:
我会从设置亮度 100% 开始，然后首先围绕原色：
FF0000, 00FF00, 0000FF
FF0000, 00FF00, 0000FF
then the combinations
然后组合
FFFF00, FF00FF, 00FFFF
FFFF00, FF00FF, 00FFFF
next for example halve the brightness and do same round. There's not too many really clearly distinct colors, after these I would start to vary the line width and do dotted/dashed lines etc.
接下来例如将亮度减半并进行相同的回合。没有太多真正明显不同的颜色，在这些之后我会开始改变线宽并做点/虚线等。
回答by Totty
You could get a random set of your 3 255 values and check it against the last set of 3 values, making sure they are each at least X away from the old values before using them.
您可以获得一组随机的 3 255 个值，并根据最后一组 3 个值进行检查，确保在使用它们之前，每个值都与旧值至少相差 X 倍。
OLD: 190, 120, 100
旧：190、120、100
NEW: 180, 200, 30
新：180、200、30
If X = 20, then the new set would be regenerated again.
如果 X = 20，则新集合将再次重新生成。
回答by aib
I think the HSV(or HSL) space has more opportunities here. If you don't mind the extra conversion, it's pretty easy to go through all the colors by just rotating the Hue value. If that's not enough, you can change the Saturation/Value/Lightness values and go through the rotation again. Or, you can always shift the Hue values or change your "stepping" angle and rotate more times.
我认为HSV（或 HSL）空间在这里有更多机会。如果您不介意额外的转换，只需旋转色调值即可轻松浏览所有颜色。如果这还不够，您可以更改饱和度/值/亮度值并再次进行旋转。或者，您可以随时移动色相值或更改“步进”角度并旋转更多次。
回答by Filip Rooms
I implemented this algorithm in a shorter way
我以更短的方式实现了这个算法
void ColorValue::SetColorValue( double r, double g, double b, ColorType myType )
{
this->c[0] = r;
this->c[1] = g;
this->c[2] = b;
this->type = myType;
}
DistinctColorGenerator::DistinctColorGenerator()
{
mFactor = 255;
mColorsGenerated = 0;
mpColorCycle = new ColorValue[6];
mpColorCycle[0].SetColorValue( 1.0, 0.0, 0.0, TYPE_RGB);
mpColorCycle[1].SetColorValue( 0.0, 1.0, 0.0, TYPE_RGB);
mpColorCycle[2].SetColorValue( 0.0, 0.0, 1.0, TYPE_RGB);
mpColorCycle[3].SetColorValue( 1.0, 1.0, 0.0, TYPE_RGB);
mpColorCycle[4].SetColorValue( 1.0, 0.0, 1.0, TYPE_RGB);
mpColorCycle[5].SetColorValue( 0.0, 1.0, 1.0, TYPE_RGB);
}
//----------------------------------------------------------
ColorValue DistinctColorGenerator::GenerateNewColor()
{
int innerCycleNr = mColorsGenerated % 6;
int outerCycleNr = mColorsGenerated / 6;
int cycleSize = pow( 2, (int)(log((double)(outerCycleNr)) / log( 2.0 ) ) );
int insideCycleCounter = outerCycleNr % cyclesize;
if ( outerCycleNr == 0)
{
mFactor = 255;
}
else
{
mFactor = ( 256 / ( 2 * cycleSize ) ) + ( insideCycleCounter * ( 256 / cycleSize ) );
}
ColorValue newColor = mpColorCycle[innerCycleNr] * mFactor;
mColorsGenerated++;
return newColor;
}
回答by Tom Hamming
You could also think of the color space as all combinations of three numbers from 0 to 255, inclusive. That's the base-255 representation of a number between 0 and 255^3, forced to have three decimal places (add zeros on to the end if need be.)
您还可以将色彩空间视为从 0 到 255（包括 0 到 255）的三个数字的所有组合。这是 0 到 255^3 之间数字的 base-255 表示，强制保留三个小数位（如果需要，在末尾添加零。）
So to generate x number of colors, you'd calculate x evenly spaced percentages, 0 to 100. Get numbers by multiplying those percentages by 255^3, convert those numbers to base 255, and add zeros as previously mentioned.
