Question

我正在尝试计算下面骨架图像中的终点数。我在这里使用矢量。我需要的是打印矢量，而不仅仅是计数。我试过很多方面。但没有奏效。我是新手打开简历，我从互联网上找到了以下代码。任何人都可以帮我打印这段代码实际得到的向量。代码如下。

// get the end points
    // Declare variable to count neighbourhood pixels
    int count, numberOFEndpoints;

    // To store a pixel intensity
    uchar pix;
    numberOFEndpoints = 0;
    // To store the ending co-ordinates
    std::vector<int> coords;

    // For each pixel in our image...
    for (int i = 1; i < CopyofSkeletionize.rows - 1; i++) {
        for (int j = 1; j < CopyofSkeletionize.cols - 1; j++) {

            // See what the pixel is at this location
            pix = CopyofSkeletionize.at<uchar>(i, j);

            // If not a skeleton point, skip
            if (pix == 0)
                continue;

            // Reset counter
            count = 0;

            // For each pixel in the neighbourhood
            // centered at this skeleton location...
            for (int y = -1; y <= 1; y++) {
                for (int x = -1; x <= 1; x++) {

                    // Get the pixel in the neighbourhood
                    pix = CopyofSkeletionize.at<uchar>(i + y, j + x);


                    // Count if non-zero
                    if (pix != 0)
                        count++;
                }
            }

            // If count is exactly 2, add co-ordinates to vector
            if (count == 2) {
                coords.push_back(i);
                coords.push_back(j);
                numberOFEndpoints = numberOFEndpoints + 1;
            }

        }
    }
     printf("numberOFEndpoints : %d \n", numberOFEndpoints);

我在互联网上使用了这段代码，但它没有用。 for（int i = 0; i＆lt; coords.size（）/ 2; i ++） cout＆lt;＆lt; “（”＆lt;

https://i.stack.imgur.com/CMZtu.png

Answer 1

首先......你想以更好的方式存储坐标。也许像两个载体：

std::vector<int> coordx;
std::vector<int> coordy;

然后你打印它：

for (int i = 0; i < coordx.size(); ++i)
    std::cout << coordx[i] << " x " << coordy[i] << std::endl;

或者使用pair（可能需要#include <pair>）或创建自定义point结构：

std::vector< std::pair<int, int> > coords;

添加到coords矢量会是这样的：

coords.push_back(std::make_pair(x, y));

然后你打印它：

for (int i = 0; i < coords.size(); ++i)
    std::cout << coords[i].first << " x " << coords[i].second << std::endl;

其次......如果您坚持使用单vector<int>，请尝试：

int i = 0;
while (i < coord.size())
    std::cout << coords[i++] << " x " << coords[i++] << std::endl;

Answer 2

// get the end points
    // Declare variable to count neighbourhood pixels
    int count, numberOFEndpoints;

    // To store a pixel intensity
    uchar pix;
    numberOFEndpoints = 0;
    // To store the ending co-ordinates
    /*std::vector<int> coords;*/
    std::vector<int> coordx;
    std::vector<int> coordy;

    // For each pixel in our image...
    for (int i = 1; i < CopyofSkeletionize.rows - 1; i++) {
        for (int j = 1; j < CopyofSkeletionize.cols - 1; j++) {

            // See what the pixel is at this location
            pix = CopyofSkeletionize.at<uchar>(i, j);

            // If not a skeleton point, skip
            if (pix == 0)
                continue;

            // Reset counter
            count = 0;

            // For each pixel in the neighbourhood
            // centered at this skeleton location...
            for (int y = -1; y <= 1; y++) {
                for (int x = -1; x <= 1; x++) {

                    // Get the pixel in the neighbourhood
                    pix = CopyofSkeletionize.at<uchar>(i + y, j + x);


                    // Count if non-zero
                    if (pix != 0)
                        count++;
                }
            }

            // If count is exactly 2, add co-ordinates to vector
            if (count == 2) {
                coordx.push_back(i);
                coordy.push_back(j);


                /*numberOFEndpoints = numberOFEndpoints + 1;*/
            }

        }


    }
    for (int i = 0; i < coordx.size(); ++i) 
        std::cout << coordx[i] << " x " << coordy[i] << std::endl;}

我甚至无法从此代码中获取矢量结果

如何打印向量内的元素

2 个答案: