使用Cython包装LAPACKE函数

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

我遇到了this previous answer,但由于dgtsv不是scipy.linalg中包含的LAPACK函数之一,我认为我不能使用这种特殊的方法.相反,我一直在努力追随this example.

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)

…这是我在_solvers.pyx中的瘦Cython包装器：

#!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,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,b = (np.ascontiguousarray(v,dtype=np.float64)
                    for v in (dl,b))

    # b will Now be modified in place to contain the solution
    info = _solvers.TDMA_lapacke(dl,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,du),(-1,1),format='csc')
    x = np.random.randn(n)
    b = M.dot(x)

    x_hat = trisolve_lapacke(dl,b)

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

不幸的是,test_trisolve只是对_solvers.TDMA_lapacke调用的段错误.
我很确定我的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n值,我倾向于不立即得到段错误,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),但偶尔我得到-7,这意味着参数7(b)具有非法值.发生的事情是,只有b的第一个值实际上被修改了.如果我继续调用test_trisolve,即使n很小,我最终也会遇到段错误.

最佳答案

好吧,我最终想通了 – 似乎我误解了在这种情况下行和列主要引用的内容.

由于C连续数组遵循行主顺序,我假设我应该将LAPACK_ROW_MAJOR指定为LAPACKE_dgtsv的第一个参数.

事实上,如果我改变

info = LAPACKE_dgtsv(LAPACK_ROW_MAJOR,...)

至

info = LAPACKE_dgtsv(LAPACK_COL_MAJOR,...)

然后我的功能工作：

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

这对我来说似乎很反直 – 有人可以解释为什么会这样吗？

使用Cython包装LAPACKE函数

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