python - 如何在不包含(或少量)背景像素的情况下调整对象内的矩形或调整其大小？

Question

在应用阈值并找到对象的轮廓后，我使用以下代码获取对象周围的直矩形(或输入其指令的旋转矩形):

img = cv2.imread('image.png')
imgray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
ret,thresh = cv2.threshold(imgray,127,255,cv2.THRESH_BINARY)
# find contours 
contours, hierarchy = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
cnt = contours[0]
# straight rectangle
x,y,w,h = cv2.boundingRect(cnt)
img= cv2.rectangle(img,(x,y),(x+w,y+h),(0,255,0),2)

看图

然后我使用以下代码计算了直矩形内的对象和背景像素的数量:

# rectangle area (total number of object and background pixels inside the rectangle)
area_rect = w*h
# white or object pixels (inside the rectangle)
obj = cv2.countNonZero(imgray)
# background pixels (inside the rectangle)
bac = area_rect - obj

现在我想根据背景像素和对象像素之间的关系来调整对象的矩形，例如，让一个矩形占据对象的大部分，而没有或只有更少的背景像素:

我如何创建它？

Answer 1

这个问题可以表述为找到非凸多边形内接的最大矩形。

近似解可以在这个 link 找到.

这个问题也可以表述为:对于每个角度，找到矩阵中仅包含零的最大矩形，在此 SO question 中进行了探索.

我的解决方案基于 this回答。这将仅找到轴对齐的矩形，因此您可以轻松地将图像旋转给定角度并将此解决方案应用于每个角度。我的解决方案是 C++，但您可以轻松地将它移植到 Python，因为我主要使用 OpenCV 函数，或者调整上述答案中的解决方案以考虑轮换。

我们在这里:

#include <opencv2\opencv.hpp>
#include <iostream>
using namespace cv;
using namespace std;


// https://stackoverflow.com/a/30418912/5008845
Rect findMinRect(const Mat1b& src)
{
    Mat1f W(src.rows, src.cols, float(0));
    Mat1f H(src.rows, src.cols, float(0));

    Rect maxRect(0,0,0,0);
    float maxArea = 0.f;

    for (int r = 0; r < src.rows; ++r)
    {
        for (int c = 0; c < src.cols; ++c)
        {
            if (src(r, c) == 0)
            {
                H(r, c) = 1.f + ((r>0) ? H(r-1, c) : 0);
                W(r, c) = 1.f + ((c>0) ? W(r, c-1) : 0);
            }

            float minw = W(r,c);
            for (int h = 0; h < H(r, c); ++h)
            {
                minw = min(minw, W(r-h, c));
                float area = (h+1) * minw;
                if (area > maxArea)
                {
                    maxArea = area;
                    maxRect = Rect(Point(c - minw + 1, r - h), Point(c+1, r+1));
                }
            }
        }
    }

    return maxRect;
}


RotatedRect largestRectInNonConvexPoly(const Mat1b& src)
{
    // Create a matrix big enough to not lose points during rotation
    vector<Point> ptz;
    findNonZero(src, ptz);
    Rect bbox = boundingRect(ptz); 
    int maxdim = max(bbox.width, bbox.height);
    Mat1b work(2*maxdim, 2*maxdim, uchar(0));
    src(bbox).copyTo(work(Rect(maxdim - bbox.width/2, maxdim - bbox.height / 2, bbox.width, bbox.height)));

    // Store best data
    Rect bestRect;
    int bestAngle = 0;

    // For each angle
    for (int angle = 0; angle < 90; angle += 1)
    {
        cout << angle << endl;

        // Rotate the image
        Mat R = getRotationMatrix2D(Point(maxdim,maxdim), angle, 1);
        Mat1b rotated;
        warpAffine(work, rotated, R, work.size());

        // Keep the crop with the polygon
        vector<Point> pts;
        findNonZero(rotated, pts);
        Rect box = boundingRect(pts);
        Mat1b crop = rotated(box).clone();

        // Invert colors
        crop = ~crop; 

        // Solve the problem: "Find largest rectangle containing only zeros in an binary matrix"
        // https://stackoverflow.com/questions/2478447/find-largest-rectangle-containing-only-zeros-in-an-n%C3%97n-binary-matrix
        Rect r = findMinRect(crop);

        // If best, save result
        if (r.area() > bestRect.area())
        {
            bestRect = r + box.tl();    // Correct the crop displacement
            bestAngle = angle;
        }
    }

    // Apply the inverse rotation
    Mat Rinv = getRotationMatrix2D(Point(maxdim, maxdim), -bestAngle, 1);
    vector<Point> rectPoints{bestRect.tl(), Point(bestRect.x + bestRect.width, bestRect.y), bestRect.br(), Point(bestRect.x, bestRect.y + bestRect.height)};
    vector<Point> rotatedRectPoints;
    transform(rectPoints, rotatedRectPoints, Rinv);

    // Apply the reverse translations
    for (int i = 0; i < rotatedRectPoints.size(); ++i)
    {
        rotatedRectPoints[i] += bbox.tl() - Point(maxdim - bbox.width / 2, maxdim - bbox.height / 2);
    }

    // Get the rotated rect
    RotatedRect rrect = minAreaRect(rotatedRectPoints);

    return rrect;
}



int main()
{
    Mat1b img = imread("path_to_image", IMREAD_GRAYSCALE);

    // Compute largest rect inside polygon
    RotatedRect r = largestRectInNonConvexPoly(img);

    // Show
    Mat3b res;
    cvtColor(img, res, COLOR_GRAY2BGR);

    Point2f points[4];
    r.points(points);

    for (int i = 0; i < 4; ++i)
    {
        line(res, points[i], points[(i + 1) % 4], Scalar(0, 0, 255), 2);
    }

    imshow("Result", res);
    waitKey();

    return 0;
}

结果图片是:

注意

我想指出这段代码没有优化，所以它可能会表现得更好。有关近似解决方案，请参阅 here , 以及在那里报道的论文。

这answer一个相关的问题让我朝着正确的方向前进。