在 C++ 中获取轮廓的有界矩形

Question

我正在尝试将 python 代码转换为 C++，但我在下面的函数中遇到了问题 得到轮廓的旋转矩形： '''

def find_rot_rectangle_for_contour(contours):
    # Find the rotated rectangles and ellipses for each contour
    minRect = [None] * len(contours)  # Initializing minRect
    print('length of contour = ',len(contours))
    count = None  #count = 0
    minLength = None # minLength = 0
    tmpLength = 0 #tempLength = 0
    for i, c in enumerate(contours): #index,c= contour
        minRect[i] = cv.minAreaRect(c)  # minRect is ((x,y), (w,h), angle), index(0, 1, 2)
        minLength = min(minRect[i][1])
        print('Minimum Length is',minLength)
        if ((minLength > 500) and (minLength > tmpLength)):
            count = i
            tmpLength = minLength
    return minRect, count
'''

我已将上述 python 代码转换为 C++，如下所示：

void find_rot_rectangle_for_contour(vector<vector<Point> > contour,vector<RotatedRect> minRect,unsigned int count)
{

    unsigned int    minLength = 0;
    unsigned int    tmpLength = 0;
    RotatedRectangleOutput rotone;

    vector<RotatedRect> minRectone(contour.size());

    cout << "Contour size is" << contour.size() << endl;

    //minRect[0] = RotatedRect(Point2f(100, 100), Size2f(100, 50), 30);

    for (size_t i = 0; i < contour.size(); i++) 
    {
        minRectone[i] = cv::minAreaRect(contour[i]);
        minLength = MIN(minRectone[i].size.width,minRectone.size.height);
    
        if ((minLength > 500) and (minLength > tmpLength))
        {
            count = i;
            tmpLength = minLength;
        }

    }

    minRect = minRectone;
}

我有几个问题：

a) for loop i python 代码有时会运行 46,38 次，具体取决于轮廓列表中轮廓的数量。 但在 C++ 代码中，contour.size() 返回 4000，因此与 python 代码不匹配。

b) c++ 代码在minLength = MIN(minRectone[i].size.width,minRectone.size.height); 中崩溃 在 python 代码中，我们得到 w,h for minRect 的行：minLength = min(minRect[i][1]) 如何在 C++ 代码中进行类似的操作。

c) 在 python 代码中，轮廓是以下代码的输出：

contours_in, _ = cv.findContours(canny_in_output_dilated, cv.RETR_TREE, cv.CHAIN_APPROX_SIMPLE)

在 C++ 代码中，轮廓是以下代码的输出：

findContours(canny_in_output_dilated, contours_in, hierarchy, RETR_TREE, CHAIN_APPROX_SIMPLE);

是在 python 和 c++ 中返回轮廓的方式不同还是 我需要在 C++ 中将轮廓作为列表传递。在 python 中它被处理了吗？

Answer 1

我已经修改了我的代码以像这样遍历所有轮廓：

void find_rot_rectangle_for_contour(vector<vector<Point> > contour, vector<RotatedRect> minRect, unsigned int count)
{

    unsigned int    minLength = 0;
    unsigned int    tmpLength = 0;
    RotatedRectangleOutput rotone;
    int loopcount = 0;

    vector<vector<Point> >::iterator itc = contour.begin();
    vector<RotatedRect> rects;

    while (itc != contour.end() && (!(*itc).empty())) {
        //Create bounding rect of object
        rects[loopcount] = minAreaRect(Mat(*itc));
        minLength = min(rects[loopcount].size.width, rects[loopcount].size.height);
        if ((minLength > 500) and (minLength > tmpLength))
        {
            count = loopcount;
            tmpLength = minLength;
        }
        loopcount++;
    }

    cout << "Count value = " << count << endl;
    cout << "LoopCount value = " << loopcount << endl;

}

但是正在给出断言：向量下标超出范围：

rects[loopcount] = minAreaRect(Mat(*itc));