如何在图像中找到局部最大值 [英] how to find local maxima in image

问题描述

问题在于特征检测的概念. 找到图像的角点后卡住了，我想知道如何在计算的角点内找到特征点.

The question is about feature detection concept. I'm stuck after I finding the corner of image and I want to know how to finding the feature point within the computed corners.

假设我有这样的灰度图像

Suppose I have grayscale image that have data like this

A = [ 1 1 1 1 1 1 1 1;
      1 3 3 3 1 1 4 1;
      1 3 5 3 1 4 4 4;
      1 3 3 3 1 4 4 4; 
      1 1 1 1 1 4 6 4;
      1 1 1 1 1 4 4 4]

如果我使用

B = imregionalmax(A);

结果将是这样

B = [ 0 0 0 0 0 0 0 0;
      0 1 1 1 0 0 1 0;
      0 1 1 1 0 1 1 1;
      0 1 1 1 0 1 1 1;
      0 0 0 0 0 1 1 1;
      0 0 0 0 0 1 1 1]

问题是如何在最大局部区域内拾取最高峰(在示例中，我是如何从3中选择5并从4中选择6)?

The question is how do I pick the highest peak inside max local region (in sample how did I chose 5 from 3 and 6 from 4)?

我的想法是使用B来检测每个区域并再次使用imregionalmax()，但是我不擅长编码，因此我需要一些建议或其他想法.

My idea was using B to detect each region and use imregionalmax() again but I'm not good at coding and I need some advice or other ideas.

推荐答案

还有其他两种简单的方法可以实现2D峰查找器:

There are a couple of other easy ways to implement a 2D peak finder: ordfilt2 or imdilate.

最直接的方法是使用ordfilt2，它对本地邻域中的值进行排序并选择第n个值. ( MathWorks示例演示了如何实现最大过滤器).您还可以使用ordfilt2来实现3x3峰值查找器，方法是:(1)使用不包括中心像素的3x3域，(2)选择最大的(第8个)值，然后( 3)与中心值比较:

The most direct method is to use ordfilt2, which sorts values in local neighborhoods and picks the n-th value. (The MathWorks example demonstrates how to implemented a max filter.) You can also implement a 3x3 peak finder with ordfilt2 by, (1) using a 3x3 domain that does not include the center pixel, (2) selecting the largest (8th) value and (3) comparing to the center value:

>> mask = ones(3); mask(5) = 0 % 3x3 max
mask =
     1     1     1
     1     0     1
     1     1     1

此掩码中考虑了8个值，因此第8个值是最大值.过滤器输出:

There are 8 values considered in this mask, so the 8-th value is the max. The filter output:

>> B = ordfilt2(A,8,mask)
B =
     3     3     3     3     3     4     4     4
     3     5     5     5     4     4     4     4
     3     5     3     5     4     4     4     4
     3     5     5     5     4     6     6     6
     3     3     3     3     4     6     4     6
     1     1     1     1     4     6     6     6

技巧是将其与每个邻域的中心值A进行比较:

The trick is compare this to A, the center value of each neighborhood:

>> peaks = A > B
peaks =
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     1     0
     0     0     0     0     0     0     0     0

imdilate

图像膨胀通常在二进制图像上完成，但是灰度图像膨胀只是一个最大滤镜(请参见). ordfilt2使用的相同技巧也适用于此:定义不包含中心邻域像素的邻域，应用滤镜并将其与未滤镜的图像进行比较:

imdilate

Image dilation is usually done on binary images, but grayscale image dilation is simply a max filter (see Definitions section of imdilate docs). The same trick used with ordfilt2 applies here: define a neighborhood that does not include the center neighborhood pixel, apply the filter and compare to the unfiltered image:

B = imdilate(A, mask);
peaks = A > B;

注意:这些方法只能找到一个像素峰值.如果任何一个邻居的值相同，那么它就不会是峰值.

NOTE: These methods only find a single pixel peak. If any neighbors have the same value, it will not be a peak.

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