OpenCv 2.4.2代码解释 - 面部识别 [英] Opencv 2.4.2 Code Explaination-Face Recognition

问题描述

我已经参考OpenCV提供的文档来做一个面部识别程序，它识别多个面孔，并正常工作。在文档中，他们使椭圆突出了脸。我不明白是他们如何计算椭圆的中心，他们计算如下

I have referred the documentation provided by OpenCV to make a face recognition program, it recognizes multiple faces and is working normally. In the documentation they have made ellipses to highlight the face. What I don't understand is how they have calculated the center of the ellipse which they have calculated as follows

for( int i = 0; i < faces.size(); i++ )
{
   Point center(faces[i].x+faces[i].width*0.5,faces[i].y+faces[i].height*0.5);
   //more code follows drawing the ellipse

它们使用的面向量如下

face_cascade.detectMultiScale(frame_gray,faces,1.1,2,0|CV_HAAR_SCALE_IMAGE,cv::Size(30,30))

文档即程序在链接中提供

The documentation i.e. the program is given in the link

http://docs.opencv.org/doc/tutorials/objdetect/ cascade_classifier / cascade_classifier.html

我想知道它们如何计算椭圆的中心，如果我想画一个矩形而不是一个圆，我需要做什么？

I want to know how they are calculating the center of the ellipse and if I want to draw a rectangle instead of a circle, what do I have to do?

推荐答案

检测到的面部作为一组矩形包围面部。如文档所述，输出矩形的向量，其中每个矩形包含检测到的对象。

Detected faces are returned as a set of rectangles surrounding the faces. As documentation says, output is Vector of rectangles where each rectangle contains the detected object.

[初始x，初始y，宽度，高度] 。所以你可以通过（x + width * 0.5，y + height * 0.5）找到它的中心。这个中心对于椭圆也是相同的。

So one rectangle is comprised of [ initial x, initial y, width, height ]. So you can find its center by ( x + width*0.5 , y + height*0.5 ). This center is same for the ellipse also.

如果要绘制矩形，请使用 rectangle 函数。请参见文档。

If you want to draw rectangles, use rectangle function. See the Documentation.

函数中的参数如下：

pt1 = ( x , y )

pt2 = ( x + width , y + height )

将线条图椭圆形更改为以下线：

Change the line drawing ellipse to following line :

rectangle(frame,Point (faces[i].x,faces[i].y),Point (faces[i].x+faces[i].width, faces[i].y+faces[i].height),Scalar(255,0,255),4,8,0);

它的结果如下：

这篇关于OpenCv 2.4.2代码解释 - 面部识别的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！