如何在MATLAB中检测这个图片中某个角度下对象的尺寸？ [英] How can I detect the dimensions of an object under an angle in this picture in MATLAB?

问题描述

我有这个电池图片：

我想确定电池的尺寸（以像素为单位）。

我遇到的问题是电池旋转未知角度。

如何检测旋转电池的尺寸？

我在想这些算法步骤：

• 以将该图像转换为黑白图像（阈值）。


• 之后，我必须找到一个中心点，并在白色像素中绘制一个矩形。


• 然后，我必须将矩形旋转360度，并找到矩形的位置（以及尺寸）。

我有点缺乏经验，我会感谢一些指导如何在Matlab中实现这些算法阶段。

谢谢

解决方案

考虑这是Matlab图像处理的初学者教程。阅读所用命令的文档，尝试并了解他们正在做什么以及原因。

1。阅读图片

使用以提高阈值结果。

您可以使用。它允许你获得各种不同的属性。您将使用它来计算 'Orientation' 您图片的白色/电池区域

 >> st = regionprops（bw，'Orientation'）
 st = 
方向：52.8694 
  

您可以看到电池旋转了52.8度。

使用 imrotate 拉直电池

 >> rbw = imrotate（bw，-st.Orientation）; 
  

一旦电池轴对准，您可以投影白色像素水平和垂直轴使用 任何 ：

 >> pc = any（rbw，2）; ％//将所有行投影到单个列中
>> pr = any（rbw，1）; ％//将所有列投影到一行
  

现在您需要找到第一个和最后一个像素集到1。使用 查找 为：

 >> fx = find（pr，1，'first'）; ％//第一个x坐标
>> Tx = find（pr，1，'last'）; ％// last x coordinat 
>> fy = find（pc，1，'first'）; ％//第一个y坐标
>> ty = find（pc，1，'last'）; ％//上次y坐标
  

一旦你有x，y坐标的角，旋转的图片：

 >> imshow（rbw，[]，'border'，'tight'）; 
>>坚持，稍等; 
>> '，'：''''，'''''''' 
  

产生：

并且坐标是：

 >> [fx fy tx ty] 
 ans = 
 406 608 866 733 
  

您可以看到您的电池是（866-406）像素长和（733-608）像素宽。

I have this image of a battery:

I would like to determine the dimensions of the battery (in pixels).
The problem I have is that the battery is rotated by an unknown angle.
How can I detect the dimensions of this rotated battery?

I was thinking of these algorithmic steps:

• First, I would have to convert this image to a black and white image (thresholding).
• After that, I would have to find a center point and draw a rectangle in the white pixels.
• Then, I have to turn the rectangle 360 degrees and locate the position of the rectangle (and so the dimensions).

I am somewhat inexperienced, and I would appreciate some guidance as to how to implement these algorithmic stages in Matlab.

Thanks

解决方案

Consider this as a beginner's tutorial to Matlab image processing. Read the documentation of the commands used and try and understand what they are doing and why.

1. Read the image

Use imread to read the image into a 3D matrix. For convenience, we convert it to double in the range [0..1] using im2double:

>> img = im2double( imread( 'path/to/battety.jpg' ) );

You can check out the size of your img using size command:

>> size( img )
ans =
    1024         768           3

You can see from the result that your image has 1024 rows, 768 columns and 3 channels (Red, Green and Blue).

2. Convert to black and white (a.k.a thresholding)

As you can see the battery is significantly brighter than the background and is colorless. We can select pixels that have large gap between the brightest channel value to darkest channel values as "battery" pixels:

>> bw = (max(img,[],3)-min(img,[],3)) > 0.2;

See max and min for more details.
There are other methods to threshold an image, see graythresh for more details.

Using imshow we can see what we got:

>> imshow(bw,[],'border','tight');

Normally one uses morphological operations to improve thresholding results.
You can use imclose:

>> bw = imclose( bw, ones(25) );

Resulting with:

3. find angle of rotation

A very useful command for processing and working with bw images is regionprops. It allows you to get all sorts of nice properties. You are going to use it to compute the 'Orientation' of the "white"/battery region of your image

>> st = regionprops( bw, 'Orientation' )
st = 
Orientation: 52.8694

As you can see the battery is rotated by 52.8 degrees.
Using imrotate to "straighten" the battery

>> rbw = imrotate( bw, -st.Orientation );

Once the battery is axis-aligned, you can "project" the white pixels onto the horizontal and vertical axes using any:

>> pc = any( rbw, 2 ); %// project all rows into a single column 
>> pr = any( rbw, 1 ); %// project all columns into a single row

Now you need to find the first and last pixels set to 1 in the projections. Use find for that:

>> fx = find( pr, 1, 'first');  %// first x coordinate
>> tx = find( pr, 1, 'last');   %// last x coordinat
>> fy = find( pc, 1, 'first');  %// first y coordinate
>> ty = find( pc, 1, 'last');   %// last y coordinate  

Once you have x,y coordinates of corners, you can plot them on the rotated image:

>> imshow(rbw,[],'border','tight');
>> hold on; 
>> plot( [fx tx tx fx fx], [fy fy ty ty fy], ':r', 'LineWidth',3);

Yields:

And the coordinates are:

>> [fx fy tx ty]
ans =
406   608   866   733

As you can see your battery is (866-406) pixels long and (733-608) pixels wide.

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