OpenCV C++ 如何知道每行的轮廓数进行排序? [英] OpenCV C++ how to know the number of contours per row for sorting?

问题描述

我有一个

排序后的轮廓:

如您所见，您也可以颠倒原始顺序，因为 cv::findContours 只是朝着相反的方向前进.;-)

一个重要的警告:如果扫描(或您获得的调查结果)稍微逆时针旋转，此例程将失败.因此，应事先检查整个扫描(或...)的角度.

I have a binary image: In this image I can easily sort the contours that I found from top to bottom and from left to right using the overloaded std::sort.

I first sort from top to bottom via:

sort(contours.begin(), contours.end(), top_to_bottom_contour_sorter());

Then I sort from left to right by:

for (int i = 0; i < contours.size(); i = i + no_of_contours_horizontally)
    {
        sort(i, i + no_of_contours_horizontally, left_to_right_contour_sorter);
    }

Where top_to_bottom and left_to_right are separate functions that I pass to the sort function. And no_of_contours_horizontally with respect to the first image is three (3).

However this only works if I know the number of contours horizontally. If the image I am using will have varying number of contours horizontally like in this image. contours_sample. The program fails. I could brute force and define for a specific index to change the no of contours found. However, it would limit the program to operate on a specific input instead of being flexible. I am thinking of creating rects or lines that I can overlay on top of the image and with that count the number of contours inside so I can get a value of the number of horizontal contours. If there is a more elegant solution I would appreciate it.

Here are my sorting functions

bool top_to_bottom_contour_sorter(const std::vector<Point> &lhs, const std::vector<Point> &rhs)
{
    Rect rectLhs = boundingRect(Mat(lhs));
    Rect rectRhs = boundingRect(Mat(rhs));

    return rectLhs.y < rectRhs.y;
}

bool left_to_right_contour_sorter(const std::vector<Point> &lhs, const std::vector<Point> &rhs)
{
    Rect rectLhs = boundingRect(Mat(lhs));
    Rect rectRhs = boundingRect(Mat(rhs));

    return rectLhs.x < rectRhs.x;
}

EDIT Here are my current outputs and desired output for each image. Using the first image and my current working code. Current_Output

My desired output for the second image. Desired_Output

解决方案

I guess, your only problem was not to respect equality for one of the coordinates!?

Here we go:

// Custom sorter.
bool sortContour(std::vector<cv::Point> a, std::vector<cv::Point> b)
{
    cv::Rect rectA = cv::boundingRect(a);
    cv::Rect rectB = cv::boundingRect(b);

    if (rectA.y == rectB.y)
        return (rectA.x < rectB.x);

    return (rectA.y < rectB.y);
}

int main()
{
    // Load image.
    cv::Mat image = cv::imread("contours.jpg", cv::IMREAD_GRAYSCALE);

    // There are some artifacts in the JPG...
    cv::threshold(image, image, 128, 255, cv::THRESH_BINARY);

    // Find contours.
    std::vector<std::vector<cv::Point>> contours;
    std::vector<cv::Vec4i> hierarchy;
    cv::findContours(image, contours, hierarchy, cv::RETR_EXTERNAL, cv::CHAIN_APPROX_NONE);

    // Output unsorted contours.
    cv::Mat imageUnsorted = image.clone();
    for (int i = 0; i < contours.size(); i++)
    {
        cv::Rect rect = cv::boundingRect(contours[i]);
        cv::putText(imageUnsorted, std::to_string(i), cv::Point(rect.x - 10, rect.y - 10), cv::FONT_HERSHEY_COMPLEX, 0.5, cv::Scalar(255));
    }
    cv::imwrite("unsorted.png", imageUnsorted);

    // Sort using custom sorter.
    std::sort(contours.begin(), contours.end(), sortContour);

    // Output sorted contours.
    cv::Mat imageSorted = image.clone();
    for (int i = 0; i < contours.size(); i++)
    {
        cv::Rect rect = cv::boundingRect(contours[i]);
        cv::putText(imageSorted, std::to_string(i), cv::Point(rect.x - 10, rect.y - 10), cv::FONT_HERSHEY_COMPLEX, 0.5, cv::Scalar(255));
    }
    cv::imwrite("sorted.png", imageSorted);
}

The unsorted contours:

The sorted contours:

As you can see, one could also just inverse the original order, since cv::findContours just goes in the opposite direction(s). ;-)

One big caveat: If the scan (or however you obtain the surveys) is even slightly rotated counterclockwise, this routine will fail. Therefore, the angle of the whole scan (or...) should be checked beforehand.

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