2D MemoryView转C指针错误(1D有效，但2D不起作用)

Question

我可以使用this StackOverflow error获得1D内存视图的指针，但将相同的方法应用于2D内存视图会给我一个“无法将类型'double *‘赋给'double **'”的错误。

cdef extern from "dgesvd.h" nogil:
    void dgesvd(double **A, int m, int n, double *S, double **U, double **VT)


cdef:
    double[:] S
    double[:,:] A, U, VT

A = np.ascontiguousarray(np.zeros((N,N)))
S = np.zeros(N)
U = np.zeros(N)
VT = np.zeros(N)

dgesvd(&A[0,0], N, N, &S[0], &U[0], &VT[0])

编辑:我通过这样做来编译它。

因此，我通过执行以下操作使其成功编译：

    cdef:
        double[:] S
        double[:,:] A, U, VT

    U = np.zeros((N,N))
    VT = np.zeros((N,N))
    A = np.zeros((N,N))
    S = np.zeros(N)

    A_p = <double *> malloc(sizeof(double) * N)
    U_p = <double *> malloc(sizeof(double) * N)
    VT_p = <double *> malloc(sizeof(double) * N)

    for i in range(N):
        A_p = &A[i, 0]
        U_p = &U[i, 0]
        VT_p = &VT[i, 0]

    dgesvd(&A_p, N, N, &S[0], &U_p, &VT_p)

    free(A_p)
    free(U_p)
    free(VT_p)

但是当我尝试运行它时，我得到了一个段错误，所以我可能做错了。

以下是"dgesvd.h“的内容(我没有写它，但我知道它是有效的)：

/*
  This file has my implementation of the LAPACK routine dgesdd for
  C++.  This program solves for the singular value decomposition of a
  rectangular matrix A.  The function call is of the form

    void dgesdd(double **A, int m, int n, double *S, double *U, double *VT)

    A: the m by n matrix that we are decomposing
    m: the number of rows in A
    n: the number of columns in A (generally, n<m)
    S: a min(m,n) element array to hold the singular values of A
    U: a [m, min(m,n)] element rectangular array to hold the right
       singular vectors of A.  These vectors will be the columns of U,
       so that U[i][j] is the ith element of vector j.
    VT: a [min(m,n), n] element rectangular array to hold the left
        singular vectors of A.  These vectors will be the rows of VT
    (it is a transpose of the vector matrix), so that VT[i][j] is
    the jth element of vector i.

  Note that S, U, and VT must be initialized before calling this
  routine, or there will be an error.  Here is a quick sample piece of
  code to perform this initialization; in many cases, it can be lifted
  right from here into your program.

    S = new double[minmn];
    U = new double*[m]; for (int i=0; i<m; i++) U[i] = new double[minmn];
    VT = new double*[minmn]; for (int i=0; i<minmn; i++) VT[i] = new double[n];

  Scot Shaw
  24 January 2000 */

void dgesvd(double **A, int m, int n, double *S, double **U, double **VT);

double *dgesvd_ctof(double **in, int rows, int cols);
void dgesvd_ftoc(double *in, double **out, int rows, int cols);

extern "C" void dgesvd_(char *jobu, char *jobvt, int *m, int *n,
            double *a, int *lda, double *s, double *u,
            int *ldu, double *vt, int *ldvt, double *work,
            int *lwork, int *info);

void dgesvd(double **A, int m, int n, double *S, double **U, double **VT)
{
  char jobu, jobvt;
  int lda, ldu, ldvt, lwork, info;
  double *a, *u, *vt, *work;

  int minmn, maxmn;

  jobu = 'S'; /* Specifies options for computing U.
         A: all M columns of U are returned in array U;
         S: the first min(m,n) columns of U (the left
            singular vectors) are returned in the array U;
         O: the first min(m,n) columns of U (the left
            singular vectors) are overwritten on the array A;
         N: no columns of U (no left singular vectors) are
            computed. */

  jobvt = 'S'; /* Specifies options for computing VT.
          A: all N rows of V**T are returned in the array
             VT;
          S: the first min(m,n) rows of V**T (the right
             singular vectors) are returned in the array VT;
          O: the first min(m,n) rows of V**T (the right
             singular vectors) are overwritten on the array A;
          N: no rows of V**T (no right singular vectors) are
             computed. */

  lda = m; // The leading dimension of the matrix a.
  a = dgesvd_ctof(A, lda, n); /* Convert the matrix A from double pointer
              C form to single pointer Fortran form. */

  ldu = m;

  /* Since A is not a square matrix, we have to make some decisions
     based on which dimension is shorter. */

  if (m>=n) { minmn = n; maxmn = m; } else { minmn = m; maxmn = n; }

  ldu = m; // Left singular vector matrix
  u = new double[ldu*minmn];

  ldvt = minmn; // Right singular vector matrix
  vt = new double[ldvt*n];

  lwork = 5*maxmn; // Set up the work array, larger than needed.
  work = new double[lwork];

  dgesvd_(&jobu, &jobvt, &m, &n, a, &lda, S, u,
      &ldu, vt, &ldvt, work, &lwork, &info);

  dgesvd_ftoc(u, U, ldu, minmn);
  dgesvd_ftoc(vt, VT, ldvt, n);

  delete a;
  delete u;
  delete vt;
  delete work;
}

double* dgesvd_ctof(double **in, int rows, int cols)
{
  double *out;
  int i, j;

  out = new double[rows*cols];
  for (i=0; i<rows; i++) for (j=0; j<cols; j++) out[i+j*rows] = in[i][j];
  return(out);
}

void dgesvd_ftoc(double *in, double **out, int rows, int cols)
{
  int i, j;

  for (i=0; i<rows; i++) for (j=0; j<cols; j++) out[i][j] = in[i+j*rows];
}

Answer 1

你不想使用“指针到指针”的形式。所有的Cython/numpy数组都存储为一个连续的数组，以及一些长度参数，以便进行2D访问。最好是在Cython语言中复制dgesvd包装器(分配工作数组，但不进行ftoc或ctof转换)。

我已经试过了，在下面，但它是未经测试的，所以可能有bug。它更多的是关于要做什么的要点，而不是被直接复制。

def dgesvd(double [:,:] A):
    """All sizes implicit in A, returns a tuple of U S V"""

    # start by ensuring we have Fortran style ordering
    cdef double[::1, :] A_f = A.copy_fortran()
    # work out the sizes - it's possible I've got this the wrong way round!
    cdef int m = A.shape[0]
    cdef int n = A.shape[1]

    cdef char jobu[] = 'S'
    cdef char jobvt[] = 'S'

    cdef double[::1,:] U
    cdef double[::1,:] Vt
    cdef double[::1] S

    cdef double[::1] work

    cdef int minnm, maxnm
    cdef int info, lwork, ldu, ldvt

    if m>=n:
       minmn = n
       maxmn = m
    else:
       minmn = m
       maxmn = n

    ldu = m;
    U = np.array((ldu,minmn), order='F')
    ldvt = minmn
    Vt = np.array((ldvt,n), order='F')
    S = np.array((minmn,)) # not absolutely sure  - check this!

    lwork = 5*maxmn
    work = np.array((lwork,))

    dgesvd_(&jobu, &jobvt, &m, &n, &A_f[0,0], &lda, &S[0], &U[0],
           &ldu, &Vt[0,0], &ldvt, &work[0], &lwork, &info);

    return U, S, Vt.T # transpose Vt on the way out

Answer 2

您调用dgesdd的方式与its prototype不一致。除此之外，这应该是可行的。例如，请参见this example，它以类似的方式执行来自Cython的dgemm调用。

还要注意的是，Scipy 0.16将包括a Cython API for BLAS/LAPACK，它可能是未来最好的方法。