在findSquares Opencv

Question

我有一幅画，里面有一张桌子，我需要提取里面有手写内容的盒子然后裁剪它们，因为我需要一个个地认出那些盒子里的手写内容，但是问题是findSquare函数给了我由于几个阈值水平而导致具有相同坐标（+ -5像素）的正方形的输出过多。我已经通过仅在一个通道中搜索将正方形的数量从300减少到90.老师告诉我过滤多余的正方形按其坐标，如果方坐标之间的差小于10个像素，则表示它是同一框。 带有桌子的图片在这里https://ibb.co/Ms8YP8n 而不是20个盒子，它让我收获了大约90个盒子，而且大多数都一样。 因此，我试图在for循环中遍历平方向量，并消除出现两次的平方。

这是我尝试使用的代码

vector<vector<Point> > same;
same.clear();
same = squares;
int sq = squares.size();

for (int i = 0; i < sq; i++) {

    for (int j = 1; j < sq-1; j++) {
        if (-5 < (same[i][0].x - same[j][0].x) < 5 && -5 < (same[i][0].y - same[j][0].y) < 5 &&
            -5 < (same[i][1].x - same[j][1].x) < 5 && -5 < (same[i][1].y - same[j][1].y) < 5 &&
            -5 < (same[i][2].x - same[j][2].x) < 5 && -5 < (same[i][2].y - same[j][2].y) < 5 &&
            -5 < (same[i][3].x - same[j][3].x) < 5 && -5 < (same[i][3].y - same[j][3].y) < 5) {
            squares.erase(same.begin() + j);
        }
    }
}

“相同”向量只是“正方形”的副本。

给我的错误是“向量擦除迭代器超出范围”。 您可以在下面的drawSquare函数中找到此代码，也可以查看完整的代码。

我需要图片以备将来使用，我需要识别手写的数字。

有人可以帮我吗？使用此代码中的其他方法或想法或一些修复程序...

谢谢！

static void findSquares(const Mat& image, vector<vector<Point> >& squares)
{
    squares.clear();
    Mat pyr, timg, timb, gray0(image.size(), CV_8UC1), gray;
    pyrDown(image, pyr, Size(image.cols / 2, image.rows / 2));
    pyrUp(pyr, timg, image.size());

    vector<vector<Point> > contours;

    // find squares in every color plane of the image
    for (int c = 0; c < 1; 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;
                //          imshow("graaay0", gray);
            }

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

            vector<Point> approx;

            // test each contour
            for (int 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 && approx.size() <= 6 &&
                    fabs(contourArea(Mat(approx))) > 5000 && fabs(contourArea(Mat(approx))) < 8000 &&
                    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 && maxCosine < 0.2)
                        squares.push_back(approx);
                }
            }
        }
    }

    }


    // the function draws all the squares in the image


 static void drawSquares(Mat& image, vector<vector<Point> > &squares)
    {

    vector<vector<Point> > same;
    same.clear();
    same = squares;
    int sq = squares.size();

    for (int i = 0; i < sq; i++) {

        for (int j = 1; j < sq; j++) {
            if (-5 < (same[i][0].x - same[j][0].x) < 5 && -5 < (same[i][0].y - same[j][0].y) < 5 &&
                -5 < (same[i][1].x - same[j][1].x) < 5 && -5 < (same[i][1].y - same[j][1].y) < 5 &&
                -5 < (same[i][2].x - same[j][2].x) < 5 && -5 < (same[i][2].y - same[j][2].y) < 5 &&
                -5 < (same[i][3].x - same[j][3].x) < 5 && -5 < (same[i][3].y - same[j][3].y) < 5) {
                squares.erase(same.begin() + j);
            }
        }
    }




    for (int 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, LINE_AA);

    }

    imshow("Plate with green", image);

    }



Mat ImageExtract(Mat& image, vector<vector<Point> >& squares)
    {

    char file_name[100];
    int x = squares.size();
    printf("squaree %d", x);


    Mat roi;
    for (int i = 0; i < squares.size(); i++)
    {

        sprintf(file_name, "cropped%d.jpg", i + 1);
        Rect r = boundingRect(squares[i]);
        roi = Mat(image, r);
        imwrite(file_name, roi);

    }
    return roi;

    }


void Preprocessing() {

    char fname[MAX_PATH];
    openFileDlg(fname);
    Mat img = imread(fname, 1);
    std::vector<std::vector<Point> > squares;

    findSquares(img, squares);
    ImageExtract(img, squares);
    drawSquares(img, squares);



    waitKey(0);
    }

int main() {
       Preprocessing();
    return 0;
    }


  [1]: https://i.stack.imgur.com/mwvxX.png