你如何使用cblas_dgemm做一个向量外积?

How do you use cblas_dgemm to do a vector outer product?

我正在尝试将列向量与行向量相乘。我可以使用 dgemm 吗?

也就是说D = A * B 其中 D 是矩阵,A 是列向量,B 是行向量。

我遵循了此处找到的文档 https://software.intel.com/en-us/node/520775。我似乎无法获得 cblas_dgemm

的正确参数

这是我的尝试。在我的例子中 m = nRows, n = nCols, k = 1

问题似乎出在lda、ldb 和ldc 上。我分别定义为nCols, k, nRows.

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <string.h>

#include <math.h>
#include <mkl.h>

#define nCols 5
#define nRows 20
#define k 1

void PrintMatrix(double* pMatrix, const size_t nR, const size_t nC, const CBLAS_ORDER Order) {
    unsigned int i, j;
    if (Order == CblasRowMajor)
    {
        for (i = 0; i < nR; i++)
        {
            for (j = 0; j < nC; j++)
            {
                printf("%f \t ", pMatrix[i * nC + j]); // !!!
            }
            printf("\n"); // !!!
        }

    }

    else

    {
        for (i = 0; i < nR; i++) {
            for (j = 0; j < nC; j++) {
                printf("%f \t ", pMatrix[i + j* nR ]); // !!!
            }
            printf("\n"); // !!!
        }

    }
    printf("\n"); // !!!

}

int main(void) {

    double A[] = { 8, 4, 7, 3, 5, 1, 1, 3, 2, 1, 2, 3, 2, 0, 1, 1 , 2, 3, 4, 1};

    double B[] = { -1, 2, -1, 1, 2 };


    double alpha = 1.0, beta = 0.0;
    int i, lda, ldb, ldc;
    double *C, *D;
    D = (double*) malloc(nRows * nCols * sizeof(double));
    C = (double*) malloc(nRows * nCols * sizeof(double));

    for (i = 0; i < nRows*nCols; i++)
        D[i] = 0.0;
    for (i = 0; i < nRows*nCols; i++)
        C[i] = 0.0;

    lda = nCols;
    ldb = k;
    ldc = nRows;

    cblas_dger(CblasRowMajor, nRows, nCols, alpha, A, 1, B, 1, C, nCols);

    PrintMatrix(C, nRows, nCols,CblasRowMajor);
    cblas_dgemm (CblasRowMajor, CblasNoTrans, CblasNoTrans, nRows,  nCols, k, alpha, A, lda, B, ldb, beta, D, ldc);

    PrintMatrix(D, nRows, nCols, CblasRowMajor);

    free(D);
    free(C);

    return 0;
}

简短的回答是,是的,您可以使用 dgemm 进行 rank-1 更新。 dger当然是建议的,因为希望能更好地优化这个操作。

至于使用 cblas_dgemm 。如您所知,领先维度的定义是:

lda: The size of the first dimension of matrix A

您尝试执行的操作是: D(20x5) = A(20x1) * B(1x5)

您正在使用 CblasRowMajor,因此前导维度是所有矩阵的列数(有关解释,请参见 )。含义:

lda = 1;
ldb = 5;
ldc = 5;

以下代码有效。我切换到 column-major 是因为在 Fortran BLAS 文档的上下文中更容易理解前导维度问题。很抱歉,我无法完全按照提出的方式解决您的问题

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <string.h>

#include <math.h>
#include <mkl.h>

void PrintMatrix(double* pMatrix, const size_t nR, const size_t nC, const CBLAS_ORDER Order) {
    unsigned int i, j;
    if (Order == CblasRowMajor) {
        for (i = 0; i < nR; i++) {
            for (j = 0; j < nC; j++) {
                printf("%f \t ", pMatrix[i * nC + j]); // !!!
            }
            printf("\n"); // !!!
        }
    } else {
        for (i = 0; i < nR; i++) {
            for (j = 0; j < nC; j++) {
                printf("%f \t ", pMatrix[i + j* nR ]); // !!!
            }
            printf("\n"); // !!!
        }
    }
    printf("\n"); // !!!
}

int main(void)
{
    const int m = 20;
    const int n = 5;
    const int k = 1;

    double A[] = { 8, 4, 7, 3, 5, 1, 1, 3, 2, 1, 2, 3, 2, 0, 1, 1, 2, 3, 4, 1};
    double B[] = { -1, 2, -1, 1, 2 };

    double alpha = 1.0, beta = 0.0;

    double * C = (double*) malloc(m * n * sizeof(double));
    double * D = (double*) malloc(m * n * sizeof(double));

    for (int i = 0; i < m*n; i++) C[i] = 0.0;
    for (int i = 0; i < m*n; i++) D[i] = 0.0;

    int lda = 20;
    int ldb = 1;
    int ldc = 20;

    cblas_dger(CblasColMajor, m, n, alpha, A, 1, B, 1, C, ldc);

    cblas_dgemm(CblasColMajor, CblasNoTrans, CblasNoTrans, 
                m, n, k, 
                alpha, A, lda, 
                B, ldb, beta, 
                D, ldc);

    PrintMatrix(C, m, n, CblasRowMajor);
    PrintMatrix(D, m, n, CblasRowMajor);

    free(D);
    free(C);

    return 0;
}