因此，要生成 x 种颜色，您需要计算 x 均匀间隔的百分比，即 0 到 100。通过将这些百分比乘以 255^3 来获取数字，将这些数字转换为以 255 为底的数字，并如前所述添加零。
Base conversion algorithm, for reference (in pseudocode that's quite close to C#):
基本转换算法，供参考（伪代码非常接近 C#）：
int num = (number to convert);
int baseConvert = (desired base, 255 in this case);
(array of ints) nums = new (array of ints);
int x = num;
double digits = Math.Log(num, baseConvert); //or ln(num) / ln(baseConvert)
int numDigits = (digits - Math.Ceiling(digits) == 0 ? (int)(digits + 1) : (int)Math.Ceiling(digits)); //go up one if it turns out even
for (int i = 0; i < numDigits; i++)
{
int toAdd = ((int)Math.Floor(x / Math.Pow((double)convertBase, (double)(numDigits - i - 1))));
//Formula for 0th digit: d = num / (convertBase^(numDigits - 1))
//Then subtract (d * convertBase^(numDigits - 1)) from the num and continue
nums.Add(toAdd);
x -= toAdd * (int)Math.Pow((double)convertBase, (double)(numDigits - i - 1));
}
return nums;
You might also have to do something to bring the range in a little bit, to avoid having white and black, if you want. Those numbers aren't actually a smooth color scale, but they'll generate separate colors if you don't have too many.
如果您愿意，您可能还需要做一些事情来稍微扩大范围，以避免出现白色和黑色。这些数字实际上并不是一个平滑的色标，但如果您没有太多颜色，它们会生成不同的颜色。
This questionhas more on base conversion in .NET.
这个问题有更多关于 .NET 中的基础转换。
回答by deancutlet
There's a flaw in the previous RGB solutions. They don't take advantage of the whole color space since they use a color value and 0 for the channels:
以前的 RGB 解决方案存在缺陷。它们不利用整个颜色空间，因为它们使用颜色值和 0 作为通道：
#006600
#330000
#FF00FF
Instead they should be using all the possible color values to generate mixed colors that can have up to 3 different values across the color channels:
相反，他们应该使用所有可能的颜色值来生成在颜色通道中最多可以有 3 个不同值的混合颜色：
#336600
#FF0066
#33FF66
Using the full color space you can generate more distinct colors. For example, if you have 4 values per channel, then 4*4*4=64colors can be generated. With the other scheme, only 4*7+1=29colors can be generated.
使用全色空间，您可以生成更多不同的颜色。例如，如果每个通道有 4 个值，则可以生成4*4*4= 64种颜色。使用另一种方案，只能生成 4*7+1= 29种颜色。
If you want N colors, then the number of values per channel required is: ceil(cube_root(N))
如果你想要 N 种颜色，那么每个通道所需的值数是：ceil(cube_root(N))
With that, you can then determine the possible (0-255 range) values (python):
这样，您就可以确定可能的（0-255 范围）值（python）：
max = 255
segs = int(num**(Decimal("1.0")/3))
step = int(max/segs)
p = [(i*step) for i in xrange(segs)]
values = [max]
values.extend(p)
Then you can iterate over the RGB colors (this is not recommended):
然后您可以遍历 RGB 颜色（不推荐这样做）：
total = 0
for red in values:
for green in values:
for blue in values:
if total <= N:
print color(red, green, blue)
total += 1
Nested loops will work, but are not recommended since it will favor the blue channel and the resulting colors will not have enough red (N will most likely be less than the number of all possible color values).
嵌套循环可以工作，但不推荐使用，因为它有利于蓝色通道，结果颜色没有足够的红色（N 很可能小于所有可能的颜色值的数量）。
You can create a better algorithm for the loops where each channel is treated equally and more distinct color values are favored over small ones.
您可以为循环创建更好的算法，其中每个通道都被平等对待，更不同的颜色值优先于小颜色值。
I have a solution, but didn't want to post it since it isn't the easiest to understand or efficient. But, you can view the solutionif you really want to.
我有一个解决方案，但不想发布它，因为它不是最容易理解或有效的。但是，如果您真的愿意，您可以查看解决方案。
Here is a sample of 64 generated colors: 64 colors
这是 64 种生成颜色的示例：64 种颜色
回答by Phrogz
I have put up a page online for procedurally generating visually distinct colors:
http://phrogz.net/css/distinct-colors.html
我在网上建立了一个页面，用于程序生成视觉上不同的颜色：http:
//phrogz.net/css/distinct-colors.html
Unlike other answers here that evenly step across RGB or HSV space (where there is a nonlinear relationship between the axis values and the perceptual differences), my page uses the standard CMI(I:c)color distance algorithm to prevent two colors from being too visually close.
与此处均匀跨 RGB 或 HSV 空间（其中轴值和感知差异之间存在非线性关系）的其他答案不同，我的页面使用标准CMI(I:c)颜色距离算法来防止两种颜色过于相似视觉上接近。
The final tab of the page allows you to sort the values in several ways, and then interleave them (ordered shuffle) so that you get very distinct colors placed next to one another.
页面的最后一个选项卡允许您以多种方式对值进行排序，然后将它们交错（有序洗牌），以便您获得彼此相邻的非常不同的颜色。
As of this writing, it only works well in Chrome and Safari, with a shim for Firefox; it uses HTML5 range input sliders in the interface, which IE9 and Firefox do not yet support natively.
在撰写本文时，它仅在 Chrome 和 Safari 中运行良好，并为 Firefox 提供了一个 shim；它在界面中使用 HTML5 范围输入滑块，IE9 和 Firefox 本身尚不支持。
回答by Tatarize
To implement a variation list where by your colors go, 255 then use all possibilities of that up, then add 0 and all RGB patterns with those two values. Then add 128 and all RGB combinations with those. Then 64. Then 192. Etc.
要实现颜色变化列表，255 然后使用所有可能性，然后添加 0 和具有这两个值的所有 RGB 模式。然后添加 128 和所有 RGB 组合。然后是 64。然后是 192。等等。
In Java,
在爪哇，
public Color getColor(int i) {
return new Color(getRGB(i));
}
public int getRGB(int index) {
int[] p = getPattern(index);
return getElement(p[0]) << 16 | getElement(p[1]) << 8 | getElement(p[2]);
}
public int getElement(int index) {
int value = index - 1;
int v = 0;
for (int i = 0; i < 8; i++) {
v = v | (value & 1);
v <<= 1;
value >>= 1;
}
v >>= 1;
return v & 0xFF;
}
public int[] getPattern(int index) {
int n = (int)Math.cbrt(index);
index -= (n*n*n);
int[] p = new int[3];
Arrays.fill(p,n);
if (index == 0) {
return p;
}
index--;
int v = index % 3;
index = index / 3;
if (index < n) {
p[v] = index % n;
return p;
}
index -= n;
p[v ] = index / n;
p[++v % 3] = index % n;
return p;
}
This will produce patterns of that type infinitely (2^24) into the future. However, after a hundred or so spots you likely won't see much of a difference between a color with 0 or 32 in the blue's place.
这将在未来无限（2^24）产生该类型的模式。但是，经过一百个左右的斑点后，您可能看不到蓝色位置为 0 或 32 的颜色之间的太大差异。
You might be better off normalizing this into a different color space. LAB color space for example with the L,A,B values normalized and converted. So the distinctness of the color is pushed through something more akin to the human eye.
您最好将其标准化为不同的色彩空间。LAB 颜色空间，例如 L、A、B 值标准化和转换。因此，颜色的独特性是通过更类似于人眼的东西来推动的。
getElement() reverses the endian of an 8 bit number, and starts counting from -1 rather than 0 (masking with 255). So it goes 255,0,127,192,64,... as the number grows it moves less and less significant bits, subdividing the number.
getElement() 反转 8 位数字的字节序，并从 -1 而不是 0（用 255 掩码）开始计数。所以它变成了 255,0,127,192,64,... 随着数字的增长，它移动的有效位越来越少，细分了数字。
getPattern() determines what the most significant element in the pattern should be (it's the cube root). Then proceeds to break down the 3N2+3N+1 different patterns that involve that most significant element.
getPattern() 确定模式中最重要的元素应该是什么（它是立方根）。然后继续分解涉及该最重要元素的 3N2+3N+1 种不同模式。
This algorithm will produce (first 128 values):
该算法将产生（前 128 个值）：
#FFFFFF
#000000
#FF0000
#00FF00
#0000FF
#FFFF00
#00FFFF
#FF00FF
#808080
#FF8080
#80FF80
#8080FF
#008080
#800080
#808000
#FFFF80
#80FFFF
#FF80FF
#FF0080
#80FF00
#0080FF
#00FF80
#8000FF
#FF8000
#000080
#800000
#008000
#404040
#FF4040
#40FF40
#4040FF
#004040
#400040
#404000
#804040
#408040
#404080
#FFFF40
#40FFFF
#FF40FF
#FF0040
#40FF00
#0040FF
#FF8040
#40FF80
#8040FF
#00FF40
#4000FF
#FF4000
#000040
#400000
#004000
#008040
#400080
#804000
#80FF40
#4080FF
#FF4080
#800040
#408000
#004080
#808040
#408080
#804080
#C0C0C0
#FFC0C0
#C0FFC0
#C0C0FF
#00C0C0
#C000C0
#C0C000
#80C0C0
#C080C0
#C0C080
#40C0C0
#C040C0
#C0C040
#FFFFC0
#C0FFFF
#FFC0FF
#FF00C0
#C0FF00
#00C0FF
#FF80C0
#C0FF80
#80C0FF
#FF40C0
#C0FF40
#40C0FF
#00FFC0
#C000FF
#FFC000
#0000C0
#C00000
#00C000
#0080C0
#C00080
#80C000
#0040C0
#C00040
#40C000
#80FFC0
#C080FF
#FFC080
#8000C0
#C08000
#00C080
#8080C0
#C08080
#80C080
#8040C0
#C08040
#40C080
#40FFC0
#C040FF
#FFC040
#4000C0
#C04000
#00C040
#4080C0
#C04080
#80C040
#4040C0
#C04040
#40C040
#202020
#FF2020
#20FF20
Read left to right, top to bottom. 729 colors (93). So all the patterns up to n = 9. You'll notice the speed at which they start to clash. There's only so many WRGBCYMK variations. And this solution, while clever basically only does different shades of primary colors.
