OpenCV C ++ / Obj-C:高级方检测

时间:2012-05-10 11:40:56

标签: c++ objective-c opencv image-processing object-detection

前段时间我问a question about square detection,而karlphillip提出了不错的结果。

现在我想更进一步,找到边缘不完全可见的方块。看一下这个例子:

example

有什么想法吗?我正在使用karlphillips代码:

void find_squares(Mat& image, vector<vector<Point> >& squares)
{
    // blur will enhance edge detection
    Mat blurred(image);
    medianBlur(image, blurred, 9);

    Mat gray0(blurred.size(), CV_8U), gray;
    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(&blurred, 1, &gray0, 1, ch, 1);

        // try several threshold levels
        const int threshold_level = 2;
        for (int l = 0; l < threshold_level; l++)
        {
            // Use Canny instead of zero threshold level!
            // Canny helps to catch squares with gradient shading
            if (l == 0)
            {
                Canny(gray0, gray, 10, 20, 3); // 

                // Dilate helps to remove potential holes between edge segments
                dilate(gray, gray, Mat(), Point(-1,-1));
            }
            else
            {
                    gray = gray0 >= (l+1) * 255 / threshold_level;
            }

            // Find contours and store them in a list
            findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);

            // Test contours
            vector<Point> approx;
            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);

                    // 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++)
                            {
                                    double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
                                    maxCosine = MAX(maxCosine, cosine);
                            }

                            if (maxCosine < 0.3)
                                    squares.push_back(approx);
                    }
            }
        }
    }
}

4 个答案:

答案 0 :(得分:43)

您可以尝试使用HoughLines来检测方块的四边。接下来,找到四个生成的线交叉点以检测角。 Hough transform对噪音和遮挡非常强大,所以它在这里很有用。此外,here是一个交互式演示,展示了Hough变换的工作原理(我认为它至少很酷:)。 Here是我之前的答案之一,可以检测到激光十字中心显示大部分相同的数学运算(除了它只找到一个角落)。

每侧可能有多条线,但定位交叉点应有助于确定内点与异常值。找到候选角后,您还可以按区域过滤这些候选项或多边形的“方形”。

编辑:所有这些代码和图片的答案让我觉得我的答案有点缺乏:)所以,这里有一个如何实现这个的实现:

#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <iostream>
#include <vector>

using namespace cv;
using namespace std;

Point2f computeIntersect(Vec2f line1, Vec2f line2);
vector<Point2f> lineToPointPair(Vec2f line);
bool acceptLinePair(Vec2f line1, Vec2f line2, float minTheta);

int main(int argc, char* argv[])
{
    Mat occludedSquare = imread("Square.jpg");

    resize(occludedSquare, occludedSquare, Size(0, 0), 0.25, 0.25);

    Mat occludedSquare8u;
    cvtColor(occludedSquare, occludedSquare8u, CV_BGR2GRAY);

    Mat thresh;
    threshold(occludedSquare8u, thresh, 200.0, 255.0, THRESH_BINARY);

    GaussianBlur(thresh, thresh, Size(7, 7), 2.0, 2.0);

    Mat edges;
    Canny(thresh, edges, 66.0, 133.0, 3);

    vector<Vec2f> lines;
    HoughLines( edges, lines, 1, CV_PI/180, 50, 0, 0 );

    cout << "Detected " << lines.size() << " lines." << endl;

    // compute the intersection from the lines detected...
    vector<Point2f> intersections;
    for( size_t i = 0; i < lines.size(); i++ )
    {
        for(size_t j = 0; j < lines.size(); j++)
        {
            Vec2f line1 = lines[i];
            Vec2f line2 = lines[j];
            if(acceptLinePair(line1, line2, CV_PI / 32))
            {
                Point2f intersection = computeIntersect(line1, line2);
                intersections.push_back(intersection);
            }
        }

    }

    if(intersections.size() > 0)
    {
        vector<Point2f>::iterator i;
        for(i = intersections.begin(); i != intersections.end(); ++i)
        {
            cout << "Intersection is " << i->x << ", " << i->y << endl;
            circle(occludedSquare, *i, 1, Scalar(0, 255, 0), 3);
        }
    }

    imshow("intersect", occludedSquare);
    waitKey();

    return 0;
}

bool acceptLinePair(Vec2f line1, Vec2f line2, float minTheta)
{
    float theta1 = line1[1], theta2 = line2[1];

    if(theta1 < minTheta)
    {
        theta1 += CV_PI; // dealing with 0 and 180 ambiguities...
    }

    if(theta2 < minTheta)
    {
        theta2 += CV_PI; // dealing with 0 and 180 ambiguities...
    }

    return abs(theta1 - theta2) > minTheta;
}

// the long nasty wikipedia line-intersection equation...bleh...
Point2f computeIntersect(Vec2f line1, Vec2f line2)
{
    vector<Point2f> p1 = lineToPointPair(line1);
    vector<Point2f> p2 = lineToPointPair(line2);

    float denom = (p1[0].x - p1[1].x)*(p2[0].y - p2[1].y) - (p1[0].y - p1[1].y)*(p2[0].x - p2[1].x);
    Point2f intersect(((p1[0].x*p1[1].y - p1[0].y*p1[1].x)*(p2[0].x - p2[1].x) -
                       (p1[0].x - p1[1].x)*(p2[0].x*p2[1].y - p2[0].y*p2[1].x)) / denom,
                      ((p1[0].x*p1[1].y - p1[0].y*p1[1].x)*(p2[0].y - p2[1].y) -
                       (p1[0].y - p1[1].y)*(p2[0].x*p2[1].y - p2[0].y*p2[1].x)) / denom);

    return intersect;
}

vector<Point2f> lineToPointPair(Vec2f line)
{
    vector<Point2f> points;

    float r = line[0], t = line[1];
    double cos_t = cos(t), sin_t = sin(t);
    double x0 = r*cos_t, y0 = r*sin_t;
    double alpha = 1000;

    points.push_back(Point2f(x0 + alpha*(-sin_t), y0 + alpha*cos_t));
    points.push_back(Point2f(x0 - alpha*(-sin_t), y0 - alpha*cos_t));

    return points;
}

注意:我调整图片大小的主要原因是我可以在屏幕上看到它并加速处理。

的Canny

这使用Canny边缘检测来帮助大大减少阈值处理后检测到的行数。

enter image description here

霍夫变换

然后使用霍夫变换来检测正方形的边。 enter image description here

交叉口

最后,我们计算所有线对的交点。 enter image description here

希望有所帮助!

答案 1 :(得分:27)

我尝试使用convex hull method非常简单。

在这里您可以找到检测到的轮廓的凸包。它消除了纸张底部的凸起缺陷。

下面是代码(在OpenCV-Python中):

import cv2
import numpy as np

img = cv2.imread('sof.jpg')
img = cv2.resize(img,(500,500))
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)

ret,thresh = cv2.threshold(gray,127,255,0)
contours,hier = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

for cnt in contours:
    if cv2.contourArea(cnt)>5000:  # remove small areas like noise etc
        hull = cv2.convexHull(cnt)    # find the convex hull of contour
        hull = cv2.approxPolyDP(hull,0.1*cv2.arcLength(hull,True),True)
        if len(hull)==4:
            cv2.drawContours(img,[hull],0,(0,255,0),2)

cv2.imshow('img',img)
cv2.waitKey(0)
cv2.destroyAllWindows()

(在这里,我没有在所有飞机上找到方格。如果你愿意,可以自己动手。)

以下是我得到的结果:

enter image description here

我希望这就是你所需要的。

答案 2 :(得分:6)

第一:开始尝试阈值技术,将白色纸张与图像的其余部分隔离开来。这是一种简单的方法:

Mat new_img = imread(argv[1]);

double thres = 200;
double color = 255;
threshold(new_img, new_img, thres, color, CV_THRESH_BINARY);

imwrite("thres.png", new_img);

但还有其他替代品可以提供更好的结果。一个是investigate inRange(),另一个是detect through color,将图像转换为HSV颜色空间。

This thread也提供了有关该主题的兴趣讨论。

第二:执行其中一个程序后,您可以尝试将结果直接反馈到find_squares()

find_squares()的替代方法是实施the bounding box technique,它有可能提供更准确的矩形区域检测(假设您有完美的阈值结果)。我用它herehere。值得注意的是,OpenCV拥有自己的bounding box tutorial

除了find_squares()之外, Abid 指出的另一种方法是使用convexHull方法。检查OpenCV的C++ tutorial on this method代码。

答案 3 :(得分:-1)

  1. 转换为实验室空间
  2. 使用kmeans进行2个群集
  3. 检测suqare一个内部集群,它将解决rgb空间中的许多事情