选择图像中的最大对象 [英] Select largest object in an image

问题描述

我正在尝试找到图像中最大的对象，并删除图像中小于它的任何其他对象。

I am trying to find the the largest object in an image and remove any other objects in the image that are smaller than it.

这就是我所拥有的但是我无法让它发挥作用。

This is what I have but I cannot get it to work.

 l=bwlabel(BW);

 %the area of all objects in the image is calculated
 stat = regionprops(l,'Area','PixelIdxList');
 [maxValue,index] = max([stat.Area]);

  %remove any connected areas smaller than the biggest object
  BW2=bwareaopen(BW,[maxValue,index],8);
  subplot(5, 5, 4);
  imshow(BW2, []);

我正在处理数字乳房X线照片，例如这些。我试图从图像中删除除乳房区域以外的所有对象。

I am working with digital mammograms such as these. I am trying to remove all objects from the image except for the breast region.

推荐答案

使用 bwconncomp ，因为它返回区域的坐标索引在一个单独的单元格中，每个单元格的大小很容易辨别：

Use bwconncomp instead since it returns the coordinate indexes for region in a separate cell, where the size of each is easily discerned:

>> BW = [1 0 0; 0 0 0; 0 1 1]; % two regions
>> CC = bwconncomp(BW)
CC = 
    Connectivity: 8
       ImageSize: [3 3]
      NumObjects: 2
    PixelIdxList: {[1]  [2x1 double]}

PixelIdxList 字段是一个单元格数组使用每个区域的坐标索引。每个数组的长度是每个区域的大小：

The PixelIdxList field is a cell array with the indexes of coordinates for each region. The length of each array is the size of each region:

>> numPixels = cellfun(@numel,CC.PixelIdxList)
numPixels =
     1     2
>> [biggestSize,idx] = max(numPixels)
biggestSize =
     2
idx =
     2

然后你可以用这个组件轻松制作一个新图像：

Then you can easily make a new image with just this component:

BW2 = false(size(BW));
BW2(CC.PixelIdxList{idx}) = true;

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编辑：来自评论，需要裁剪输出图像，以便区域到达边缘可以用 regionprops 使用'BoundingBox'选项：

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EDIT: From the comments, the need to crop the output image so that the region comes to the edges can be addressed with regionprops using the 'BoundingBox' option:

s  = regionprops(BW2, 'BoundingBox');

给你一个矩形 s.BoundingBox 可用于裁剪 BW3 = imcrop（BW2，s.BoundingBox）; 。

which gives you a rectangle s.BoundingBox which you can use to crop with BW3 = imcrop(BW2,s.BoundingBox);.

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