OpenCV shape detection -- C++版本

最新推荐文章于 2024-12-15 08:45:00 发布

原创最新推荐文章于 2024-12-15 08:45:00 发布 · 3.7k 阅读

3 ·

CC 4.0 BY-SA版权

文章标签：

#opencv

OpenCV 同时被 2 个专栏收录

83 篇文章

订阅专栏

C++/C

63 篇文章

订阅专栏

部署运行你感兴趣的模型镜像

本文主要参考OpenCV shape detection ，用C++版的OpenCV API进行重写。

源码

ShapeDetector.h

#pragma once
#ifndef SHAPEDETECTOR_H_
#define SHAPEDETECTOR_H_

#include <iostream>
#include <vector>
#include <string>

using namespace std;

#include "opencv2/core/core.hpp"
#include "opencv2/highgui/highgui.hpp"
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/features2d/features2d.hpp"

using namespace cv;

class ShapeDetector
{
public:
    ShapeDetector();
    ~ShapeDetector();
    void detect(const Mat &curve);
    string get_shape_type();

private:
    string m_shape;
};

#endif

ShapeDetector.cpp

#include "ShapeDetector.h"

ShapeDetector::ShapeDetector()
{
}

ShapeDetector::~ShapeDetector()
{
}

void ShapeDetector::detect(const Mat &curve)
{
    string shape = "unidentified";
    double peri = arcLength(curve, true);
    Mat approx;
    approxPolyDP(curve, approx, 0.04 * peri, true); // 0.01~0.05
    const int num_of_vertices = approx.rows;

    // if the shape is a triangle, it will have 3 vertices
    if (num_of_vertices == 3)
    {
        shape = "triangle";
    }
    else if (num_of_vertices == 4)
    {// if the shape has 4 vertices, it is either a square or a rectangle
        // Compute the bounding box of the contour and
        // use the bounding box to compute the aspect ratio
        Rect rec = boundingRect(approx);
        double ar = 1.0 * rec.width / rec.height;

        // A square will have an aspect ratio that is approximately
        // equal to one, otherwise, the shape is a rectangle
        if (ar >= 0.95 && ar <= 1.05)
        {
            shape = "square";
        }
        else
        {
            shape = "rectangle";
        }
    }
    else if (num_of_vertices == 5)
    {// if the shape is a pentagon, it will have 5 vertices
        shape = "pentagon";
    }
    else
    {// otherwise, we assume the shape is a circle
        shape = "circle";
    }
    m_shape = shape;
}

string ShapeDetector::get_shape_type()
{
    return m_shape;
}

main.cpp

#include "ShapeDetector.h"

int main(int argc, char* argv[])
{
    const string imgpath = "D:/TMP/test.png";
    Mat image = imread(imgpath);
    if (image.empty())
    {
        return -1;
    }
    Mat gray;
    if (image.channels()==3)
    {
        cvtColor(image, gray, COLOR_BGR2GRAY);
    }
    else
    {
        gray = image.clone();
    }

    Mat blurred, thresh;
    GaussianBlur(gray, blurred, Size(5, 5), 0.0);
    threshold(blurred, thresh, 60, 255, THRESH_BINARY);

    vector< vector<Point> > contours;
    vector<Vec4i> hierarchy;
    findContours(thresh, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE);

    ShapeDetector sd;
    vector<Point> c;
    for (size_t i = 0; i < contours.size(); i++)
    {
        c = contours[i];
        Rect crect = boundingRect(c);
        // compute the center of the contour, then detect the name of the
        // shape using only the contour
        Moments M = moments(c);
        int cX = static_cast<int>(M.m10 / M.m00);
        int cY = static_cast<int>(M.m01 / M.m00);
        sd.detect(Mat(c));
        string shape = sd.get_shape_type();
        drawContours(image, contours, i, Scalar(0, 255, 0), 2);
        Point pt(cX, cY);
        putText(image, shape, pt, FONT_HERSHEY_SIMPLEX, 0.5, Scalar(255, 255, 255), 2);
        imshow("Image", image);
        waitKey(0);
    }
    return 0;
}