Question

我想在Java包装器中使用partition function of Opencv。但我找不到它的包装。

修改
这是opencv上的open issue。
还有其他选择吗？

Answer 1

我知道这个问题大约有半年的历史了，但是我最近在搜索opencv分区时遇到了这个问题。该方法仍未在Java中实现，因此我复制了C ++代码并将其更改为可在Java中进行编译。在这里：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.function.BiPredicate;

public class Partition {

    public static void main(String[] args) {
        List<String> list = new LinkedList<>(Arrays.asList("aa", "ab", "aa", "bb", "bb", "a", "b", "a", "b", "b"));
        List<Integer> list2 = new ArrayList<>(Arrays.asList(1, 2, 10, 11, 10, 42, 3, 2, 3));

        List<Integer> labels = new ArrayList<>();
        int classes = partition(list, labels);
        System.out.println("There are " + classes + " classes of equal strings:");
        System.out.println(labels);

        classes = partition(list2, labels);
        System.out.println("There are " + classes + " classes of equal integers:");
        System.out.println(labels);

        classes = partition(list, labels, (str1, str2) -> str1.length() == str2.length());
        System.out.println("There are " + classes + " classes of strings with same length:");
        System.out.println(labels);

        classes = partition(list2, labels, (i1, i2) -> Math.abs(i1-i2) <= 1);
        System.out.println("There are " + classes + " classes of integers within 1 range:");
        System.out.println(labels);
    }

    /**
     * Overloaded method with default equality predicate
     * 
     * @param _vec vector of elements to be partitioned
     * @param labels output list of labels
     * @return number of classes
     */
    public static <E> int partition(final List<E> _vec, List<Integer> labels) {
        return partition(_vec, labels, E::equals);
    }

    /**
     * Port of C++ partition function
     * 
     * @param _vec list of elements to be partitioned
     * @param labels output list of labels
     * @param predicate predicate to test whether two elements belong to the same class
     * @return number of classes
     * @see https://github.com/opencv/opencv/blob/master/modules/core/include/opencv2/core/operations.hpp
     * @see https://docs.opencv.org/2.4/modules/core/doc/clustering.html
     */
    //template<typename _Tp, class _EqPredicate> int
    //partition( const std::vector<_Tp>& _vec, std::vector<int>& labels,
    //          _EqPredicate predicate=_EqPredicate())
    public static <E> int partition(final List<E> _vec, List<Integer> labels, BiPredicate<E, E> predicate)
    {

        int i, j, N = (int)_vec.size();

        //const _Tp* vec = &_vec[0];
        final ArrayList<E> vec = new ArrayList<>(_vec);

        final int PARENT=0;
        final int RANK=1;

        //std::vector<int> _nodes(N*2);
        //int (*nodes)[2] = (int(*)[2])&_nodes[0];
        int[][] nodes = new int[N*2][2];

        // The first O(N) pass: create N single-vertex trees
        for(i = 0; i < N; i++)
        {
            nodes[i][PARENT]=-1;
            nodes[i][RANK] = 0;
        }

        // The main O(N^2) pass: merge connected components
        for( i = 0; i < N; i++ )
        {
            int root = i;

            // find root
            while( nodes[root][PARENT] >= 0 )
                root = nodes[root][PARENT];

            for( j = 0; j < N; j++ )
            {
                //if( i == j || !predicate(vec[i], vec[j]))
                if(i == j || !predicate.test(vec.get(i), vec.get(j)))
                    continue;
                int root2 = j;

                while( nodes[root2][PARENT] >= 0 )
                    root2 = nodes[root2][PARENT];

                if( root2 != root )
                {
                    // unite both trees
                    int rank = nodes[root][RANK], rank2 = nodes[root2][RANK];
                    if( rank > rank2 )
                        nodes[root2][PARENT] = root;
                    else
                    {
                        nodes[root][PARENT] = root2;
                        nodes[root2][RANK] += (rank == rank2 ? 1 : 0);
                        root = root2;
                    }
                    //CV_Assert( nodes[root][PARENT] < 0 );
                    assert(nodes[root][PARENT] < 0);

                    int k = j, parent;

                    // compress the path from node2 to root
                    while( (parent = nodes[k][PARENT]) >= 0 )
                    {
                        nodes[k][PARENT] = root;
                        k = parent;
                    }

                    // compress the path from node to root
                    k = i;
                    while( (parent = nodes[k][PARENT]) >= 0 )
                    {
                        nodes[k][PARENT] = root;
                        k = parent;
                    }
                }
            }
        }

        // Final O(N) pass: enumerate classes
        //labels.resize(N);
        Integer[] _labels = new Integer[N];
        int nclasses = 0;

        for( i = 0; i < N; i++ )
        {
            int root = i;
            while( nodes[root][PARENT] >= 0 )
                root = nodes[root][PARENT];
            // re-use the rank as the class label
            if( nodes[root][RANK] >= 0 )
                nodes[root][RANK] = ~nclasses++;
            _labels[i] = ~nodes[root][RANK];
        }
        labels.clear();
        labels.addAll(Arrays.asList(_labels));
        return nclasses;
    }
}

输出如下：

There are 5 classes of equal strings:
[0, 1, 0, 2, 2, 3, 4, 3, 4, 4]
There are 6 classes of equal integers:
[0, 1, 2, 3, 2, 4, 5, 1, 5]
There are 2 classes of strings with same length:
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1]
There are 3 classes of integers within 1 range:
[0, 0, 1, 1, 1, 2, 0, 0, 0]

我试图保留尽可能多的原始代码，并注释掉必须更改的行。要查看原始代码（在C ++中），您可能需要查看here。可以在here中找到该文档。

如何在Java包装器中使用Opencv的分区功能

1 个答案: