A Depth Image Segmentation Method Based on Mean Shift Algorithm and Mathematical Morphology
A mean shift algorithm and mathematical morphology technology, applied in the field of depth map segmentation, can solve problems such as background segmentation with high degree of automation, and achieve the effect of high degree of automation
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specific Embodiment approach 1
[0027] Specific implementation mode 1. Combination Figure 4 to Figure 9 This specific embodiment will be described. This specific implementation mode is:
[0028] Step 1: Generate a grayscale histogram hist(k) of the depth map according to the depth image whose grayscale is f(x, y); said L is grayscale, L=0, 1...255;
[0029] The abscissa of the gray histogram is the gray level L, and the ordinate is the frequency of occurrence of the gray level;
[0030] Step 2: Smooth the gray histogram by using a one-dimensional Gaussian window function, and set the foreground threshold T;
[0031] The foreground threshold T is the first minimum value that includes a certain semantic range according to the gray value from high to low, and the certain semantic range is that the number of pixels contained in this area is greater than 10% of the number of pixels in the entire image, That is, hist(k>T)>10%*hist(0...L-1); the minimum value is hist(k)=min(hist(k-4), hist(k-3)...hist(k +4)); ...
specific Embodiment approach 2
[0048] Specific embodiment 2. The difference between this specific embodiment and specific embodiment 2 is that the process of generating the gray histogram hist(k) of the depth map according to the depth image whose gray level is f(x, y) in step 1 is:
[0049] Step 1A: Initialize hist(k)=0; k=0,1...L-1;
[0050] Step 1B: Statistics f(x, y)=k; hist(k+1), x=0, 1...M-1; y=0, 1...N-1.
specific Embodiment approach 3
[0051] Specific embodiment three, the difference between this specific embodiment and specific embodiment three is that the grayscale histogram is smoothed using the one-dimensional Gaussian modulus described in step 2, and the Gaussian modulus used is [0.10.20.40.20.1]; after smoothing The result is hist(k)=0.1hist(k-2)+0.2hist(k-1)+0.4hist(k)+0.2hist(k+1)+0.1hist(k+2); k=0, 1...L-1.
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