Question

因此，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的数组，并用它填充uniqueCombinations。

我尝试过使用此函数，但是它只有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，即为每个矩阵向量计算哈希值，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：

// stores height h, side1 a, side2 b
double a, b, h;

//Reads out b from a textbox
private void txbB_TextChanged(object sender, EventArgs e)
{
    string parseB = txbB.Text;
    b = double.Parse(parseB);
}

//Reads out a from a textbox
private void txbA_TextChanged(object sender, EventArgs e)
{
    string parseA = txbA.Text;
    a = double.Parse(parseA);
}

//Reads out h from a textbox
private void txbH_TextChanged(object sender, EventArgs e)
{
    string parseH = txbH.Text;
    h = double.Parse(parseH);
}

//button which calculates the volume of the pyramid 
//when clicked and prints it to the label "LblErgebnis"
private void Cmdrechnen_Click(object sender, EventArgs e)
{
    double Total = 1/3 * h * a * b;
    string Result = Total.ToString();
    LblErgebnis.Text = Result;
}

在大型矩阵上仍然可能发生哈希未命中，因此使用了Google Guava提供的MurMur3A代替了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 ); } } }

从Java中的2D数组中删除重复项

1 个答案: