从Java中的二维数组中删除重复项

Question

所以，我有两个多维数组。

    double[][] combinations = new double[10000][3];
    double[][] uniqueCombinations = new double[100][3];

数组值示例：

[[1.233, 1.333, 0.76], [1.1, 1.333, 1.333], [0.9, 1.1, 0.9], [1.1, 1.333, 1.333]]

这就是我想要的

[[1.233, 1.333, 0.76], [1.1, 1.333, 1.333], [0.9, 1.1, 0.9]]

我想从组合中获取所有唯一数组并用它填充 uniqueCombinations。

我试过这个函数，但它只填充了 5 个数组，很奇怪！

public static double[][] removeDuplicate(double[][] matrix) {
    double[][] newMatrix = new double[matrix.length][matrix[0].length];
    int newMatrixRow = 1;

    for (int i = 0; i < matrix[0].length; i++)
        newMatrix[0][i] = matrix[0][i];

    for (int j = 1; j < matrix.length; j++) {
        List<Boolean> list = new ArrayList<>();
        for (int i = 0; newMatrix[i][0] != 0; i++) {
            boolean same = true;
            for (int col = 2; col < matrix[j].length; col++) {
                if (newMatrix[i][col] != matrix[j][col]) {
                    same = false;
                    break;
                }
            }
            list.add(same);
        }

        if (!list.contains(true)) {
            for (int i = 0; i < matrix[j].length; i++) {
                newMatrix[newMatrixRow][i] = matrix[j][i];
            }
            newMatrixRow++;
        }
    }

    int i;
    for(i = 0; newMatrix[i][0] != 0; i++);

    double finalMatrix[][] = new double[i][newMatrix[0].length];
    for (i = 0; i < finalMatrix.length; i++) {
        for (int j = 0; j < finalMatrix[i].length; j++)
            finalMatrix[i][j] = newMatrix[i][j];
    }

    return finalMatrix;
}

Answer 1

您可以尝试基于哈希表的算法，即计算每个矩阵向量的哈希值并使用哈希键将向量索引保存在哈希图中。然后根据哈希表索引值构造一个结果矩阵。例如：

   import static org.junit.Assert.assertArrayEquals;

import java.util.Arrays;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

import org.junit.Test;

import com.google.common.hash.HashFunction;
import com.google.common.hash.Hasher;
import com.google.common.hash.Hashing;

public class ArraysCombination {

    private static double[][] COMBINATIONS = { 
            {1.233, 1.333, 0.76 }, 
            { 1.1, 1.333, 1.333 }, 
            { 0.9, 1.1, 0.9 },
            { 1.1, 1.333, 1.333 } };


    private static double[][] uniqieCombinations(double[][] all) {
        final Map<Integer,Integer> uniqueIdx = new HashMap<>();
        // hashing can be replaced with Arrays.hashCode(all[i])
        final HashFunction hashFunction = Hashing.murmur3_32(all.length);
        for (int i = 0; i < all.length; i++) {
            final Hasher hasher = hashFunction.newHasher();
            for (int j = 0; j < all[i].length; j++) {
                hasher.putDouble(all[i][j]);
            }
            final Integer hash = hasher.hash().asInt();
            if( !uniqueIdx.containsKey(hash) ) {
                uniqueIdx.put(hash, Integer.valueOf(i));
            } 
        }
        double[][] arr = new double[uniqueIdx.size()][];
        Iterator<Integer> it = uniqueIdx.values().iterator();
        for (int i=0; i < arr.length; i++ ) {
            int idx = it.next();
            arr[i] = Arrays.copyOf( all[ idx ], all[idx].length  );
        }
        return arr;
    }



    @Test
    public void shouldFindUniqueCombinations() {
        double [][] uniqueCombination = uniqieCombinations(COMBINATIONS);
        for (double[] ds : uniqueCombination) {
            System.out.println(Arrays.toString(ds));
        }
        double[][] expected  = {{1.233, 1.333, 0.76}, {1.1, 1.333, 1.333}, {0.9, 1.1, 0.9}};
        for (int i = 0; i < expected.length; i++) {
            assertArrayEquals("Wrong unique combinations", expected[i] , uniqueCombination[i], 0 );
        }
    }

}

在巨大的矩阵上hash miss还是有可能的，所以谷歌提供的MurMur3A用Guava代替Arrays.hashCode(all[i])