从左到右，从上到下阅读。729 色 (93)。所以直到 n = 9 的所有模式。你会注意到它们开始冲突的速度。只有这么多 WRGBCYMK 变体。而这个解决方案虽然聪明，但基本上只处理不同深浅的原色。
Much of the clashing is due to green and how similar most greens look to most people. The demand that each be maximally different at start rather than just different enough to not be the same color. And basic flaws in the idea resulting in primary colors patterns, and identical hues.
大部分冲突是由于绿色以及大多数果岭对大多数人的相似程度。要求每个人在开始时最大程度地不同，而不仅仅是不同到不同的颜色。这个想法的基本缺陷导致了原色图案和相同的色调。
Using CIELab2000 Color Space and Distance Routine to randomly select and try 10k different colors and find the maximally-distant minimum-distance from previous colors, (pretty much the definition of the request) avoids clashing longer than the above solution:
使用 CIELab2000 颜色空间和距离例程随机选择并尝试 10k 种不同的颜色，并找到与先前颜色的最大距离最小距离，（几乎是请求的定义）避免了比上述解决方案更长的冲突：
Which could be just called a static list for the Easy Way. It took an hour and a half to generate 729 entries:
这可以称为 Easy Way 的静态列表。生成 729 个条目花了一个半小时：
#9BC4E5
#310106
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#B2387B
#888C6F
#314B5F
#E5B678
#38A3C6
#586148
#5C515B
#CDCCE1
#C8977F
Using brute force to (testing all 16,777,216 RGB colors through CIELab Delta2000 / Starting with black) produces a series. Which starts to clash at around 26 but could make it to 30 or 40 with visual inspection and manual dropping (which can't be done with a computer). So doing the absolute maximum one can programmatically only makes a couple dozen distinct colors. A discrete list is your best bet. You will get more discrete colors with a list than you would programmatically. The easy way is the best solution, start mixing and matching with other ways to alter your data than color.
使用蛮力（通过 CIELab Delta2000 测试所有 16,777,216 种 RGB 颜色/从黑色开始）产生一个系列。它在 26 左右开始发生冲突，但可以通过目视检查和手动掉落（无法用计算机完成）使其达到 30 或 40。因此，以编程方式执行绝对最大值只能生成几十种不同的颜色。离散列表是您最好的选择。与以编程方式相比，您将通过列表获得更多离散颜色。简单的方法是最好的解决方案，开始与其他方法混合和匹配来改变你的数据而不是颜色。
#000000
#00FF00
#0000FF
#FF0000
#01FFFE
#FFA6FE
#FFDB66
#006401
#010067
#95003A
#007DB5
#FF00F6
#FFEEE8
#774D00
#90FB92
#0076FF
#D5FF00
#FF937E
#6A826C
#FF029D
#FE8900
#7A4782
#7E2DD2
#85A900
#FF0056
#A42400
#00AE7E
#683D3B
#BDC6FF
#263400
#BDD393
#00B917
#9E008E
#001544
#C28C9F
#FF74A3
#01D0FF
#004754
#E56FFE
#788231
#0E4CA1
#91D0CB
#BE9970
#968AE8
#BB8800
#43002C
#DEFF74
#00FFC6
#FFE502
#620E00
#008F9C
#98FF52
#7544B1
#B500FF
#00FF78
#FF6E41
#005F39
#6B6882
#5FAD4E
#A75740
#A5FFD2
#FFB167
#009BFF
#E85EBE
Update:
I continued this for about a month so, at 1024 brute force.
更新：我以 1024 的蛮力持续了大约一个月。
public static final String[] indexcolors = new String[]{
"#000000", "#FFFF00", "#1CE6FF", "#FF34FF", "#FF4A46", "#008941", "#006FA6", "#A30059",
"#FFDBE5", "#7A4900", "#0000A6", "#63FFAC", "#B79762", "#004D43", "#8FB0FF", "#997D87",
"#5A0007", "#809693", "#FEFFE6", "#1B4400", "#4FC601", "#3B5DFF", "#4A3B53", "#FF2F80",
"#61615A", "#BA0900", "#6B7900", "#00C2A0", "#FFAA92", "#FF90C9", "#B903AA", "#D16100",
"#DDEFFF", "#000035", "#7B4F4B", "#A1C299", "#300018", "#0AA6D8", "#013349", "#00846F",
"#372101", "#FFB500", "#C2FFED", "#A079BF", "#CC0744", "#C0B9B2", "#C2FF99", "#001E09",
"#00489C", "#6F0062", "#0CBD66", "#EEC3FF", "#456D75", "#B77B68", "#7A87A1", "#788D66",
"#885578", "#FAD09F", "#FF8A9A", "#D157A0", "#BEC459", "#456648", "#0086ED", "#886F4C",
"#34362D", "#B4A8BD", "#00A6AA", "#452C2C", "#636375", "#A3C8C9", "#FF913F", "#938A81",
"#575329", "#00FECF", "#B05B6F", "#8CD0FF", "#3B9700", "#04F757", "#C8A1A1", "#1E6E00",
"#7900D7", "#A77500", "#6367A9", "#A05837", "#6B002C", "#772600", "#D790FF", "#9B9700",
"#549E79", "#FFF69F", "#201625", "#72418F", "#BC23FF", "#99ADC0", "#3A2465", "#922329",
"#5B4534", "#FDE8DC", "#404E55", "#0089A3", "#CB7E98", "#A4E804", "#324E72", "#6A3A4C",
"#83AB58", "#001C1E", "#D1F7CE", "#004B28", "#C8D0F6", "#A3A489", "#806C66", "#222800",
"#BF5650", "#E83000", "#66796D", "#DA007C", "#FF1A59", "#8ADBB4", "#1E0200", "#5B4E51",
"#C895C5", "#320033", "#FF6832", "#66E1D3", "#CFCDAC", "#D0AC94", "#7ED379", "#012C58",
"#7A7BFF", "#D68E01", "#353339", "#78AFA1", "#FEB2C6", "#75797C", "#837393", "#943A4D",
"#B5F4FF", "#D2DCD5", "#9556BD", "#6A714A", "#001325", "#02525F", "#0AA3F7", "#E98176",
"#DBD5DD", "#5EBCD1", "#3D4F44", "#7E6405", "#02684E", "#962B75", "#8D8546", "#9695C5",
"#E773CE", "#D86A78", "#3E89BE", "#CA834E", "#518A87", "#5B113C", "#55813B", "#E704C4",
"#00005F", "#A97399", "#4B8160", "#59738A", "#FF5DA7", "#F7C9BF", "#643127", "#513A01",
"#6B94AA", "#51A058", "#A45B02", "#1D1702", "#E20027", "#E7AB63", "#4C6001", "#9C6966",
"#64547B", "#97979E", "#006A66", "#391406", "#F4D749", "#0045D2", "#006C31", "#DDB6D0",
"#7C6571", "#9FB2A4", "#00D891", "#15A08A", "#BC65E9", "#FFFFFE", "#C6DC99", "#203B3C",
"#671190", "#6B3A64", "#F5E1FF", "#FFA0F2", "#CCAA35", "#374527", "#8BB400", "#797868",
"#C6005A", "#3B000A", "#C86240", "#29607C", "#402334", "#7D5A44", "#CCB87C", "#B88183",
"#AA5199", "#B5D6C3", "#A38469", "#9F94F0", "#A74571", "#B894A6", "#71BB8C", "#00B433",
"#789EC9", "#6D80BA", "#953F00", "#5EFF03", "#E4FFFC", "#1BE177", "#BCB1E5", "#76912F",
"#003109", "#0060CD", "#D20096", "#895563", "#29201D", "#5B3213", "#A76F42", "#89412E",
"#1A3A2A", "#494B5A", "#A88C85", "#F4ABAA", "#A3F3AB", "#00C6C8", "#EA8B66", "#958A9F",
"#BDC9D2", "#9FA064", "#BE4700", "#658188", "#83A485", "#453C23", "#47675D", "#3A3F00",
"#061203", "#DFFB71", "#868E7E", "#98D058", "#6C8F7D", "#D7BFC2", "#3C3E6E", "#D83D66",
"#2F5D9B", "#6C5E46", "#D25B88", "#5B656C", "#00B57F", "#545C46", "#866097", "#365D25",
"#252F99", "#00CCFF", "#674E60", "#FC009C", "#92896B", "#1E2324", "#DEC9B2", "#9D4948",
"#85ABB4", "#342142", "#D09685", "#A4ACAC", "#00FFFF", "#AE9C86", "#742A33", "#0E72C5",
"#AFD8EC", "#C064B9", "#91028C", "#FEEDBF", "#FFB789", "#9CB8E4", "#AFFFD1", "#2A364C",
"#4F4A43", "#647095", "#34BBFF", "#807781", "#920003", "#B3A5A7", "#018615", "#F1FFC8",
"#976F5C", "#FF3BC1", "#FF5F6B", "#077D84", "#F56D93", "#5771DA", "#4E1E2A", "#830055",
"#02D346", "#BE452D", "#00905E", "#BE0028", "#6E96E3", "#007699", "#FEC96D", "#9C6A7D",
"#3FA1B8", "#893DE3", "#79B4D6", "#7FD4D9", "#6751BB", "#B28D2D", "#E27A05", "#DD9CB8",
"#AABC7A", "#980034", "#561A02", "#8F7F00", "#635000", "#CD7DAE", "#8A5E2D", "#FFB3E1",
"#6B6466", "#C6D300", "#0100E2", "#88EC69", "#8FCCBE", "#21001C", "#511F4D", "#E3F6E3",
"#FF8EB1", "#6B4F29", "#A37F46", "#6A5950", "#1F2A1A", "#04784D", "#101835", "#E6E0D0",
"#FF74FE", "#00A45F", "#8F5DF8", "#4B0059", "#412F23", "#D8939E", "#DB9D72", "#604143",
"#B5BACE", "#989EB7", "#D2C4DB", "#A587AF", "#77D796", "#7F8C94", "#FF9B03", "#555196",
"#31DDAE", "#74B671", "#802647", "#2A373F", "#014A68", "#696628", "#4C7B6D", "#002C27",
"#7A4522", "#3B5859", "#E5D381", "#FFF3FF", "#679FA0", "#261300", "#2C5742", "#9131AF",
"#AF5D88", "#C7706A", "#61AB1F", "#8CF2D4", "#C5D9B8", "#9FFFFB", "#BF45CC", "#493941",
"#863B60", "#B90076", "#003177", "#C582D2", "#C1B394", "#602B70", "#887868", "#BABFB0",
"#030012", "#D1ACFE", "#7FDEFE", "#4B5C71", "#A3A097", "#E66D53", "#637B5D", "#92BEA5",
"#00F8B3", "#BEDDFF", "#3DB5A7", "#DD3248", "#B6E4DE", "#427745", "#598C5A", "#B94C59",
"#8181D5", "#94888B", "#FED6BD", "#536D31", "#6EFF92", "#E4E8FF", "#20E200", "#FFD0F2",
"#4C83A1", "#BD7322", "#915C4E", "#8C4787", "#025117", "#A2AA45", "#2D1B21", "#A9DDB0",
"#FF4F78", "#528500", "#009A2E", "#17FCE4", "#71555A", "#525D82", "#00195A", "#967874",
"#555558", "#0B212C", "#1E202B", "#EFBFC4", "#6F9755", "#6F7586", "#501D1D", "#372D00",
"#741D16", "#5EB393", "#B5B400", "#DD4A38", "#363DFF", "#AD6552", "#6635AF", "#836BBA",
"#98AA7F", "#464836", "#322C3E", "#7CB9BA", "#5B6965", "#707D3D", "#7A001D", "#6E4636",
"#443A38", "#AE81FF", "#489079", "#897334", "#009087", "#DA713C", "#361618", "#FF6F01",
"#006679", "#370E77", "#4B3A83", "#C9E2E6", "#C44170", "#FF4526", "#73BE54", "#C4DF72",
"#ADFF60", "#00447D", "#DCCEC9", "#BD9479", "#656E5B", "#EC5200", "#FF6EC2", "#7A617E",
"#DDAEA2", "#77837F", "#A53327", "#608EFF", "#B599D7", "#A50149", "#4E0025", "#C9B1A9",
"#03919A", "#1B2A25", "#E500F1", "#982E0B", "#B67180", "#E05859", "#006039", "#578F9B",
"#305230", "#CE934C", "#B3C2BE", "#C0BAC0", "#B506D3", "#170C10", "#4C534F", "#224451",
"#3E4141", "#78726D", "#B6602B", "#200441", "#DDB588", "#497200", "#C5AAB6", "#033C61",
"#71B2F5", "#A9E088", "#4979B0", "#A2C3DF", "#784149", "#2D2B17", "#3E0E2F", "#57344C",
"#0091BE", "#E451D1", "#4B4B6A", "#5C011A", "#7C8060", "#FF9491", "#4C325D", "#005C8B",
"#E5FDA4", "#68D1B6", "#032641", "#140023", "#8683A9", "#CFFF00", "#A72C3E", "#34475A",
"#B1BB9A", "#B4A04F", "#8D918E", "#A168A6", "#813D3A", "#425218", "#DA8386", "#776133",
