如何计算从0到360的色调值的平均值和标准偏差？ [英] How to calculate mean and standard deviation for hue values from 0 to 360?

问题描述

假设使用用于颜色的简单HSV模型获取5个色调样本，具有值355,5,5,5,5，所有的红色色调和关于彼此的下一个。但是简单的平均值是75，远离0或360，接近黄绿色。

什么是更好的方法来计算这个平均值和相关的std？

解决方案

简单的解决方案是将这些角度转换为一组向量，从极坐标到笛卡尔坐标。

因为你使用颜色，所以认为这是一个转换（a *，b *）平面。然后取这些坐标的平均值，然后再次恢复为极坐标形式。在matlab中完成，

  theta = [355,5,5,5,5]; 
 x = cosd（theta）; ％余弦以度数表示
 y = sind（theta）;具有度参数的％正弦
  

现在，取x和y的平均值，

  meanangle = atan2（mean（y），mean（x））* 180 / pi 
 averageangle = 
 3.0049 
  

当然，只对平均角。正如你所看到的，它直接产生与角度的平均值一致的结果，其中我认识到355度真的包裹到-5度。

  mean（[ -  5 5 5 5 5]）
 ans = 
 3 
  

b $ b

要计算标准差，最简单的做法是

  std（[ -  5 5 5 5 5]）
 ans = 
 4.4721 
  

明确做包装。

Suppose 5 samples of hue are taken using a simple HSV model for color, having values 355, 5, 5, 5, 5, all a hue of red and "next" to each other as far as perception is concerned. But the simple average is 75 which is far away from 0 or 360, close to a yellow-green.

What is a better way to calculate this mean and associated std?

解决方案

The simple solution is to convert those angles to a set of vectors, from polar coordinates into cartesian coordinates.

Since you are working with colors, think of this as a conversion into the (a*,b*) plane. Then take the mean of those coordinates, and then revert back into polar form again. Done in matlab,

theta = [355,5,5,5,5];
x = cosd(theta); % cosine in terms of degrees
y = sind(theta); % sine with a degree argument

Now, take the mean of x and y, compute the angle, then convert back from radians to degrees.

meanangle = atan2(mean(y),mean(x))*180/pi
meanangle =
       3.0049

Of course, this solution is valid only for the mean angle. As you can see, it yields a consistent result with the mean of the angles directly, where I recognize that 355 degrees really wraps to -5 degrees.

mean([-5 5 5 5 5])
ans =
     3

To compute the standard deviation, it is simplest to do it as

std([-5 5 5 5 5])
ans =
       4.4721

Yes, that requires me to do the wrap explicitly.

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