使用Cython包装LAPACKE函数

Question

我正在尝试dgtsv使用Cython 包装LAPACK函数(三对角方程组的求解器).

我遇到了这个前面的答案,但由于dgtsv不是LAPACK功能之一,scipy.linalg我不认为我可以使用这种特殊的方法.相反,我一直试图遵循这个例子.

这是我的lapacke.pxd文件的内容:

ctypedef int lapack_int

cdef extern from "lapacke.h" nogil:

    int LAPACK_ROW_MAJOR
    int LAPACK_COL_MAJOR

    lapack_int LAPACKE_dgtsv(int matrix_order,
                             lapack_int n,
                             lapack_int nrhs,
                             double * dl,
                             double * d,
                             double * du,
                             double * b,
                             lapack_int ldb)

...这是我的瘦Cython包装_solvers.pyx:

#!python

cimport cython
from lapacke cimport *

cpdef TDMA_lapacke(double[::1] DL, double[::1] D, double[::1] DU,
                   double[:, ::1] B):

    cdef:
        lapack_int n = D.shape[0]
        lapack_int nrhs = B.shape[1]
        lapack_int ldb = B.shape[0]
        double * dl = &DL[0]
        double * d = &D[0]
        double * du = &DU[0]
        double * b = &B[0, 0]
        lapack_int info

    info = LAPACKE_dgtsv(LAPACK_ROW_MAJOR, n, nrhs, dl, d, du, b, ldb)

    return info

...这是一个Python包装器和测试脚本:

import numpy as np
from scipy import sparse
from cymodules import _solvers


def trisolve_lapacke(dl, d, du, b, inplace=False):

    if (dl.shape[0] != du.shape[0] or dl.shape[0] != d.shape[0] - 1
            or b.shape != d.shape):
        raise ValueError('Invalid diagonal shapes')

    if b.ndim == 1:
        # b is (LDB, NRHS)
        b = b[:, None]

    # be sure to force a copy of d and b if we're not solving in place
    if not inplace:
        d = d.copy()
        b = b.copy()

    # this may also force copies if arrays are improperly typed/noncontiguous
    dl, d, du, b = (np.ascontiguousarray(v, dtype=np.float64)
                    for v in (dl, d, du, b))

    # b will now be modified in place to contain the solution
    info = _solvers.TDMA_lapacke(dl, d, du, b)
    print info

    return b.ravel()


def test_trisolve(n=20000):

    dl = np.random.randn(n - 1)
    d = np.random.randn(n)
    du = np.random.randn(n - 1)

    M = sparse.diags((dl, d, du), (-1, 0, 1), format='csc')
    x = np.random.randn(n)
    b = M.dot(x)

    x_hat = trisolve_lapacke(dl, d, du, b)

    print "||x - x_hat|| = ", np.linalg.norm(x - x_hat)

不幸的是,test_trisolve只是在调用段错误_solvers.TDMA_lapacke.我很确定我setup.py是正确的 - ldd _solvers.so显示_solvers.so在运行时链接到正确的共享库.

我不确定如何从这里开始 - 任何想法？

---

简要更新:

对于较小的nI 值,我不会立即得到段错误,但我确实得到了无意义的结果(|| x - x_hat ||应该非常接近0):

In [28]: test_trisolve2.test_trisolve(10)
0
||x - x_hat|| =  6.23202576396

In [29]: test_trisolve2.test_trisolve(10)
-7
||x - x_hat|| =  3.88623414288

In [30]: test_trisolve2.test_trisolve(10)
0
||x - x_hat|| =  2.60190676562

In [31]: test_trisolve2.test_trisolve(10)
0
||x - x_hat|| =  3.86631743386

In [32]: test_trisolve2.test_trisolve(10)
Segmentation fault

通常LAPACKE_dgtsv返回代码0(应该表示成功),但偶尔会得到-7,这意味着参数7(b)具有非法值.发生的事情是只有第一个值b实际上被修改了.如果我继续打电话,test_trisolve即使n很小,我也最终会遇到一个段错误.

Answer 1

好吧，我最终想通了 - 看来我误解了在这种情况下行和列主要指的是什么。

由于 C 连续数组遵循行主序，我认为我应该指定LAPACK_ROW_MAJOR为 的第一个参数LAPACKE_dgtsv。

事实上，如果我改变

info = LAPACKE_dgtsv(LAPACK_ROW_MAJOR, ...)

到

info = LAPACKE_dgtsv(LAPACK_COL_MAJOR, ...)

然后我的函数起作用了：

test_trisolve2.test_trisolve()
0
||x - x_hat|| =  6.67064747632e-12

这对我来说似乎非常违反直觉 - 谁能解释为什么会这样？