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本文介紹了正方形檢測沒有找到正方形的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！

問題描述

我正在使用 OpenCV 庫示例中提供的程序 squares.c.它適用于每個圖像，但我真的不明白為什么它不能識別該圖像中繪制的正方形

DILATE 之后:

RESULT 圖像(紅色)http://img267.imageshack.us/img267/8016/resultuq.jpg

如您所見，未檢測到正方形.

檢測后我需要提取正方形中包含的區域......沒有ROI怎么可能?

解決方案

下面的源代碼展示了 Square Detector 程序的一個小變化.它并不完美，但它說明了解決您的問題的一種方法.

您可以將此代碼與原始代碼進行比較并檢查所做的所有更改，但主要的更改是:

將閾值級別的數量減少到 2.
在findSquares()的開頭，擴大圖像檢測細小的白色方塊，然后模糊整個圖像，因此算法不會將大海和天空檢測為單個方塊.

編譯后，使用以下語法運行應用程序:./app

//平方檢測器"程序.//它順序加載多個圖像并嘗試在其中找到正方形//每張圖片#include "highgui.h"#include "cv.h"#include #include <math.h>#include 使用命名空間 cv；使用命名空間標準；無效的幫助(){cout<<"
一個使用金字塔縮放、Canny、輪廓、輪廓簡化和
的程序"內存存儲(所有人都可以找到)
""圖像列表中的方塊 pic1-6.png
""返回在圖像上檢測到的正方形序列.
""序列存儲在指定的內存中
"呼叫:
""./平方
""使用 OpenCV 版本 %s
" <<CV_VERSION <<"
" <<結束;}整數閾值 = 50，N = 2；//karlphillip:將 N 減少到 2，是 11.const char* wndname = "平方檢測演示";//輔助函數://找到向量之間夾角的余弦值//從 pt0->pt1 和從 pt0->pt2雙角(點pt1，點pt2，點pt0){雙 dx1 = pt1.x - pt0.x;雙 dy1 = pt1.y - pt0.y;雙 dx2 = pt2.x - pt0.x;雙 dy2 = pt2.y - pt0.y;返回 (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);}//返回在圖像上檢測到的正方形序列.//序列存儲在指定的內存中void findSquares( const Mat& image, vector<vector<Point> >& squares ){squares.clear();Mat pyr, timg, gray0(image.size(), CV_8U), gray;//karlphillip: 擴大圖像，以便此技術可以檢測白色方塊，墊出(圖像)；擴張(出，出，墊()，點(-1，-1))；//然后模糊它，使海洋/大海成為一大段，以避免將它們檢測為 2 個大方塊.中值模糊(出，出，7)；//縮小和放大圖像以濾除噪聲pyrDown(out, pyr, Size(out.cols/2, out.rows/2));pyrUp(pyr, timg, out.size());矢量<矢量<點>>輪廓；//在圖像的每個顏色平面中找到正方形for( int c = 0; c <3; c++ ){int ch[] = {c, 0};mixChannels(&timg, 1, &gray0, 1, ch, 1);//嘗試幾個閾值級別for( int l = 0; l = (l+1)*255/N；}//找到輪廓并將它們全部存儲為列表findContours(灰色，輪廓，CV_RETR_LIST，CV_CHAIN_APPROX_SIMPLE)；矢量<點>約;//測試每個輪廓for( size_t i = 0; i 1000 & &isContourConvex(Mat(approx)) ){雙最大余弦 = 0;for( int j = 2; j <5; j++ ){//找到關節邊緣之間夾角的最大余弦值雙余弦 = fabs(angle(approx[j%4],approx[j-2],approx[j-1]));maxCosine = MAX(maxCosine, cosine);}//如果所有角度的余弦都很小//(所有角度都是~90度)然后寫quandrange//頂點到結果序列如果(最大余弦＜0.3)squares.push_back(大約)；}}}}}//該函數繪制圖像中的所有方塊void drawSquares( Mat& image, const vector<vector<Point> >& squares ){for( size_t i = 0; i < squares.size(); i++ ){const Point* p = &squares[i][0];int n = (int)squares[i].size();polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);}imshow(wndname，圖像)；}int main(int argc, char** argv){如果 (argc <2){cout<<用法:./program <文件>"<<結束;返回-1；}//static const char* names[] = { "pic1.png", "pic2.png", "pic3.png",//"pic4.png", "pic5.png", "pic6.png", 0 };靜態常量字符*名稱[] = { argv[1], 0 };幫助();命名窗口( wndname, 1 );矢量<矢量<點>>正方形；for( int i = 0; names[i] != 0; i++ ){Mat image = imread(names[i], 1);如果(圖像.空()){cout<<無法加載" <<名稱[i] <<結束;繼續;}findSquares(圖像，正方形)；drawSquares(圖像，正方形)；imwrite("out.jpg", image);int c = waitKey();如果((字符)c == 27 )休息;}返回0；}

輸出:

I'm using the program squares.c available in the samples of OpenCV libraries. It works well with every image, but I really can't figure it out why it doesn't recognize the square drawn in that image

http://desmond.imageshack.us/Himg12/scaled.php?server=12&filename=26725680.jpg&res=medium

After CANNY:

After DILATE:

The RESULT image (in red) http://img267.imageshack.us/img267/8016/resultuq.jpg

As you can see, the square is NOT detected.

After the detection I need to extract the area contained in the square...How is it possible without a ROI?

