在Matlab中计算中位数图像 [英] Calculate Median Image in Matlab
问题描述
我是matlab的新手，所以请原谅我，如果我在这里要求明显的：我所拥有的是一系列彩色摄影图像（所有相同的尺寸）。我想要做的是计算每个像素的中位数颜色值。
I am new to matlab, so forgive me if i am asking for the obvious here: what i have is a collection of color photographic images (all the same dimensions). What i want to do is calculate the median color value for each pixel.
我知道matlab中有一个中值滤波器，但据我所知它并没有完全符合我的要求。因为我想计算整个图像集合之间的中间值，对于每个单独的像素。
I know there is a median filter in matlab, but as far as i know it does not do exactly what i want. Because i want to calculate the median value between the entire collection of images, for each separate pixel.
所以例如，如果我有三个图像，我想要matlab来计算（对于每个像素）这三个图像中的哪个颜色值是中值。我怎么会这样做，有人知道吗？
So for example, if i have three images, i want matlab to calculate (for each pixel) which colorvalue out of those three images is the median value. How would i go about doing this, does anyone know?
编辑：从我能想到的，我将不得不将所有图像加载到一个矩阵中。矩阵必须具有4个维度（高度，宽度，rgb，图像），并且对于每个像素，每个颜色在第4维度（图像之间）中找到中值。
这是正确的（也可能）？我该怎么做？
推荐答案
将我的评论扩展为完整答案;
Expanding my comments into a full answer;
@ prototoast的答案很优雅，但由于每个像素的R，G和B值的中位数是单独计算的，输出图像看起来会很奇怪。
@prototoast's answer is elegant, but since medians for the R, G and B values of each pixel are calculated separately, the output image will look very strange.
要获得具有视觉意义的明确定义的中位数，最简单的方法是在尝试取中位数之前将图像转换为黑白图像。
To get a well-defined median that makes visual sense, the easiest thing to do is cast the images to black-and-white before you try to take the median.
来自图像处理工具箱的rgb2gray（）将以保留每个像素的亮度的方式执行此操作丢弃色调和饱和度。
rgb2gray() from the Image Processing toolbox will do this in a way that preserves the luminance of each pixel while discarding the hue and saturation.
编辑：
如果你想定义RGB中位数为笛卡尔坐标中的中间值这对于三幅图像来说很容易做到。
If you want to define the "RGB median" as "the middle value in cartesian coordinates" this is easy enough to do for three images.
考虑单个像素，中位数有三种可能的选择颜色， C1 =（r1，g1，b1）， C2 =（r2，g2，b2）， C3 =（R3，G3，B3）。通常这些在3D空间中形成一个三角形。
Consider a single pixel with three possible choices for the median colour, C1=(r1,g1,b1), C2=(r2,g2,b2), C3=(r3,g3,b3). Generally these form a triangle in 3D space.
取三种颜色之间的毕达哥拉斯距离： D1_2 = abs（C2-C1）， D2_3 = abs（C3-C2）， D1_3 = abs（C3-C1）。
Take the Pythagorean distance between the three colours: D1_2=abs(C2-C1), D2_3=abs(C3-C2), D1_3=abs(C3-C1).
选择中位数作为与其他两个距离最小的颜色。定义 D1 = D1_2 + D1_3 等，并且 min（D1，D2，D3）应该有效，礼貌 Triangle Inequality 。注意退化情况：等边三角形（C1，C2，C3等距），线（C1，C2，C3彼此线性）或点（C1 = C2 = C3）。
Pick the "median" to be the colour that has lowest distance to the other two. Defining D1=D1_2+D1_3, etc. and taking min(D1,D2,D3) should work, courtesy of the Triangle Inequality. Note the degenerate cases: equilateral triangle (C1, C2, C3 equidistant), line (C1, C2, C3 linear with each other), or point (C1=C2=C3).
请注意，这种考虑3D中值的简单方法很难扩展到三个以上的图像，因为一组四个或更多3D点的中位数有点难以定义。
Note that this simple way of thinking about a 3D median is hard to extend to more than three images, because "the median" of a set of four or more 3D points is a bit harder to define.
编辑2
用于定义N点的中位数作为最小点的中心在3D空间中包围它们的球体，您可以尝试：
For defining the "median" of N points as the centre of the smallest sphere that encloses them in 3D space, you could try:
- 找到{N}中相距最远的两个点N1和N2。 N1和N2之间的距离是包围所有点的最小球体的直径。 （证明：任何较小的球体都不能同时包围N1和N2。）
- 中位数在N1和N2之间：
M =（N1 + N2）/ 2。
- Find the two points N1 and N2 in {N} that are furthest apart. The distance between N1 and N2 is the diameter of the smallest sphere that encloses all the points. (Proof: Any smaller and the sphere would not be able to enclose both N1 and N2 at the same time.)
- The median is then halfway between N1 and N2:
M = (N1+N2)/2.
编辑3 ：以上只有在没有三个等距的情况下才有效。也许你需要问math.stackexchange.com？
Edit 3: The above only works if no three points are equidistant. Maybe you need to ask math.stackexchange.com?
编辑4 ：维基百科再次发布！ 最小的圈子问题， Bounding sphere 。
Edit 4: Wikipedia delivers again! Smallest circle problem, Bounding sphere.
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