一种检验颜色相似性的算法

我正在寻找一个算法，比较两个 RGB 颜色和生成它们的相似性(其中相似性意味着“相似的平均人类感知”)的值。

有什么想法吗？

编辑:

因为我不能再回答了，我决定把我的“解决方案”作为一个编辑的问题。

我决定在我的应用程序中使用一个(非常)小的真彩色子集，这样我就可以自己处理颜色的比较。我使用大约30种颜色，并使用它们之间的硬编码距离。

因为它是一个 iPhone 应用程序，我使用 Objective-C，实现多少是一个表示下面表格的矩阵，它显示颜色之间的距离。

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Color similarity in the RGB cube is measured by the euclidean distance (use pythagoras formula).

EDIT: On a second thought, this should be true for most other color spaces too.

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最佳答案

RGB distance in the euclidean space is not very similar to "average human perception".

You can use YUV color space, it takes into account this factor :

 |  Y' |     |  0.299     0.587    0.114   | | R |
|  U  |  =  | -0.14713  -0.28886  0.436   | | G |
|  V  |     |  0.615    -0.51499 -0.10001 | | B |

You can also use the CIE color space for this purpose.

EDIT:

I shall mention that YUV color space is an inexpensive approximation that can be computed via simple formulas. But it is not perceptually uniform. Perceptually uniform means that a change of the same amount in a color value should produce a change of about the same visual importance. If you need a more precise and rigourous metric you must definitely consider CIELAB color space or an another perceptually uniform space (even if there are no simple formulas for conversion).

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Human perception is weaker in chroma than intensity.

For example, in commercial video, the YCbCr/YPbPr color spaces (also called Y'UV) reduces the resolution of the chroma info but preserves the luma (Y). In digital video compression such as 4:2:0 and 4:2:2 reduces the chroma bitrate due to relatively weaker perception.

I believe that you can calculate a distance function giving higher priority over luma (Y) and less priority over chroma.

Also, under low intensity, human vision is practically black-and-white. Therefore, the priority function is non-linear in that for low luma (Y) you put less and less weight on chroma.

More scientific formulas: http://en.wikipedia.org/wiki/Color_difference

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Color perception is not Euclidean. Any distance formula will be both good enough and terrible at the same time. Any measure based on Euclidean distance (RGB, HSV, Luv, Lab, ...) will be good enough for similar colors, showing aqua being close to teal. But for non-close values it gets to be arbitrary. For instance, is red closer to green or to blue?

From Charles Poynton's Color FAQ:

The XYZ and RGB systems are far from exhibiting perceptual uniformity. Finding a transformation of XYZ into a reasonably perceptually-uniform space consumed a decade or more at the CIE and in the end no single system could be agreed.

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I would recommend using CIE94 (DeltaE-1994), it's said to be a decent representation of the human color perception. I've used it quite a bit in my computer-vision related applications, and I am rather happy with the result.

It's however rather computational expensive to perform such a comparison:

  1. RGB to XYZ for both colors
  2. XYZ to LAB for both colors
  3. Diff = DeltaE94(LABColor1,LABColor2)

Formulas (pseudocode):

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There's an excellent write up on the subject of colour distances here: http://www.compuphase.com/cmetric.htm

In case that resource disappears the author's conclusion is that the best low-cost approximation to the distance between two RGB colours can be achieved using this formula (in C code).

typedef struct {
unsigned char r, g, b;
} RGB;


double ColourDistance(RGB e1, RGB e2)
{
long rmean = ( (long)e1.r + (long)e2.r ) / 2;
long r = (long)e1.r - (long)e2.r;
long g = (long)e1.g - (long)e2.g;
long b = (long)e1.b - (long)e2.b;
return sqrt((((512+rmean)*r*r)>>8) + 4*g*g + (((767-rmean)*b*b)>>8));
}

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