解決方案

The source code below presents a small variation of the Square Detector program. It's not perfect, but it illustrates one way to approach your problem.

You can diff this code to the original and check all the changes that were made, but the main ones are:

Decrease the number of threshold levels to 2.
In the beginning of findSquares(), dilate the image to detect the thin white square, and then blur the entire image so the algorithm doesn't detect the sea and the sky as individual squares.

Once compiled, run the application with the following syntax: ./app <image>

// The "Square Detector" program.
// It loads several images sequentially and tries to find squares in
// each image

#include "highgui.h"
#include "cv.h"

#include <iostream>
#include <math.h>
#include <string.h>

using namespace cv;
using namespace std;

void help()
{
        cout <<
        "
A program using pyramid scaling, Canny, contours, contour simpification and
"
        "memory storage (it's got it all folks) to find
"
        "squares in a list of images pic1-6.png
"
        "Returns sequence of squares detected on the image.
"
        "the sequence is stored in the specified memory storage
"
        "Call:
"
        "./squares
"
    "Using OpenCV version %s
" << CV_VERSION << "
" << endl;
}


int thresh = 50, N = 2; // karlphillip: decreased N to 2, was 11.
const char* wndname = "Square Detection Demo";

// helper function:
// finds a cosine of angle between vectors
// from pt0->pt1 and from pt0->pt2
double angle( Point pt1, Point pt2, Point pt0 )
{
    double dx1 = pt1.x - pt0.x;
    double dy1 = pt1.y - pt0.y;
    double dx2 = pt2.x - pt0.x;
    double dy2 = pt2.y - pt0.y;
    return (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
}

// returns sequence of squares detected on the image.
// the sequence is stored in the specified memory storage
void findSquares( const Mat& image, vector<vector<Point> >& squares )
{
    squares.clear();

    Mat pyr, timg, gray0(image.size(), CV_8U), gray;

    // karlphillip: dilate the image so this technique can detect the white square,
    Mat out(image);
    dilate(out, out, Mat(), Point(-1,-1));
    // then blur it so that the ocean/sea become one big segment to avoid detecting them as 2 big squares.
    medianBlur(out, out, 7);

    // down-scale and upscale the image to filter out the noise
    pyrDown(out, pyr, Size(out.cols/2, out.rows/2));
    pyrUp(pyr, timg, out.size());
    vector<vector<Point> > contours;

    // find squares in every color plane of the image
    for( int c = 0; c < 3; c++ )
    {
        int ch[] = {c, 0};
        mixChannels(&timg, 1, &gray0, 1, ch, 1);

        // try several threshold levels
        for( int l = 0; l < N; l++ )
        {
            // hack: use Canny instead of zero threshold level.
            // Canny helps to catch squares with gradient shading
            if( l == 0 )
            {
                // apply Canny. Take the upper threshold from slider
                // and set the lower to 0 (which forces edges merging)
                Canny(gray0, gray, 0, thresh, 5);
                // dilate canny output to remove potential
                // holes between edge segments
                dilate(gray, gray, Mat(), Point(-1,-1));
            }
            else
            {
                // apply threshold if l!=0:
                //     tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 0
                gray = gray0 >= (l+1)*255/N;
            }

            // find contours and store them all as a list
            findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);

            vector<Point> approx;

            // test each contour
            for( size_t i = 0; i < contours.size(); i++ )
            {
                // approximate contour with accuracy proportional
                // to the contour perimeter
                approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);

                // square contours should have 4 vertices after approximation
                // relatively large area (to filter out noisy contours)
                // and be convex.
                // Note: absolute value of an area is used because
                // area may be positive or negative - in accordance with the
                // contour orientation
                if( approx.size() == 4 &&
                    fabs(contourArea(Mat(approx))) > 1000 &&
                    isContourConvex(Mat(approx)) )
                {
                    double maxCosine = 0;

                    for( int j = 2; j < 5; j++ )
                    {
                        // find the maximum cosine of the angle between joint edges
                        double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
                        maxCosine = MAX(maxCosine, cosine);
                    }

                    // if cosines of all angles are small
                    // (all angles are ~90 degree) then write quandrange
                    // vertices to resultant sequence
                    if( maxCosine < 0.3 )
                        squares.push_back(approx);
                }
            }
        }
    }
}


// the function draws all the squares in the image
void drawSquares( Mat& image, const vector<vector<Point> >& squares )
{
    for( size_t i = 0; i < squares.size(); i++ )
    {
        const Point* p = &squares[i][0];
        int n = (int)squares[i].size();
        polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);
    }

    imshow(wndname, image);
}


int main(int argc, char** argv)
{
    if (argc < 2)
    {
        cout << "Usage: ./program <file>" << endl;
        return -1;
    }

//    static const char* names[] = { "pic1.png", "pic2.png", "pic3.png",
//        "pic4.png", "pic5.png", "pic6.png", 0 };
    static const char* names[] = { argv[1], 0 };

    help();
    namedWindow( wndname, 1 );
    vector<vector<Point> > squares;

    for( int i = 0; names[i] != 0; i++ )
    {
        Mat image = imread(names[i], 1);
        if( image.empty() )
        {
            cout << "Couldn't load " << names[i] << endl;
            continue;
        }

        findSquares(image, squares);
        drawSquares(image, squares);
        imwrite("out.jpg", image);

        int c = waitKey();
        if( (char)c == 27 )
            break;
    }

    return 0;
}

Outputs:

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正方形檢測沒有找到正方形

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