我在这里定义了计算矩阵行列式的函数.但有时我会得到错误的信号.我从
this answer模拟了我的功能.
from scipy.linalg.cython_lapack cimport dgetrf
cpdef double det_c(double[:, ::1] A, double[:, ::1] work, double[::1] ipiv):
'''obtain determinant of float type square matrix A
Notes
-----
As is, this function is not yet computing the sign of the determinant
correctly, help!
Parameters
----------
A : memoryview (numpy array)
n x n array to compute determinant of
work : memoryview (numpy array)
n x n array to use within function
ipiv : memoryview (numpy array)
length n vector use within function
Returns
-------
detval : float
determinant of matrix A
'''
cdef int n = A.shape[0], info
work[...] = A
dgetrf(&n, &n, &work[0,0], &n, &ipiv[0], &info)
cdef double detval = 1.
cdef int j
for j in range(n):
if j != ipiv[j]:
detval = -detval*work[j, j]
else:
detval = detval*work[j, j]
return detval
当我测试这个函数并将它与np.linalg.det进行比较时,有时我得到了错误的符号.
>>> a = np.array([[1,2],[3,5.]])
>>> np.linalg.det(a)
>>> -1.0000000000000004
>>> det_c(a, np.zeros((2, 2)), np.zeros(2, dtype=np.int32))
>>> 1
其他时候,正确的标志.
>>> b = np.array([[1,2,3],[1,2,1],[5,6,1.]])
>>> np.linalg.det(b)
>>> -7.999999999999998
>>> det_c(a, np.zeros((3, 3)), np.zeros(3, dtype=np.int32))
>>> -8.0
最佳答案 dgetrf是Fortran子例程,Fortran使用基于1的索引,因此ipiv中的值介于1和n之间(包括1和n).为了解决这个问题,请更改循环中的测试
if j != ipiv[j]:
至
if j != ipiv[j] - 1: