RGB值基色名称

时间:2016-05-27 05:48:11

标签: algorithm colors rgb

我想找到颜色名称,给出它的rgb值。

示例RGB值为:(237,65,83)

预定义值

阵列(11,' Red','#FF0000',' 255,0,0'),

阵列(3,'布朗','#A52A2A',' 165,42,42')

如果我使用此方法distance calculation

我的颜色变成棕色。

但是如果我们测试rgb值here

,实际颜色是红色的

已编辑1

<?php 

 $colors = array(
 array(1, 'Black', '#000000', '0,0,0'),
 array(2, 'Blue',  '#0000FF', '0,0,255'),
 array(3, 'Brown', '#A52A2A', '165,42,42'),    
 array(4, 'Cream', '#FFFFCC', '255,255,204'),   
 array(5, 'Green', '#008000', '0,128,0'),        
 array(6, 'Grey',  '#808080', '128,128,128'),
 array(7, 'Yellow', '#FFFF00', '255,255,0'),
 array(8, 'Orange', '#FFA500', '255,165,0'),          
 array(9, 'Pink', '#FFC0CB', '255,192,203'), 
 array(11, 'Red', '#FF0000', '255,0,0'),  
 array(10, 'Purple', '#800080', '128,0,128'),
 array(12, 'Tan', '#d2b48c', '210,180,140'),
 array(13, 'Turquoise', '#40E0D0', '64,224,208'),
 array(14, 'White', '#FFFFFF', '255,255,255')
   );



 $miDist = 99999999999999999 ;
 $loc = 0 ; 



 $findColor = RGBtoHSV(72, 70, 68);
 for( $i = 0 ; $i < 14 ; $i++){
  $string =  $colors[$i][3];
  $pieces = explode(',' , $string);
  $r0 = $pieces[0];
  $g0 = $pieces[1];
  $b0 = $pieces[2];

  $storedColor = RGBtoHSV($r0,$g0,$b0);

 echo $storedColor[0] ."-" . $storedColor[1] ;
 // distance between colors (regardless of intensity)


 $d = sqrt( ($storedColor[0]-$findColor[0])
      *($storedColor[0]-$findColor[0])
      + 
      ($storedColor[1]-$findColor[1])
      *($storedColor[1]-$findColor[1])



      );



  echo $colors[$i][1] ."=" .$d;
  //echo $d ;
     if( $miDist >= $d )
       {
       $miDist = $d;
       $loc = $i ;    
       } 
      echo "<br>" ;


 }

 echo $colors[$loc][1];









 function RGBtoHSV($R, $G, $B)    // RGB values:    0-255, 0-255, 0-255
 {                                // HSV values:    0-360, 0-100, 0-100
// Convert the RGB byte-values to percentages
$R = ($R / 255);
$G = ($G / 255);
$B = ($B / 255);

// Calculate a few basic values, the maximum value of R,G,B, the
//   minimum value, and the difference of the two (chroma).
$maxRGB = max($R, $G, $B);
$minRGB = min($R, $G, $B);
$chroma = $maxRGB - $minRGB;

// Value (also called Brightness) is the easiest component to calculate,
//   and is simply the highest value among the R,G,B components.
// We multiply by 100 to turn the decimal into a readable percent value.
$computedV = 100 * $maxRGB;

// Special case if hueless (equal parts RGB make black, white, or grays)
// Note that Hue is technically undefined when chroma is zero, as
//   attempting to calculate it would cause division by zero (see
//   below), so most applications simply substitute a Hue of zero.
// Saturation will always be zero in this case, see below for details.
if ($chroma == 0)
    return array(0, 0, $computedV);

// Saturation is also simple to compute, and is simply the chroma
//   over the Value (or Brightness)
// Again, multiplied by 100 to get a percentage.
$computedS = 100 * ($chroma / $maxRGB);

// Calculate Hue component
// Hue is calculated on the "chromacity plane", which is represented
//   as a 2D hexagon, divided into six 60-degree sectors. We calculate
//   the bisecting angle as a value 0 <= x < 6, that represents which
//   portion of which sector the line falls on.
if ($R == $minRGB)
    $h = 3 - (($G - $B) / $chroma);
elseif ($B == $minRGB)
    $h = 1 - (($R - $G) / $chroma);
else // $G == $minRGB
    $h = 5 - (($B - $R) / $chroma);

// After we have the sector position, we multiply it by the size of
//   each sector's arc (60 degrees) to obtain the angle in degrees.
$computedH = 60 * $h;

return array($computedH, $computedS, $computedV);
 }

 ?>

1 个答案:

答案 0 :(得分:3)

因此,如果您希望两种颜色(r0,g0,b0)(r1,g1,b1)之间的距离检测最接近的颜色,无论其强度如何(在这种情况下,基色是什么意思),您应该

  1. 将颜色矢量标准化为通用尺寸
  2. 计算距离
  3. 缩放结果
  4. // variables
    int r0,g0,b0,c0;
    int r1,g1,b1,c1,d;
    // color sizes
    c0=sqrt(r0*r0+g0*g0+b0*b0);
    c1=sqrt(r1*r1+g1*g1+b1*b1);
    // distance between normalized colors
    d = sqrt((r0*c1-r1*c0)^2 + (g0*c1-g1*c0)^2 + (b0*c1-b1*c0)^2) / (c0*c1);
    

    这种方法比较深色会变得不稳定,因此您可以添加简单的条件,如

    if (c0<treshold)  color is dark 
    

    比较这种颜色只会再次出现灰色阴影或返回未知颜色。我们的愿景同样有效,我们无法安全识别暗色......

    无论如何,HSV色彩空间对于色彩比较要好得多,因为它更像人类色彩识别。所以转换RGB -> HSV并计算距离忽略值V这是颜色的强度......

    HSV 中,您需要将H作为周期性的全圆值处理,因此更改只能是圆圈的一半。 S告诉您是否需要单独处理彩色或灰度,V是强度。

    // variables
    int h0,s0,v0;
    int h1,s1,v1,d,q;
    q=h1-h0;
    if (q<-128) q+=256; // use shorter angle
    if (q>+128) q-=256; // use shorter angle
             q*=q; d =q;
    q=s1-s0; q*=q; d+=q;
    if (s0<32)          // grayscales
        {
        d=0;            // ignore H,S
        if (s1>=32) continue; // compare only to gray-scales so ignore this color
        }
    q=v1-v0; q*=q; d+=q;
    

    要比较的一些事情......

    您应该对您的来源进行目视检查,以便真正了解发生了什么,否则您将进入圈子而没有任何结果。例如,我刚刚在 C ++ / VCL / mine图像类

    中对此进行了编码
    union color
        {
        DWORD dd; WORD dw[2]; byte db[4];
        int i; short int ii[2];
        color(){}; color(color& a){ *this=a; }; ~color(){}; color* operator = (const color *a) { dd=a->dd; return this; }; /*color* operator = (const color &a) { ...copy... return this; };*/
        };
    
        enum{ // this is inside my picture:: class
            _x=0,   // dw
            _y=1,
    
            _b=0,   // db
            _g=1,
            _r=2,
            _a=3,
    
            _v=0,   // db
            _s=1,
            _h=2,
            };
    
    void rgb2hsv(color &c)
        {
        double r,g,b,min,max,del,h,s,v,dr,dg,db;
        r=c.db[picture::_r]; r/=255.0;
        g=c.db[picture::_g]; g/=255.0;
        b=c.db[picture::_b]; b/=255.0;
        min=r; if (min>g) min=g; if(min>b) min=b;
        max=r; if (max<g) max=g; if(max<b) max=b;
        del=max-min;
        v=max;
        if (del<=1e-10) { h=0; s=0; }   // grayscale
        else{
            s=del/max;
            dr=(((max-r)/6.0)+(del/2.0))/del;
            dg=(((max-g)/6.0)+(del/2.0))/del;
            db=(((max-b)/6.0)+(del/2.0))/del;
            if      (fabs(r-max)<1e-10) h=db-dg;
            else if (fabs(g-max)<1e-10) h=(1.0/3.0)+dr-db;
            else if (fabs(b-max)<1e-10) h=(2.0/3.0)+dg-dr;
            if (h<0.0) h+=1.0;
            if (h>1.0) h-=1.0;
            }
        c.db[picture::_h]=h*255.0;
        c.db[picture::_s]=s*255.0;
        c.db[picture::_v]=v*255.0;
        }
    
    void hsv2rgb(color &c)
        {
        int i;
        double r,g,b,h,s,v,vh,v1,v2,v3,f;
        h=c.db[picture::_h]; h/=255.0;
        s=c.db[picture::_s]; s/=255.0;
        v=c.db[picture::_v]; v/=255.0;
        if (s<=1e-10) { r=v; g=v; b=v; }    // grayscale
        else{
            vh=h*6.0;
            if (vh>=6.0) vh=0.0;
            f=floor(vh); i=f;
            v1=v*(1.0-s);
            v2=v*(1.0-s*(    vh-f));
            v3=v*(1.0-s*(1.0-vh+f));
                 if (i==0) { r=v ; g=v3; b=v1; }
            else if (i==1) { r=v2; g=v ; b=v1; }
            else if (i==2) { r=v1; g=v ; b=v3; }
            else if (i==3) { r=v1; g=v2; b=v ; }
            else if (i==4) { r=v3; g=v1; b=v ; }
            else           { r=v ; g=v1; b=v2; }
            }
        c.db[picture::_r]=r*255.0;
        c.db[picture::_g]=g*255.0;
        c.db[picture::_b]=b*255.0;
        }
    
    struct _base_color
        {
        DWORD rgb,hsv;
        const char *nam;
        _base_color(DWORD _rgb,const char *_nam){ nam=_nam; rgb=_rgb; color c; c.dd=rgb; rgb2hsv(c); hsv=c.dd; }
    
        _base_color(){};
        _base_color(_base_color& a){};
        ~_base_color(){};
        _base_color* operator = (const _base_color *a){};
        //_base_color* operator = (const _base_color &a);
        };
    const _base_color base_color[]=
        {
        //          0x00RRGGBB
        _base_color(0x00000000,"Black"),
        _base_color(0x00808080,"Gray"),
        _base_color(0x00C0C0C0,"Silver"),
        _base_color(0x00FFFFFF,"White"),
        _base_color(0x00800000,"Maroon"),
        _base_color(0x00FF0000,"Red"),
        _base_color(0x00808000,"Olive"),
        _base_color(0x00FFFF00,"Yellow"),
        _base_color(0x00008000,"Green"),
        _base_color(0x0000FF00,"Lime"),
        _base_color(0x00008080,"Teal"),
        _base_color(0x0000FFFF,"Aqua"),
        _base_color(0x00000080,"Navy"),
        _base_color(0x000000FF,"Blue"),
        _base_color(0x00800080,"Purple"),
        _base_color(0x00FF00FF,"Fuchsia"),
        _base_color(0x00000000,"")
        };
    
    void compare_colors()
        {
        picture pic0;
        int h0,s0,v0,h1,s1,v1,x,y,i,d,i0,d0;
        int r0,g0,b0,r1,g1,b1,c0,c1,q,xx;
        color c;
        pic0.resize(256*4,256);
        pic0.pf=_pf_rgba;
        for (y=0;y<256;y++)
         for (x=0;x<256;x++)
            {
            // get color from image
            c=pic0.p[y][x];
            xx=x;
            r0=c.db[picture::_r];
            g0=c.db[picture::_g];
            b0=c.db[picture::_b];
            rgb2hsv(c);
            h0=c.db[picture::_h];
            s0=c.db[picture::_s];
            v0=c.db[picture::_v];
            // naive RGB
            xx+=256;
            for (i0=-1,d0=-1,i=0;base_color[i].nam[0];i++)
                {
                // compute distance
                c.dd=base_color[i].rgb;
                r1=c.db[picture::_r];
                g1=c.db[picture::_g];
                b1=c.db[picture::_b];
                // no need for sqrt
                d=((r1-r0)*(r1-r0))+((g1-g0)*(g1-g0))+((b1-b0)*(b1-b0));
                // remember closest match
                if ((d0<0)||(d0>d)) { d0=d; i0=i; }
                }
            pic0.p[y][xx].dd=base_color[i0].rgb;
            // normalized RGB
            xx+=256;
            c0=sqrt((r0*r0)+(g0*g0)+(b0*b0));
            if (!c0) i0=0; else
             for (i0=-1,d0=-1,i=1;base_color[i].nam[0];i++)
                {
                // compute distance
                c.dd=base_color[i].rgb;
                r1=c.db[picture::_r];
                g1=c.db[picture::_g];
                b1=c.db[picture::_b];
                c1=sqrt((r1*r1)+(g1*g1)+(b1*b1));
                // no need for sqrt
                q=((r0*c1)-(r1*c0))/4; q*=q; d =q;
                q=((g0*c1)-(g1*c0))/4; q*=q; d+=q;
                q=((b0*c1)-(b1*c0))/4; q*=q; d+=q;
                d/=c1*c0; d<<=16; d/=c1*c0;
                // remember closest match
                if ((d0<0)||(d0>d)) { d0=d; i0=i; }
                }
            pic0.p[y][xx].dd=base_color[i0].rgb;
            // HSV
            xx+=256;
            for (i0=-1,d0=-1,i=0;base_color[i].nam[0];i++)
                {
                // compute distance
                c.dd=base_color[i].hsv;
                h1=c.db[picture::_h];
                s1=c.db[picture::_s];
                v1=c.db[picture::_v];
                // no need for sqrt
                q=h1-h0;
                if (q<-128) q+=256; // use shorter angle
                if (q>+128) q-=256; // use shorter angle
                         q*=q; d =q;
                q=s1-s0; q*=q; d+=q;
                if (s0<32)          // grayscales
                    {
                    d=0;            // ignore H,S
                    if (s1>=32) continue; // compare only to grayscales
                    }
                q=v1-v0; q*=q; d+=q;
                // remember closest match
                if ((d0<0)||(d0>d)) { d0=d; i0=i; }
                }
            pic0.p[y][xx].dd=base_color[i0].rgb;
            }
        pic0.bmp->Canvas->Brush->Style=bsClear;
        pic0.bmp->Canvas->Font->Color=clBlack;
        x =256; pic0.bmp->Canvas->TextOutA(5+x,5,"Naive RGB");
        x+=256; pic0.bmp->Canvas->TextOutA(5+x,5,"Normalized RGB");
        x+=256; pic0.bmp->Canvas->TextOutA(5+x,5,"HSV");
        pic0.bmp->Canvas->Brush->Style=bsSolid;
        //pic0.save("colors.png");
        }
    

    您可以忽略pic0这只是像素访问图像的东西。我在RGB距离方程中添加了一些怪癖来移动子结果,使它们适合32位int以避免溢出。作为输入,我使用此图像:

    in

    然后,对于每个像素,找到来自 LUT 的相应基色。这是结果:

    out

    在左边是源图像,然后是天真的 RGB 比较,然后是规范化的 RGB 比较(不能区分相同的颜色阴影)和右边的< strong> HSV 比较。

    对于标准化的 RGB ,找到的颜色始终是 LUT 中的第一个,颜色相同但强度不同。比较选择较暗,因为它们是 LUT 中的第一个。

    正如我之前提到的,黑暗和灰度颜色是这个问题,应该单独处理。如果您得到类似的结果并且仍然检测错误,那么您需要添加更多基色来弥补差距。如果您根本没有类似的结果,那么您很可能遇到以下问题:

      每个频道
    1. 错误的 HSV RGB 范围为<0,255>
    2. 溢出某处

      当乘以数字时,使用的位总和!所以

      8bit * 8bit * 8bit * 8bit = 32bit
      

      如果数字已经签名,那么你就麻烦了......如果在上面的例子中就像我这样的32位变量那么你需要稍微改变范围或者在{{1上使用 FPU 间隔。

    3. 为了确保我还添加了 HSV / RGB 转换,以防您遇到问题。

      这里原来的HSV产生了转换:

      out