"#563930", "#8498AE", "#90C1D3", "#B5666B", "#9B585E", "#856465", "#AD7C90", "#E2BC00",
"#E3AAE0", "#B2C2FE", "#FD0039", "#009B75", "#FFF46D", "#E87EAC", "#DFE3E6", "#848590",
"#AA9297", "#83A193", "#577977", "#3E7158", "#C64289", "#EA0072", "#C4A8CB", "#55C899",
"#E78FCF", "#004547", "#F6E2E3", "#966716", "#378FDB", "#435E6A", "#DA0004", "#1B000F",
"#5B9C8F", "#6E2B52", "#011115", "#E3E8C4", "#AE3B85", "#EA1CA9", "#FF9E6B", "#457D8B",
"#92678B", "#00CDBB", "#9CCC04", "#002E38", "#96C57F", "#CFF6B4", "#492818", "#766E52",
"#20370E", "#E3D19F", "#2E3C30", "#B2EACE", "#F3BDA4", "#A24E3D", "#976FD9", "#8C9FA8",
"#7C2B73", "#4E5F37", "#5D5462", "#90956F", "#6AA776", "#DBCBF6", "#DA71FF", "#987C95",
"#52323C", "#BB3C42", "#584D39", "#4FC15F", "#A2B9C1", "#79DB21", "#1D5958", "#BD744E",
"#160B00", "#20221A", "#6B8295", "#00E0E4", "#102401", "#1B782A", "#DAA9B5", "#B0415D",
"#859253", "#97A094", "#06E3C4", "#47688C", "#7C6755", "#075C00", "#7560D5", "#7D9F00",
"#C36D96", "#4D913E", "#5F4276", "#FCE4C8", "#303052", "#4F381B", "#E5A532", "#706690",
"#AA9A92", "#237363", "#73013E", "#FF9079", "#A79A74", "#029BDB", "#FF0169", "#C7D2E7",
"#CA8869", "#80FFCD", "#BB1F69", "#90B0AB", "#7D74A9", "#FCC7DB", "#99375B", "#00AB4D",
"#ABAED1", "#BE9D91", "#E6E5A7", "#332C22", "#DD587B", "#F5FFF7", "#5D3033", "#6D3800",
"#FF0020", "#B57BB3", "#D7FFE6", "#C535A9", "#260009", "#6A8781", "#A8ABB4", "#D45262",
"#794B61", "#4621B2", "#8DA4DB", "#C7C890", "#6FE9AD", "#A243A7", "#B2B081", "#181B00",
"#286154", "#4CA43B", "#6A9573", "#A8441D", "#5C727B", "#738671", "#D0CFCB", "#897B77",
"#1F3F22", "#4145A7", "#DA9894", "#A1757A", "#63243C", "#ADAAFF", "#00CDE2", "#DDBC62",
"#698EB1", "#208462", "#00B7E0", "#614A44", "#9BBB57", "#7A5C54", "#857A50", "#766B7E",
"#014833", "#FF8347", "#7A8EBA", "#274740", "#946444", "#EBD8E6", "#646241", "#373917",
"#6AD450", "#81817B", "#D499E3", "#979440", "#011A12", "#526554", "#B5885C", "#A499A5",
"#03AD89", "#B3008B", "#E3C4B5", "#96531F", "#867175", "#74569E", "#617D9F", "#E70452",
"#067EAF", "#A697B6", "#B787A8", "#9CFF93", "#311D19", "#3A9459", "#6E746E", "#B0C5AE",
"#84EDF7", "#ED3488", "#754C78", "#384644", "#C7847B", "#00B6C5", "#7FA670", "#C1AF9E",
"#2A7FFF", "#72A58C", "#FFC07F", "#9DEBDD", "#D97C8E", "#7E7C93", "#62E674", "#B5639E",
"#FFA861", "#C2A580", "#8D9C83", "#B70546", "#372B2E", "#0098FF", "#985975", "#20204C",
"#FF6C60", "#445083", "#8502AA", "#72361F", "#9676A3", "#484449", "#CED6C2", "#3B164A",
"#CCA763", "#2C7F77", "#02227B", "#A37E6F", "#CDE6DC", "#CDFFFB", "#BE811A", "#F77183",
"#EDE6E2", "#CDC6B4", "#FFE09E", "#3A7271", "#FF7B59", "#4E4E01", "#4AC684", "#8BC891",
"#BC8A96", "#CF6353", "#DCDE5C", "#5EAADD", "#F6A0AD", "#E269AA", "#A3DAE4", "#436E83",
"#002E17", "#ECFBFF", "#A1C2B6", "#50003F", "#71695B", "#67C4BB", "#536EFF", "#5D5A48",
"#890039", "#969381", "#371521", "#5E4665", "#AA62C3", "#8D6F81", "#2C6135", "#410601",
"#564620", "#E69034", "#6DA6BD", "#E58E56", "#E3A68B", "#48B176", "#D27D67", "#B5B268",
