OpenCV从正方形的向量提取图像的区域 [英] OpenCV extract area of an image from a vector of squares

问题描述

我有一个包含正方形的图片，我需要提取该正方形中包含的区域。
应用 squares.c 脚本（在每个OpenCV分发的示例中可用），我获得一个正方形的向量，然后我需要为每个对象保存一个图像。

用户 karlphillip 建议：

  （size_t x = 0; x< squares.size（）; x ++）
 {
 Rect roi（squares [x] [0] .x，squares [x] [0] .y，
 square [x] [1] .x-squares [x] [0] .x，
 squares [x] [3] .y-squares [x] [0]。 
 Mat subimage（image，roi）; 
} 
  

以生成一个新的Mat原始图像

当karl记住我时，图像中检测到的点可能不是完美的正方形（如上图所示），但是我刚才建议的代码

其实我得到这个错误：

  OpenCV Error：Assertion failed（0< = roi.x&&< = roi.width&& 
 roi.x + roi.width< = m.cols&& ; 0 <= roi.y&& 0 <= roi.height&& 
 roi.y + roi.height <= m.rows）in mat，file / usr / include / opencv / cxmat.hpp，
 line 187 
 
在抛出'cv :: Exception'实例后调用终止
 what（）：/ usr / include / opencv / cxmat。 hpp：187：error：（-215）0 <= roi.x&& 
 0< = roi.width&& roi.x + roi.width <= m.cols&& 0< = roi.y&& 
 0< = roi.height&&& roi.y + roi.height <= m.rows in function Mat 
 
中止
  

$ b b

建议让脚本接受非完美的正方形吗？

解决方案

我觉得我需要澄清一些事关于该代码。

首先，假设检测到的区域是一个完美的正方形，因为它忽略了 x] 创建一个新的 Mat 。

第二，它还假定以顺时针方向检测到该区域的点，从 p0

$ p $ （p0）第一个 - 第二个图片的左上角

< （p1）
| |
| |
（p3）4th ---- 3rd（p2）

对于所有检测到的区域。这意味着这个代码：

  Rect roi（squares [x] [0] .x，squares [x] .y，
 squares [x] [1] .x  -  squares [x] [0] .x，
 squares [x] [3] .y- squares [x] [0] .y ）; 
  

可能会产生一个ROI无效的ROI，例如负宽度和高度值，这就是为什么OpenCV在 Mat子图像（image，roi）; 上抛出 cv :: Exception
$ b

你应该做的是编写一个代码，识别该区域的左上角，并调用 p0 在右侧的neightbor， p1 ，然后找到该区域的右下角，并调用 p2 那么剩下的是 p3 。之后，组装ROI很容易：

  Rect roi（p0.x，p0.y，
 p1。 x  -  p0.x，
 p3.y  -  p0.y）; 
  

EDIT ：

我在阅读文档时发现了优秀解决方案的OpenCV v2.3。它自动化我前面描述的过程，它使事情变得更容易和干净。你可以使用这个技巧将向量中的4个点排序成一个有意义的 Rect 结构：

  //由findSquares（）返回并填充的数据。检查示例squares.cs有关此功能的更多信息。 
 vector< vector< Point> >正方形; 
 
 for（size_t i = 0; i< squares.size（）; i ++）
 {
 Rect rectangle = boundingRect（Mat（squares [i]）） 
 cout<< ＃<< i<< rectangle x：<< rectangle.x<< y：< rectangle.y<< < rectangle.width<< x< rectangle.height< endl; 
} 
  

I have an image that contains a square, and I need to extract the area contained in that square. After applying the squares.c script (available in the samples of every OpenCV distribution) I obtain a vector of squares, then I need to save an image for each of them.

The user karlphillip suggested this:

for (size_t x = 0; x < squares.size(); x++) 
{
    Rect roi(squares[x][0].x, squares[x][0].y, 
             squares[x][1].x - squares[x][0].x, 
             squares[x][3].y - squares[x][0].y);
    Mat subimage(image, roi);
}

in order to generate a new Mat called subimage for all the squares detected in the original image

As karl remembered me, the points detected in the image may not represent a perfect square (as you can see in the image above) but the code I just suggested to you assumes they do.

In fact I get this error:

OpenCV Error: Assertion failed (0 <= roi.x && 0 <= roi.width &&
      roi.x + roi.width <= m.cols && 0 <= roi.y && 0 <= roi.height &&
      roi.y + roi.height <= m.rows) in Mat, file /usr/include/opencv/cxmat.hpp, 
      line 187

terminate called after throwing an instance of 'cv::Exception'
what():  /usr/include/opencv/cxmat.hpp:187: error: (-215) 0 <= roi.x && 
       0 <= roi.width && roi.x + roi.width <= m.cols && 0 <= roi.y &&
       0 <= roi.height && roi.y + roi.height <= m.rows in function Mat

Aborted

Suggestion for make the script accept also non perfect squares?

解决方案

I feel like I need to clarify a few things about that code.

First, it assumes that the region detected is a perfect square because it ignores some of the points inside squares[x] to create a new Mat.

Second, it also assumes that the points that make the region were detected in the clockwise direction, starting with p0 in the top-left corner of the image:

(p0)  1st----2nd  (p1)
       |      |
       |      |
(p3)  4th----3rd  (p2)

which might not be true for all the regions detected. That means that this code:

Rect roi(squares[x][0].x, squares[x][0].y, 
         squares[x][1].x - squares[x][0].x, 
         squares[x][3].y - squares[x][0].y);

probably will generate a ROI with invalid dimensions, such as negative width and height values, and that's why OpenCV throws a cv::Exception at you on Mat subimage(image, roi);.

What you should do, is write a code that will identify the top-left point of the region and call it p0, then it's nearest neightbor on the right side, p1, then find the bottom-right point of the region and call it p2, and then what's left is p3. After this, assembling the ROI is easy:

Rect roi(p0.x, p0.y, 
         p1.x - p0.x, 
         p3.y - p0.y);

EDIT:

I found an excellent solution while reading the documentation of the v2.3 of OpenCV. It automates the process I described earlier and it make things so much easier and clean. You can use this trick to order the 4 Points in the vector to a meaningful Rect structure:

// Data returned and filled by findSquares(). Check the example squares.cpp for more info on this function.
vector<vector<Point> > squares;

for (size_t i = 0; i < squares.size(); i++)
{
    Rect rectangle = boundingRect(Mat(squares[i]));
    cout << "#" << i << " rectangle x:" << rectangle.x << " y:" << rectangle.y << " " << rectangle.width << "x" << rectangle.height << endl;
}

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