矩形中的最大和最小顶点 [英] Max and min vertices in a rectangle
问题描述
我是在用更好的解释来问这个问题.
I am re asking this question with a better explanation.
我已经获得了一个矩形的四个点，我希望对其执行扭曲透视变换.由于我手动选择了点并为变换指定了它们的位置，因此我已经成功地对其中一张图像进行了变换.
I have obtained the four points of a rectangle that I wish to perform a warp perspective transformation on. I have successfully achieved a transform on one of my images as I manually selected the points and assigned their location for the transform.
cv::Point2f src_vertices[4];
src_vertices[0] = corners[3];
src_vertices[1] = corners[1];
src_vertices[2] = corners[0];
src_vertices[3] = corners[2];
Point2f dst_vertices[4];
dst_vertices[0] = Point(0, 0);
dst_vertices[1] = Point(box.boundingRect().width-1, 0);
dst_vertices[2] = Point(0, box.boundingRect().height-1);
dst_vertices[3] = Point(box.boundingRect().width-1, box.boundingRect().height-1);
如您所见，我手动将角分配给转换顶点.
As you can see I assigned the corners to the transformation vertices manually.
因为我希望在多个略有不同的图像上使用此代码，所以我想要一种更准确的角分配方法.
As I wish to use this code on multiple slightly different images I would like a more accurate way of assigning the corners.
所有图像看起来都类似于:
All the images will look similar to this:
这张图片的角如下:左上角 - (1106, 331),右上角 - (810, 747),左下角- (825, 187),右下角 - (510, 537)
The corners for this image are as follows: Top left - (1106, 331), Top right- (810, 747), Bottom left- (825, 187), Bottom right- (510, 537)
我找到了最大和最小 x 和 y 值，如下所示:
I have found the max and min x and y values as follows:
float X, Y;
float maxX= 0;
float minX = 10000;
float maxY= 0;
float minY = 10000;
for(int i=0; i< 4; i++){
if( corners[i].x > maxX){
maxX = corners[i].x;
}
if( corners[i].x < minX){
minX = corners[i].x;
}
if( corners[i].y > maxY){
maxY = corners[i].y;
}
if( corners[i].y < minY){
minY = corners[i].y;
}
}
这给了我:
maxX - 1106，minX - 510，maxY - 747，minY - 187.
maxX - 1106, minX - 510, maxY - 747, minY - 187.
我想知道如何将最大值和最小值重新组合为其各自的值，以便我可以使用这些点来执行变换.我对 opencv 还很陌生，如果这很明显，我很抱歉.
I would like to know how to recombine the max and min values to their respective values so I can use the points to perform the transform. I am quite new to opencv so sorry if this is very obvious.
推荐答案
要在不丢失其他坐标信息的情况下获得角点，您可以做的是找到整点.而不是只存储 float maxX= 0; 你可以存储 Point2f maxX(0,0); 并修改你 for 循环以跟踪点本身，而不仅仅是 x 坐标.
What you can do to get the corners without losing information on the other coordinate is to find the whole point. Instead of storing just float maxX= 0; you can store Point2f maxX(0,0); and modify you for loop to keep track of the point itself, not only the x coordinate.
更好的是，摆脱那些循环并使用 STL
Better yet, get rid of those loops and make use of STL
Point2f maxX =*std::max_element(corners, corners+4, [](Point2f a, Point2f b){return a.x < b.x;});
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