Calculate Median Image in Matlab

Question

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.

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.

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?

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Edit: From what i can come up with, i would have to load all the images into a single matrix. The matrix would have to have 4 dimensions (height, width, rgb, images), and for each pixel and each color find the median in the 4th dimension (between the images). Is that correct (and possible)? And how can i do this?

Answer 1

Expanding my comments into a full answer;

@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.

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.

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.

EDIT:

If you want to define the "RGB median" as "the middle value in cartesian coordinates" this is easy enough to do for three images.

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.

Take the Pythagorean distance between the three colours: D1_2=abs(C2-C1), D2_3=abs(C3-C2), D1_3=abs(C3-C1).

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).

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.

Edit 2

For defining the "median" of N points as the centre of the smallest sphere that encloses them in 3D space, you could try:

1. 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.)
2. The median is then halfway between N1 and N2: M = (N1+N2)/2.

Edit 3: The above only works if no three points are equidistant. Maybe you need to ask math.stackexchange.com?

Edit 4: Wikipedia delivers again! Smallest circle problem, Bounding sphere.

Answer 2

Your intuition is correct. If you have images image_1, image_2, image_3, for example, you can assign them to a 4 dimensional matrix:

X(:,:,:,1) = image_1;
X(:,:,:,2) = image_2;
X(:,:,:,3) = image_3;

Then use:

Y=median(X,4);

To get the median.