Question

我在条形码图像中检测到了轮廓，并为每个条形图计算了最小值。图像中还有其他对象，因此我希望通过找到类似角度矩形的簇来隔离条形码。

我是C ++ opencv的新手，所以我不确定我应该采用什么方法。有什么建议吗？

Answer 1

vector<vector<int>> RAPC(vector<RotatedRect> minRect, float radius) {

    vector<vector<int>> clusters; // vector that will contain identify who belongs to which cluster
    vector<bool> touched(minRect.size(), false); // initializing vector that will identify whether or not a rectangle has been assigned to a cluster

    // for every rect
    for (int i = 0; i < minRect.size()-1; i++) 
    {
        // compare to every other rect
        for (int n = (i + 1); n < minRect.size(); n++) 
        {
            // distance from rect i to rect n
            float distance = sqrt(pow((minRect[i].center.x - minRect[n].center.x), 2) 
            + pow((minRect[i].center.y - minRect[n].center.y), 2));

            //criteria to determine whether the two rectangles belong in the same cluster
            if ( distance < radius)
            {
                // if neither rectangle has been assigned to a cluster yet
                if (touched[i] == false && touched[n] == false) {

                    // create new cluster in clusters with i and n as members
                    clusters.emplace_back(initializer_list<int>{i, n});

                    touched[i] = true; // make note of i being placed

                    touched[n] = true; // make note of n being placed
                }

                // if i has already been assigned to a cluster
                else if (touched[i] == true && touched[n] == false) {

                    int loc = findLoc(i, clusters); // find i in clusters

                    clusters[loc].emplace_back(n); // place n in that cluster

                    touched[n] = true; // make note of n being placed
                }

                // if n has already been assigned to a cluster
                else if (touched[n] == true && touched[i] == false) {

                    int loc = findLoc(n, clusters); // find n in clusters

                    clusters[loc].emplace_back(i); // place n in that cluster

                    touched[i] = true; // make note of i being placed
                }
                // if both rectangles have already been assigned
                else (touched[i] == true && touched[n] == true); {

                    int locI = findLoc(i, clusters); // search for location of i
                    int locN = findLoc(n, clusters); // search for location of n

                    // if both rectangles are already part of the same cluster, ignore
                    if (locI == locN) {
                        break;
                    }

                    // concatenate vector n with vector i
                    clusters[locI].insert(clusters[locI].end(), clusters[locN].begin(), clusters[locN].end());

                    // erase original location of vector n
                    clusters[locN].erase(clusters[locN].begin(), clusters[locN].end()); 
                }

            }
        }
    }
    return clusters;
}


    // function used by RAPC mainly just to make code more concise 
int  findLoc(int numSearchedFor, vector<vector<int>> clusters) {
    bool searchComplete = false;
    int ans; // initializing function output ie the cluster ID where the value is found
    for (int i = 0; i < clusters.size(); i++) {
        for (int n = 0; n < clusters[i].size(); n++) {
            if (clusters[i][n] == numSearchedFor) {
                ans = i;
                searchComplete = true;
                break;
            }
        }
        if (searchComplete == true) {
            break;
        }
    }
    return ans;
}

根据标准查找轮廓的集群

1 个答案: