我能够使用this StackOverflow问题获取1D内存视图的指针，但是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 *'分配给'double **'”错误。< / p>
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])
所以我通过以下方式成功编译：
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)
但是当我尝试运行时遇到段错误，所以我可能做错了。
以下是“dgesvd.h”的内容（我没有写，但我知道它有效）：
/*
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];
}
答案 0 :(得分：2)
您不希望使用“指向指针”表单。所有Cython / numpy数组都存储为单个连续数组以及一些长度参数，以便进行2D访问。您可能最好在Cython中复制dgesvd包装器（以分配工作数组，但不进行ftoc或ctof转换。）
我有一个去，下面，但它未经测试，所以可能有错误。除了被彻底复制之外，更多的是要做什么的要点。
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
答案 1 :(得分：1)
您拨打dgesdd的方式与its prototype不一致。除此之外，这应该工作。例如，请参阅this example，它以类似的方式执行Cython的dgemm调用。
另请注意，Scipy 0.16将包含a Cython API for BLAS/LAPACK，它可能是未来的最佳方法。