Info

此问题已关闭。 请重新打开它进行编辑或回答。

the problem of image segementation

1 次查看（过去 30 天）
Y
Y 2017-10-12
关闭： MATLAB Answer Bot 2021-8-20
Hello, everyone. I read the demo from help about color-based segmentation using L*a*b space. First, it acquires the image and then it calculate color value in L*a*b space for each region. But I can not understand the code for classification.
if true
%for count = 1:nColors
distance(:,:,count) = ( (a - color_markers(count,1)).^2 + ...
(b - color_markers(count,2)).^2 ).^0.5;
[~, label] = min(distance,[],3);
label = color_labels(label);
end
end
Could U explain the function of min here? I know it needs to find the nearest distance, but why the first parameter in the [ ] is ~? Thanks in advance. Best, Helena.

此问题已关闭。

回答（1 个）

Nicolas Schmit
Nicolas Schmit 2017-10-12
In this problem, you are not interested in the value of the minimum distance itself, but in where the minimum distance occurs in your distance array. The command
[~, label] = min(distance,[],3);
returns in 'label' the index of the minimum value in the distance array. ~ is a command to ignore the first value returned by the min() function, which is the minimum value of the distance array. Here, you ignore this value since you do not need it.
  3 个评论
显示 1更早的评论
Image Analyst
Image Analyst 2017-10-12
A better way than squaring and using min() would be to use a discriminant classifier, classify(). If you have the Statistics and Machine Learning Toolbox and would like a demo, let me know. Another way is to use rgb2ind().
Y
Y 2017-10-13
I would like a demo about that. Tx!!

此问题已关闭。

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!