"#7F8427", "#FF84E6", "#435740", "#EAE408", "#F4F5FF", "#325800", "#4B6BA5", "#ADCEFF",
"#9B8ACC", "#885138", "#5875C1", "#7E7311", "#FEA5CA", "#9F8B5B", "#A55B54", "#89006A",
"#AF756F", "#2A2000", "#576E4A", "#7F9EFF", "#7499A1", "#FFB550", "#00011E", "#D1511C",
"#688151", "#BC908A", "#78C8EB", "#8502FF", "#483D30", "#C42221", "#5EA7FF", "#785715",
"#0CEA91", "#FFFAED", "#B3AF9D", "#3E3D52", "#5A9BC2", "#9C2F90", "#8D5700", "#ADD79C",
"#00768B", "#337D00", "#C59700", "#3156DC", "#944575", "#ECFFDC", "#D24CB2", "#97703C",
"#4C257F", "#9E0366", "#88FFEC", "#B56481", "#396D2B", "#56735F", "#988376", "#9BB195",
"#A9795C", "#E4C5D3", "#9F4F67", "#1E2B39", "#664327", "#AFCE78", "#322EDF", "#86B487",
"#C23000", "#ABE86B", "#96656D", "#250E35", "#A60019", "#0080CF", "#CAEFFF", "#323F61",
"#A449DC", "#6A9D3B", "#FF5AE4", "#636A01", "#D16CDA", "#736060", "#FFBAAD", "#D369B4",
"#FFDED6", "#6C6D74", "#927D5E", "#845D70", "#5B62C1", "#2F4A36", "#E45F35", "#FF3B53",
"#AC84DD", "#762988", "#70EC98", "#408543", "#2C3533", "#2E182D", "#323925", "#19181B",
"#2F2E2C", "#023C32", "#9B9EE2", "#58AFAD", "#5C424D", "#7AC5A6", "#685D75", "#B9BCBD",
"#834357", "#1A7B42", "#2E57AA", "#E55199", "#316E47", "#CD00C5", "#6A004D", "#7FBBEC",
"#F35691", "#D7C54A", "#62ACB7", "#CBA1BC", "#A28A9A", "#6C3F3B", "#FFE47D", "#DCBAE3",
"#5F816D", "#3A404A", "#7DBF32", "#E6ECDC", "#852C19", "#285366", "#B8CB9C", "#0E0D00",
"#4B5D56", "#6B543F", "#E27172", "#0568EC", "#2EB500", "#D21656", "#EFAFFF", "#682021",
"#2D2011", "#DA4CFF", "#70968E", "#FF7B7D", "#4A1930", "#E8C282", "#E7DBBC", "#A68486",
"#1F263C", "#36574E", "#52CE79", "#ADAAA9", "#8A9F45", "#6542D2", "#00FB8C", "#5D697B",
"#CCD27F", "#94A5A1", "#790229", "#E383E6", "#7EA4C1", "#4E4452", "#4B2C00", "#620B70",
"#314C1E", "#874AA6", "#E30091", "#66460A", "#EB9A8B", "#EAC3A3", "#98EAB3", "#AB9180",
"#B8552F", "#1A2B2F", "#94DDC5", "#9D8C76", "#9C8333", "#94A9C9", "#392935", "#8C675E",
"#CCE93A", "#917100", "#01400B", "#449896", "#1CA370", "#E08DA7", "#8B4A4E", "#667776",
"#4692AD", "#67BDA8", "#69255C", "#D3BFFF", "#4A5132", "#7E9285", "#77733C", "#E7A0CC",
"#51A288", "#2C656A", "#4D5C5E", "#C9403A", "#DDD7F3", "#005844", "#B4A200", "#488F69",
"#858182", "#D4E9B9", "#3D7397", "#CAE8CE", "#D60034", "#AA6746", "#9E5585", "#BA6200"
};
回答by sai siva sundar
for getting nth colour. Just this kind of code would be enough. This i have use in my opencv clustering problem. This will create different colours as col changes.
获得第n种颜色。只需这种代码就足够了。这是我在我的 opencv 聚类问题中使用的。这将随着 col 变化创建不同的颜色。
for(int col=1;col<CLUSTER_COUNT+1;col++){
switch(col%6)
{
case 1:cout<<Scalar(0,0,(int)(255/(int)(col/6+1)))<<endl;break;
case 2:cout<<Scalar(0,(int)(255/(int)(col/6+1)),0)<<endl;break;
case 3:cout<<Scalar((int)(255/(int)(col/6+1)),0,0)<<endl;break;
case 4:cout<<Scalar(0,(int)(255/(int)(col/6+1)),(int)(255/(int)(col/6+1)))<<endl;break;
case 5:cout<<Scalar((int)(255/(int)(col/6+1)),0,(int)(255/(int)(col/6+1)))<<endl;break;
case 0:cout<<Scalar((int)(255/(int)(col/6)),(int)(255/(int)(col/6)),0)<<endl;break;
}
}
