使用Strepe乘以2 double [] [] Matrices

Question

我想知道double[][]使用流的多个2 阵列矩阵最紧凑和有效的方法.该方法应遵循此处所示的矩阵乘法规则:http: //www.mathwarehouse.com/algebra/matrix/multiply-matrix.php

这是使用for循环的一种方法('this'是第一个矩阵'):

final int nRows = this.getRowDimension();
final int nCols = m.getColumnDimension();
final int nSum = this.getColumnDimension();

final double[][] outData = new double[nRows][nCols];
// Will hold a column of "m".
final double[] mCol = new double[nSum];
final double[][] mData = m.data;

// Multiply.
for (int col = 0; col < nCols; col++) {
    // Copy all elements of column "col" of "m" so that
    // will be in contiguous memory.
    for (int mRow = 0; mRow < nSum; mRow++) {
        mCol[mRow] = mData[mRow][col];
    }

    for (int row = 0; row < nRows; row++) {
        final double[] dataRow = data[row];
        double sum = 0;
        for (int i = 0; i < nSum; i++) {
            sum += dataRow[i]
                    * mCol[i];
        }
        outData[row][col] = sum;
    }
}

该程序应符合以下测试数据:

    double[][] md1 = { { 4d, 8d }, { 0d, 2d }, { 1d, 6d } };
    double[][] md2 = { { 5d, 2d, 5d, 5d }, { 9d, 4d, 5d, 5d } };

    double[][] md1 = { { 4d, 8d }, { 0d, 2d }, { 1d, 6d } };
    double[][] md2 = { { 5d }, { 9d } };

Answer 1

更紧凑和可读的解决方案是在第一个矩阵的行上创建Stream,将每行映射到将其与第二个矩阵列相乘的结果,并将其收集回到a double[][].

public static void main(String[] args) {
    double[][] m1 = { { 4, 8 }, { 0, 2 }, { 1, 6 } };
    double[][] m2 = { { 5, 2 }, { 9, 4 } };

    double[][] result = Arrays.stream(m1).map(r -> 
        IntStream.range(0, m2[0].length).mapToDouble(i -> 
            IntStream.range(0, m2.length).mapToDouble(j -> r[j] * m2[j][i]).sum()
    ).toArray()).toArray(double[][]::new);

    System.out.println(Arrays.deepToString(result));
}

这将计算m1 * m2,结果将在result.对于每一行的乘法,我们不能用Arrays.stream第二个矩阵创建一个Stream,因为当我们在列上需要Stream时,这会在行上创建一个Stream.为了抵消这种情况,我们只需回过头来使用IntStream索引.